Week 6 Prompt

how do you determine concavity? How do you know if it is concave up or down?

another throwback blog

differentiability is the ability to take a derivative from a function

there are many ways that a function cannot be differentiable
those ways are:

1) if there is a curve
curves occur when absolute values and/or piecewises are found in the function that you are trying to take the derivative of
an example of a function with a curve is:
1) |3-x| you set what's inside of the absolute value equal to 0 and solve for x, so |3-x| is not differentiable at x=3

2) if there is a vertical asymptote
vertical asymptotes occur when x is raised to the (odd/odd)
x^5/9 would be a vertical asymptote, so x would not be differentiable at x=1

3) if it's not continuous
if a function is not continuous, then it's not differentiable, it's as simple as that

4) if there's a cusp
a cusp is when you have x^(even/odd)
x^2/7 would not be differentiable at x=0

taylor Rodriguez blog #14

I am going to use this blog to attempte to explain what we have learned so far concerning optimization.

Optimization is basically an extended application of what we have learned thus far in chapter 3 and some of chapter 5 if im not mistaken..

THE STEPS TO OPTIMIZATION: (are also listed on page 219 of our text book)

#1
list what is given to you in the problem

#2

list the primary equation

#3
List the secondary equation


#4

Solve for y in the secondary equation

#5

plug y= into the primary equation for y

#6

set = 0 and solve

#7

perform the first derivative test

meaning:
take the derivative
set = 0
solve for x
set up intervals
plug in for F(x) between intervals into the derivative
determine if result is positive or negative and thereby increasing or decreasing respectivley on the intervals
therefore enabling you to determine whther x= is a max or a min






EXAMPLE:
A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. what dimensions will produce a maximum volume?


#1
given: SA=108in^2 SA= x^2+4xy
Primary Equation: V= x^2y
Secondary equation: x^2+4xy=108
4xy=108-x^2
y= 108-x^2/ 4x

Plug in: V= X^2(108-x^2/4x)
V= 108x-x^3/4
27x-1/4x^3


first derivative test:
27-3/4x^2=0
-3/4x^2= -27
x^2=36
x= + - 6

Set up intervals

(0,6) U (6,sqrt108)
F(1) is positive and thereby increasing
F(7) is negative and thereby decreasing

Therefore X=6 is a maximum

Therefore V= LXWXH
V= 6X6XH
6^2 x y = 108
y= 108/ 36
y=3
V= 6X6X3

11/28

I'm going to review some simple but important things we have learned so far:

Derivatives

1. Constant rule: constants = 0 f = 3 f'= 0
2. Power rule: exponent at front raised to exponent-1 f=x^3 f'=3x^2
3. Product rule: first x f'second + second x f'first
4. Quotient rule: bottom f'top - [top (f'bottom)]
5. Chain rule: f'outside recopy inside x f'inside
6. Natural Logs: 1 / recopy x f'bottom

Limits at Infinity

1. top > bottom = infinity
2. top = bottom = fraction
3. top < bottom = 0

Extrema
1. points that are positive are maxes
2. negative points are mins (bottom of curve)
3. points of inflection are where the concavity changes

...

Review Blog. So for this blog I'm gonna review the product and quotient rules as well as the definition of a derivative. They are all pretty simple and everyone should know how to do these.

The first thing I will review is the definition of a derivative. This is what your are doing when you take a derivative. All you do is plug in your equation to the following formula:

lim [F(x+dx)-F(x)]/(dx)
dx->0

For example: Derive x^2+2

Plug in: (x+dx)^2+2-(x^2+2)/dx

Simplify: x^2+2xdx+dx^2+2-x^2-2/dx

Further Simplify: 2xdx+dx^2/dx

Factor: dx(2x+dx)/dx

Simplify and plug in 0 for dx: 2x

For the product rule: (x^2)(2x-1)

Derive the first, leave the second: 2x(2x-1)
+Derive the second, leave the first: +2(x^2)

Simplify: 4x^2-2x+2x^2

Further Simplify: 6x^2-2x

And the quotient rule: (x^2)/2x

[Bottom(Top Derivative)-Top(Bottom Derivative)]/(Bottom)^2

2x(2x)-(x^2)2/4x^2

Simplify: 4x^2-2x^2/4x^2

Further Simplify: 2x^2/4x^2

Further Simplify: 1/2

And thats it for this blog.

Blog for 11/28/2010

For the 2nd review blog of the Thanksgiving holidays, I will shortly describe how to effectively use the rules of concavity and probably the shortcut to the first derivative test, amongst other things.

Concavity

You use the same steps as the first derivative test, but you're using the 2nd derivative.

1. Take first derivative and set it = 0 then solve for x.

2. Take 2nd derivative and set it = 0 then solve for x.

3. Check for differentiability or non-differentiability.

4. Set up your intervals.

5. Pick numbers on the intervals and plug in to 2nd derivative.

6. If it is positive then it is concave up, if it is negative, then it is concave down.

7. If there is a change it is a point of inflection.

Shortcut to the First Derivative Test

1. Take the first derivative, set it equal = 0, then solve for x.

2. Check for differentiability or non-differentiability.

3. Plug critical points into 2nd derivative.

4. If it is positive then it is a min, if negative then it is a max, if 0, then the test fails.

Blog for 11/21/2010

Due to me being basically hospitalized all week, I was unable to post a blog for last Sunday, so I'm posting it now. I understand we have to do 2 review blogs over the Thanksgiving holidays, this is numero uno. I will describe the first derivative test and how to find relative extrema such as maxes and/or mins.

*Critical numbers - are maxes or mins

First Derivative Test-

1. Take derivative and set it = 0

2. solve for x

3. Set up intervals

(-infinity, __) u (__, infinity)

a. Plug a value on the interval in to the derivative.

i. If you get a positive number then the function is increasing.

ii. If you get a negative number then the function is decreasing.

b. A max, which is increasing then decreasing, the graph will look like an upside-down, or negative parabola.

c. A min, which is decreasing then increasing, the graph will look like a rightside-up, or positive parabola.

Blog #14

For this review blog, I’ll go over some of the derivative rules we learned.

THE CONSTANT RULE:
The constant rule: the derivative of a constant is always 0

Ex. 1) d/dx 7 = 0


THE POWER RULE:
The power rule: whenever you take the derivative of something with an exponent, you lose a power. The formula for this shortcut is d/dx (x^n) = nx^(n-1)

Ex. 2) d/dx 3x^2
First, bring the exponent to the front (and if there is a constant in front of the x, multiply the two). Then, subtract one from the exponent.
= 2(3)x^(2-1) = 6x

Ex. 3) d/dx (x^3 + 9x)
When there are multiple terms in an equation, take the derivative of each term individually.
= 3x^2 + 9


THE PRODUCT RULE:
The product rule formula: f(x)g’(x) + g(x)f’(x)
When using the product rule, the problem must state that you take the derivative of two separate equations multiplied by each other.
*note: g’(x) or f’(x) means take the derivative of g(x) or f(x)

Ex. 4) (x+3)(x^2+6x)
(x+3)(2x+6) + (x^2+6x)(1)
2x^2 + 6x + 6x + 18 + x^2 + 6x
3x^2 + 18x + 18
= 3(x^2 + 6x +6)
**note: Make sure the answer is as fully simplified as possible.


THE QUOTIENT RULE:
The quotient rule formula: (g(x)f’(x) – f(x)g’(x)) / (g(x))^2
When using the quotient rule, the problem must state that you take the derivative of two equations divided by each other.
***note: if the denominator is just a single term, it is not considered an equation and you do not use the quotient rule; you simply rewrite the problem and take the derivative normally

Ex. 5) (x^3+4) / (x-2)
((x-2)(3x^2) – (x^3+4)(1)) / ((x-2)^2)
((3x^3-6x^2) – (x^3+4)) / ((x-2)^2)
****note: to keep things in simplest terms, do not factor out the bottom
(3x^3 – 6x^2 – x^3 – 4) / ((x-2)^2)
*****note: be sure to distribute the negative throughout the second equation
(2x^3 – 6x^2 – 4) / ((x – 2)^2)
(2(x^3 – 3x^2 – 2)) / ((x – 2)^2)


THE CHAIN RULE:
You use the chain rule when you have a function inside of a function, f(g(x)).
Ex. 6) (x^2 + 3)^4
First, take the exponent of the entire function, bring it to the front, then subtract one from the exponent, and then just copy the inside.
4(x^2 + 3)^3
Now, multiply that by the derivative of the inside equation.
4(x^2 + 3)^3 (2x)
Put in simplest terms.
= 8x(x^2 + 3)^3

You can also use the chain rule with the product rule or quotient rule. If this is the case, then you first need to determine whether the chain rule is on the outside or the inside; this is to figure out what order you do things in. If the chain rule is on the outside, you do that first. If not, you do the product rule or quotient rule first.
Ex. 7) (x + 3)(x^2 – 1)^2
In this problem, the product rule comes first. When taking the derivative of the second equation, use the chain rule.
(x + 3)(2(x^2 – 1)(2x)) + (x^2 – 1)^2 (1)
= 4x(x + 3)(x^2 – 1) + (x^2 – 1)^2

Ex. 8) ((x + 2)/(x – 1))^2
For this problem, the chain rule is on the outside, therefore it goes first.
2((x + 2)/(x – 1)) (((x – 1)(1) – (x + 2)(1)) / (x – 1)^2))
2((x + 2)/(x – 1)) ((x – 1 – x – 2) / (x – 1)^2)
2((x + 2)/(x – 1)) ((-3)/(x – 1)^2)
= -3(2x + 4) / (x – 1) (x – 1)^2

11/27/10

Well since we didn't have school this week, I think I'll just review some simple things like Rolle's Theorem and the Mean Value Theorem..
**First, let's start off with Rolle's Theorem. When using it, it tells you IF there is a max or a min on an interval..and if you DO have one, it is going to be AT LEAST 1 max OR min. Typically, Rolle's Theorem uses x-intercepts. Butttttttt, before you can even apply the theorem, you first have to check the function to make sure it is both continuous and differentiable. Then you have to check to make sure the y-values match (check this by plugging both numbers of your interval into the function you're given). Once you've checked all that, you take the derivative of the function and set it equal to zero to find your x-value(s). Once you get those x-values you have to check to make sure they are on the interval you were given. If not, then you just don't include them in your answer.
Ex 1.) Determine whether Rolle's Theorem can be applied to the function f(x)=x^2-5x+4 on the closed interval [1,4]...and if it can be applied, find all values of c.
*Okay, first check the function for differentiability and continuity....it is continuous and differentiable. Now check to make sure the y-values match, so plug 1 and 4 into the function:
f(1)=1-5+4=0
f(4)=16-20+4=0
andddddddd the y-values do match, so now you can move on to the next step and take the derivative of the function which is:
2x-5
Now set it equal to zero and solve for x:
2x-5=0
x=5/2 ...and that's your value of c
**Now let's move on to the Mean Value Theorem...It is actually very similar to Rolle's Theorem except that it involves slope. When you use this theorem, it means on an interval there must be some x-value where the derivative equals the slope between the numbers of that interval. But before you use the theorem, you have to check the function for differentiability and continuity (but NOT if the y-values are the same; that doesn't matter). Once you do that, you find the slope using this formula: f'(c)=f(b)-f(a)/b-a...where the interval is [a,b]
Once you do that, you take the derivative of the function and then set it equal to the slope and solve for x
Ex 2.) Determine whether the Mean Value Theorem can be applied to f(x)=x^3+2x on the closed interval [-1,1]...if it can be applied, find all values of c.
*Yessssssss, the function is continuous and differentiable so we can move on with our lives...and find the slope of the function!
*So you should get (3+3)/2
=3
*Now take the derivative of the function......and you should get 3x^2+2
*Now set the derivative equal to the slope and solve for x:
3x^2+2=3
x^2=1/3
x=+/-squareroot of 3/3 ....and that's your value of c

**And I SERIOUSLY NEED HELP with drawing the graphs..(like when you're given the graph of f and they ask you questions about the graph of f' and f''..) Those are really tough; I could use a lot more practice with them.

Blog 14

Thanksgiving holidays are almost over :( they flew by so fast! Hopefully everyone had a good holiday! Since we didn’t have school this week I’ll just explain 3.5 which is horizontal asymptotes, and I’ll also explain the short cut to the first derivative test. Horizontal asymptote: To find a horizontal asymptote you take the limit as x is approaching infinity. Y=# is the horizontal asymptote. To take the limit you just simply use the limit rules. If the degree of the top is equal to the degree of the bottom the answer is the coefficients. Example: x^2+2/x^2-1=1, If the degree of the top is greater than the degree of the bottom then the horizontal asymptote is either negative infinity or infinity. Example: x^2+2/x-1=infinity, If the degree of the bottom is greater than the degree of the top then the horizontal asymptote is zero. Example: x^2+2/x^3-1=0.

Now the short cut to the first derivative test:

There are four steps:
1. First derivative, set equal to zero, solve.
2. Is it differentiable?
3. Take second derivative; Plug critical points into second derivative.
4. If positive number then it’s a MIN, if negative number then it’s a MAX (May seem odd, but it’s the rule so memorize it) and If 0 then the test fails.

Example:
Find the relative extrema for f(x)=-3x^5+5x^3
1. -15x^4+15x^2=0
15x^2(-x^2+1)
X=0,+/- 1

2. All polynomials are differentiable!
3. -60x^2+30x
4. -60(0)^3+30(0)=0;FAILS
-60(-1)^3+30(-1)=Positive number;MIN
-60(1)^3+30(1)=Negative number;MAX

I pretty much understand everything; I just need more practice on it all. Minus the graphs, I don’t get that at all.

Week 6 Prompt

Due to the fact that the internet was not dependable on my laptop I was unable to post the blog prompt for the holidays. Therefore it is a freebie for everyone. Just make sure to post your 2 regular blogs. They can come from any review topic.

Alaina's blog, 21 Nov. 2010

Chapter 3 Section 5
limits at infinity

-to find a horizontal asymptote, you take the lim x->infinity
y=ans is an asymptote

degree of top = degree of bottom => coefficient
degree of top > degree of bottom => -inifinity or infinity
degree of top < degree of bottom => 0

Ex 1: y=2x+5/(3x^2+1)
lim x->infinity 2x+5/(3x^2+1)=0
lim x-> infinity is 0
0 is a horizontal asymptote.

Chapter 3 Section 6
Curve Sketching

I can tell you how to work problems to get all of the critical information to sketch the graph but i will not show the graph.

STEPS
1. identify the domain and range
2. find the x and y intercepts, identify the vertical and horizontal asymptotes
3. do the first derivative test
4. do the second derivative test
5. plot the information and sketch

Ex: y=2(x^2-9)/(x^2=4)
1. domain: x^2-4=0; x=+/- 2
domain=(-infinity, -2)u(-2,2)u(2,infinity)
range: find horizontal asymptotes (limits approaching infinity) and set up intervals
y=2; (-infinity,2)u(2,infinity)

2. vertical asymptotes: x=+/-2
horizontal asymptotes: y=2
x intercepts: (3,0)(-3,0)
y intercepts: (0,9/2)

3.a) use quotient rule
f'(x)=20x/(x^2-4)^2=0
20x=0
x=0
b)x=+/-2
c)(-infinity,-2)u(-2,0)u(0,2)u(2,infinity)
d) f'(-3)=-ve; f'(-1)=-ve; f'(1)=+ve; f'(3)=+ve
e) decreasing; decreasing; increasing; increasing
f) x=0 is a min

4. a) second derivative(quotient rule)
f''(x)=-20(3x^2+4)/(x^2-4)^3=0
x=+/- squareroot (-4/3)
b)x=+/-2
c)(-infinity,-2)u(-2,2)u(2,infinity)
d)f''(-3)=-ve; f''(0)=+ve; f''(3)=-v3
e)decreasing; increasing; decreasing
f)concave down; concave up; concave down
x=-2,2 are points of inflection

Throwback Blog

well, it looks like I'm doin a throwback blog for now

I'll just talk about limits, because I didn't really understand the stuff from the end of the week.

let's say you wanna find the limit of an equation, such as lim ((x-3)^2)/x
x->7

You've got to plug in whatever x approaches into the equation, as long as it doesn't result in a 0 on the bottom of the equation.

So if it were x -> 0 instead of x -> 7, the answer to this problem would most likely be Ø.
If that happens, you need to do one of two things.
*note* You do this whenever the bottom would equal Ø when you plug in whatever x approaches, not just when x -> 0.

One way would be by plugging the function into a table and using numbers that are close to the x -> to find what both sides approach.
In this case, the table would be: -.1 -.01 -.001 0 .001 .01 .1
-96.1 -906 -9006 8994 894.01 84.1
So as you can see, from the left, the graph/equation is approaching -∞, and from the right, the graph/equation is approaching ∞.

Another way you could figure it out is by using arithmetic.
((x-3)^2)/x
(x^2-6x+9)/x = 0
x^2-6x+9=0
x^2-6x=-9
x(x-6)=-9
x=-9 x=6
And then you figure out which is the limit, which it usually is hard to figure out.

Blog #13

This week we learned about horizontal asymptotes. First of all, to find a horizontal asymptote, you take the limit as x approaches infinity. Your answer should be set up as… y = #.

Second of all, how to find the limit is quite simple:
If the degree of the top is greater than the degree of the bottom, the limit is infinity.
If the degree of the bottom is greater than the degree of the top, the limit is 0.
If the degree of the top and bottom are the same, then you take the coefficient of the variable raised to the greatest degree of the top over the coefficient of the variable raised to the greatest degree of the bottom. And that's your limit.

Now, if you limit is a number, then that is also your horizontal asymptote. But if you limit is infinity, then you have no horizontal asymptotes.

Ex. 1) Find the horizontal asymptote(s) of f(x) = (x + 4) / (2x^2 + 5).
The degree of the bottom is greater than the top, so the limit is 0.
Therefore, y = 0 is a horizontal asymptote.

Ex. 2) Find the horizontal asymptote(s) of f(x) = (x^2 + 2x + 4) / x.
The degree of the top is greater than the bottom, so the limit is infinity.
Therefore, there are not horizontal asymptotes.

Ex. 3) Find the horizontal asymptote(s) of f(x) = (3x + 4x^2 – 6) / (2x^2 – 1).
The degree of the top and bottom are the same, so the limit is the coefficients, which are 4/2 (which simplifies to 2.)
Therefore, y = 2 is a horizontal asymptote.

11/21

So the holidays are finally here yay! I'm going to make this a review blog on stuff from chapters 3 and 5

Limits at infinity:

There are simple rules that apply here:

1. Look at the degree of the top and bottom of the fraction
2. If the top is Larger than the bottom, the limit is either infinity or negative infinity
3. If the degrees are equal, the limit is determined by the coefficients
ex: 4x^2+1 / x^2 = 4
4. If the bottom is Larger than the top, the limit is 0
ex: 4x^2 / x^6 = 0

There is a FUDGE method that we learned stating if there is a square root on the top or bottom just take the root of the number and divide the exponent

Taking the derivatives of natural logs:

the derivative of a natural log is 1/recopy x derivative of inside

ex: f(x)= lnx^3
f'(x)= 1/x^3 x 3x2 = 3/x

that should be a brief reminder of what we are up to so enjoy the week off and all!

11/21/10

Soooo this past week we started off by learning how to find horizontal asymptotes, which are in fact the same as vertical asymptotes except the fact that they are horizontal...duh. But anyway, you find those by taking the limit as x goes to infinity of whatever function you are given. And, you use your limit rules as x goes to infinity: if the degree of the top = degree of bottom, then the answer is the coefficients; if the degree of the top > degree of the bottom, then the answer is either -infinity or infinity; and if the degree of the top < degree of the bottom, then the answer is zero. Then once you get what the limit is, your horiztonal asymptote would be written as y="whatever the limit is"..Then we worked with curve sketching which involves an INSANELY LONG process before you even draw the graph...Then on Thursday we started learning optimization, which is kind of like related rate but a little different. Here are some examples of what we did this week:

Ex. 1) y=x(sqrt of 4-x) .."x times the square root of 4-x"
Directions: Analyze and sketch a graph of the function. Label any intercepts, relative extrema, points of inflection, and asymptotes.
*Well obviously I won't be sketching a graph on here, but to find all these things they're asking for you're going to have to do it in steps..
Step 1.) Find the domain and range of the function.
Domain: Back to advanced math..you take the inside of the square root and set it equal to zero, solve for x, and then set up a number line and plug in numbers from each side.
So when you solve for x you get x=4...plug in numbers on both sides of 4 on the number line...numbers to the left work, numbers to the right do not.
So your domain should be (-infinity,4]
Range: I'm not so sure about this since there's a square root..If I remember from advanced math correctly I think the range would be the shift..which there is none..
Sooo maybe the range is (-infinity,infinity)? or possibly (-infinity,0)u(0,infinity)?
Step 2.) Find vertical and horizontal asymptotes (if any) and find x and y intercepts.
*Well there are no vertical asymptotes because the function is not a fraction and I'm pretty sure there aren't any horizontal asymptoes either..so that step is done
*X-intercepts: Set the function equal to zero and solve for x
x=4
x-intercept is (4,0)
*Y-intercepts: Plug in 0 for x and solve for y
y-intercept is (0,0)
Step 3.) First Derivative Test!!!!!!!!!!
*We should all know how to take a derivative by now so here's what you get:
y'=(-3x+8)/2sqrt of 4-x
*Now take the numerator of the fraction and set it equal to zero and solve for x
-3x+8=0
x=8/3
*Now check for differentiability...and the function is in fact differentiable everywhere
*Now set up your intervals like this:
(-infinity,8/3)u(8/3,infinity)
*Now plug numbers in between those intervals into the 1st derivative
*for the first interval you can plug in 0 and you should get a positive number; for the second interval you can plug in 3 and you should get a negative number
-So that means you have a function that is increasing then decreasing...which means there is a max at x=8/3
^To find the y-value for that, all you do is plug 8/3 into the original equation and you should get something like 3.079
Step 4.) Second Derivative Test!!!!!!!!!!
*And again, I think we all know how to do that so here's what you get:
y''=(3x-16)/4(4-x)^3/2
*Now you set the top of the fraction equal to zero and solve for x
3x-16=0
x=16/3
*Now set up your intervals
(-infinity, 16/3)u(16/3, infinity)
*Now plug in values
*You can plug 0 into the 1st interval and you should get a negative number; and you can plug 6 into the second interval and you should get a negative number again...which means it's concave down and concave down
***And that's all the steps! Then you have to plot all the points and sketch the graph.

**Now for what I didn't understand this week...OPTIMIZATION! I honestly don't get how to find the equations first off..it seems like you pretty much pull them from thin air. I don't know, I guess I just need some work with it..

Blog 13

Before the holidays we learned optimization, which is in chapter three (section 3.7). It isn’t too hard it is basically a recap of everything we learned in chapter three, just put together. Most of the problems are word problems, which can be difficult. However, there are steps which help make solving the problem easier. The first thing you’re going to do is find the “given”, or the information the problem provides for you. Then you’re going to find a primary equation using the information they gave you. Next you’re going to find a secondary equation which is basically the given equation solved for either x or y. Then you are going to plug back into the primary equation. Then you’re going to take the derivative of the answer you just got from plugging in, and solve it for x. After doing that, you set up intervals to find if the function is increasing or decreasing. **Hint: if you make y 0 in the secondary equation it will tell you what the endpoint of your interval should be, which makes it easier to pick a point to plug into the derivative. If the interval isn’t going from –infinity, infinity then you’re going to have to check intervals by plugging all interval points into the given equation, to find the highest point. The highest point will be the answers. And that’s all there is too it, it may seem confusing, but it is really simple, just follow the steps.
Example: Find 2 non-negative numbers whose sum is 9 and so that the product of 1 number and the square of the other is a maximum.
1. Given: x+y=9
2. Primary: P=xy^2
3. Secondary: x+y=9
Solved: y=9-x
*Making y 0, x=9
4. Plug in: P=x(9-x)^2;=x^3-18x^2+81x
5. Maximize: 3x^2-36x+81; x=9,x=3
(0,3)u(3,9);f’(1)=+,f’(4)=-
Check:
X=0, y=9 P=(0)(9)=0;x=3,y=6 P=(3)(6)=18; x=9,y=0 P=(9)(0)
X=3, y=6 yields the highest product
The only thing that kind of confuses me is how to know what the equation is for the given part.

Taylor Blog #13

This week we covered two major topics
these topic included:
Horizontal asymptotes
and
curve sketching

To find a horizontal asymptote there are three rules to follow
these rules apply to the degree of the leading coefficients of the polynomials in both the numerator and denominator
These rules are:

* If the top degree of the leading coefficient is equal to the bottom degree of the leading coefficient then you take the coefficient of the highest degree in the numerator and put it over the coefficient of the highest degree in the denominator

Example:
What is the horizontal asymptote of
2x^2+5 / 3x^2+1

because the degrees of the leading coefficients are equivalent there is a horizontal asymptote at 2/3


*If the top degree of the leading coefficient is greater than bottom degree of the leading coefficient then there is a horizontal asymptote at - infinity , infinity. In other words there is no horizontal asymptote.


Example:
What is the horizontal asymptote of
2x^2+5 / 3x+1

because the degree of the leading coefficient in the top is greater than the leading coefficient in the bottom the result is
-infinity, infinity and therefore there are no horizontal asymptotes.




*If the top degree of the leading coefficient is less than bottom degree of the leading coefficient then there is a horizontal asymptote at 0.


Example:
What is the horizontal asymptote of
2x+5 / 3x^2+1

because the degree of the leading coefficient in the top is less than the leading coefficient in the bottom the horizontal asymptote is 0.




For curve sketching problems there are five steps to follow

these steps are:

#1
find the domain and find the range
The domain is found by setting the bottom equal to zero, solving for x, and setting up intervals
The range is found by finding the horizontal asymptotes and setting up intervals

#2
Find the x intercepts, y intercepts, vertical asymptotes, and horizontal asymptotes.
X intercepts are found by setting the original equal to zero and solving for x to get the point (result,0)
Y intercepts are found by plugging in 0 for x and solving to get the point (0,result)
Vertical asymptotes and horizontal asymptotes have already been found at this point
your vertical asymptote is the x= from when you found the domain before you set up your intervals
your horizontal asymptote is the y= from when you found the range before you set up your intervals

#3
Do the first derivative test by
Taking the derivative and setting it equal to zero
Solve for x
check differentiability
Set up intervals (-infinity, pt)U(pt, infinity)
Plug in value on the interval into the derivative and solve
((if you get a positive number the function is increasing, If you get a negative number the function is decreasing))
(( A max is increasing then decreasing, A min is decreasing then increasing))

#4
Do second derivative test
take the derivative of the first derivative
set it equal to zero
solve for x
check differentiability
set up intervals (-infinity, pt)U(pt,infinity)
plug into the second derivative and solve
((if you get a positive number the function is concave up, If you get a negative number the function is concave down))
((any point on the interval where a number switches from concave up to concave down or vise versa there is a point of inflection))

#5
Plot all important information on a sketched graph.

IMPORTANT INFORMATION INCLUDES:
domain intervals
range intervals
x intercepts
y intercepts
vertical asymptotes
horizontal asymptotes
whether the intervals of the first derivative test are increasing or decreasing
where there is a max or a min
whether the intervals of the second derivative test are concave up or concave down



Example:
y= 2(x^2-9)/x^2-4


#1
find the domain and find the range

The domain:
x^2-4=0
x^2=4
x = +/- 2
(-infinity,-2)(-2,2)(2,infinity)

The range is found by finding the horizontal asymptotes and setting up intervals
y= 2
(-infinity,2)(2,infinity)

#2

X intercepts are found by setting the original equal to zero and solving for x to get the point (result,0)
2(x^2-9)=0 therefore: x=+/-3
Y intercepts are found by plugging in 0 for x and solving to get the point (0,result)
2(0^2-9)/0^2-4
2(-9)/-4
-18/-4
9/2
(0,9/2)

your vertical asymptote is the x= from when you found the domain before you set up your intervals
your horizontal asymptote is the y= from when you found the range before you set up your intervals

#3
Do the first derivative test by
Taking the derivative and setting it equal to zero
x^2-4(4x)-[2x^2-18(2x)]/(2^2-4)^2
20x/(x^2-4)^2
Solve for x
20x=0
x=0
check differentiability
vertical asymptotes at +/- 2
Set up intervals (-infinity, pt)U(pt, infinity)
(-infinity,-2)(-2,0)(0,2)(2,infinity)
Plug in value on the interval into the derivative and solve
f(-3) negative therefore decreasing
f(-1) negative therefore decreasing
f(1) positive therefore increasing
f(3) positive therefore increasing

MIN @ x=0
((if you get a positive number the function is increasing, If you get a negative number the function is decreasing))
(( A max is increasing then decreasing, A min is decreasing then increasing))

#4
Do second derivative test
take the derivative of the first derivative
-20(3x^2+4)/(x^2-4)^3

set it equal to zero
-20(3x^2+14)=0
solve for x
x=sqrt -4/3 therefore does not apply

check differentiability
vertical asymptotes at +/-2
set up intervals (-infinity, pt)U(pt,infinity)
(-infinity, -2)(-2,2)(2, infinity)
plug into the second derivative and solve
f(-3) negative concave down
f(0) positive concave up
f(3) negative concave down


Then you would proceed to plot all important information on a sketched graph

Week 5 Blog Prompt

Using Calculus how can you find a max or a min? How do you determine if the value is a max or min?

Blog 11?

Alrightttt, so this week we went back to chapter three, and we are on section 3.4 which is continuity. It is verrry easy surprisingly! The whole thing is just taking first derivatives and second derivatives, while using the first derivative test. **Doing this will help you determine whether the graph is concave up or down, and if you do all 7 steps, it will help you to find the points of inflection.
Here are the 7 steps:
1. Take 1st derivative=0 Solve.
2. Take 2nd derivative=0 Solve.
3. Differentiable (Remember what makes a graph differentiable)
4. Set up intervals
5. Pick number’s on an interval, and plug into **2nd derivative
6. If the numbers positive the graph is concave up, if negative concave down
7. If there is a change it is a point of inflection

Example: Determine the open intervals on which the graph is concave up or down
**Notice this problem isn’t looking for points of inflection, so you wouldn’t have to work step 7.
-x^3+6x^2-9x-1
F’(x)=-3^2+12x-9
F”(x)=-6x+12
X=2
(-infinity,2) u (2,infinity)
F(1)=-6(1)+12=6
F(3)=-6(3)+12=-6
(-infinity,2) is concave up (infinity, 2) is concave down

Example 2:
X^2+1/x^2-1
F’(x)=-4x/x^2-1^2
F”(x)=4(-1+4x^2)/(x^2-1)^2
(-infinity,-1),(1, infinity) Concave up; (-1,1) Concave down
**With fractions, if you use quotient rule to solve for second derivative numbers will always cancel out from the top and the bottom making it easier to solve.

Now for something I don’t understand! I don’t understand at all how to find points of inflection, I know what it is, I just don’t know how to find it. I’m not sure if you have to graph the function then look at it? Or your just suppose to know where the graph changes?

31 810g0 d3 m47h3m471c0

Well, this week we learned about the second derivative test
basically, it can be summed up like this:
"*In an Xzibit accent* Yo dawg, we heard you like math, so we put a derivative in your derivative, so you can derive after you derive."
hahahaha, good old internet memes B)
anyways, the second derivative test is just the process of taking the second derivative after the first derivative test; I know, i was in shock after I found out the ugly truth too

anyways, there's a very simple way to solving it/using it
1) take the first derivative set equal to zero and solve.
2) take the second derivative set = 0 and solve
3) Is it differentiable?
4) set up intervals
5) pick numbers on the interval and plug them into the last derivative
6) if its positive then the function is concave up on that interval and if its negative then it is the opposite.
7) Where is there is a change it is a point of inflection?

11/14

so we learned about the second derivative test...

take the first derivative and set = 0
then take the second derivative set = 0 and solve for x.
check for differentiability; if it isnt differentiable, the test failed.

set up intervals with the x values from the second derivative
plug values from intervals into the second derivative
if the value is positive, the graph is concave up; if the value is negative, it is concave down

there is a shortcut:

take the first derivative and set = 0
check differentiability
plug in critical numbers into the second derivative
if it is positive, its a min. if it is negative, its a max. if it is zero, it fails.

Alaina's blog, 14 November 2010

This week, we talked about the second derivative test. It has most of the same steps as the first derivative test, and then some of its own.
First, you take the first derivative, set it equal to zero, and solve for x.
Then, you take the second derivative, set it equal to zero, and solve for x.
Next, you check for differentiability. If it is not differentiable, the second derivative test fails.
Then, you set up intervals with your x value from the second derivative.
Next, you plug values from your intervals into the second derivative.
If the output value is positive, it is concave up. If the output value is negative, it is concave down.

Ex: Determine the intervals on which the graph is concave up or concave down for f(x)=(-x^3)+(6x^2)-9x-1
**because it is only asking for the intervals on which the graph is concave up or down, you do not need to complete step 1 (solve the first derivative for x).
1. f’(x)=(-3x^2)+(12x)-9
2. f”(x)=(-6x)+12 = 0
-6x=-12
x=2
3. yes it is differentiable
4. (-infinity, 2)u(2, infinity)
5. f”(0)=(-6(0)^2)+12= positive, concave up.
f”(3)=(-6(3)^2)+12=negative, concave down.
The function is concave up on the interval (-infinity, 2) and concave down on the interval (2, infinity).

11/14

So we learned the 2nd derivative test to go with some concave related problems so not that bad

Our 7 Steps:

1. Take the 1st Derivative and set = 0 Solve for x
2. Take the 2nd Derivative and set = 0 Solve
3. Check for Differentiability
4. Set up your interval
5. Plug in numbers on the intervals into the 2nd derivative
6. Positive results go up, Negative results go down, If there is a shift, then there is a point of inflection
7. Changes are the points
Also if a point causes it to = 0 then the test fails (no surprise there)

We use this test to find points of inflection and concavity, as well as a shortcut to the first derivative test.

THE SHORTCUT:
1. Take the 1st Derivative and set = 0
2. Differentiability check
3. Plug in your critical numbers into the 2nd derivative
4. Positive is a min, negative is a max, 0 is a fail

This should wrap up this week, not so bad i hope this one will be good too :D

Blog 11/14/2010

This week wasn't that bad, it passed by pretty fast, and this coming week will pass even faster. This week we learned the 2nd derivative test, amongst other math stuffs like the shortcut to the 1st derivative test. I will explain the 2nd derivative test and give an example or two.

The steps for the 2nd derivative test:

1. Take the 1st derivative and set it equal to zero, then solve for x.

2. Take the 2nd derivative, and set it equal to zero, then solve for x.

3. Now you have to check for differentiability or non-differentiability.

4. Set up your intervals.

5. Pick numbers on the intervals then plug them into the 2nd derivative.

6. If it comes out positive, the function is concave up. If negative, then the function is concave down.

7. There is a point of inflection if there is a change. i.e. concave up to down, or down to up, increasing to decreasing, decreasing to increasing, etc.

The steps for the shortcut to the 1st derivative test:

1. Find the first derivative, set it equal to zero, then solve for x.

2. Check for differentiability or non-differentiability.

3. Plug in critical numbers to the 2nd derivative.

4. If it is positive, there is a minimum. If it is negative, there is a maximum. If it is 0, then the test fails.

Bloggy Bloggy Bloggedy

So, this week we learned the second derivative test and the first derivative shortcut(a little late?).

So, to do the second derivative test you basically do the same thing as the long way of the first one just with the second derivative.
First: Take the derivative and set =0(haha a face) and solve
Next: Take the 2nd derivative and set =0 and solve
Thirdlyful: Take the 3rd derivat... I mean check to see if its differentiable and find out where its not

AfterStep3: Set up Intervalmajiggers
Fifth: Pick some numbers from your intervals and plug into the second derivative
Next: If you get positive, you have ai...i mean its concave up, if negative you aren't preg...i mean its concave down
Lastly: Where it goes from up to down, or down to up, or up to not up, there is a point of inflection, which is apparently important.

First derivative test shortcut oh my
First, take the derivative set =0(theres that face again) and solve
next, find your not differentiable areas
third, plug in critical points to the second derivative
last: if positive you get a min, negative its not a min, 0 then you suc... i mean the test doesn't work

blog #keeweyb

this week we learned all aout the second deivative test.

first yu take the first derivative and set it equal to zero and solve.

then you take the second derivative and do the same.

then you check for differentiaility

the yo et up your intervals

then yo pik numbers on that interval and plug them into the second derivative.

chec for points of inflection.

if youz gotz a positive numba its concave up

if it be negative then its concave down

n ifs it zero false

It's a bloggg. BLOG!

This week we learned the second derivative test and the shortcut to the first derivative test. The second derivative test is the exact same thing as the first derivative test but you take a second derivative. Easy aye?

1:take the first derivative set equal to zero and solve.
2:take the second derivative set = 0 and solve
3:Is it differentiable?

4:set up intervals
5:pick numbers on the interval and plug them into the last derivative
6:if its positive then the function is concave up on that interval and if its negative then it is the opposite.
7:Where is there is a change it is a point of inflection?

The shortcut to the first derivative test makes your life oh so much easier.

1:take the derivative, set equal to zero, and solve
2:Check to see if differentiable
3:Plug in critical points into the second derivative
4:If + there is a min
If - there is a max
If 0 the test is false

Blog #12

This week, we continued ch. 3 and learned about concavity and the 2nd Derivative Test.
Okay, concavity. To test concavity, you use the same steps as the first derivative test, but with the 2nd derivative. To review, here are the steps:
1. Take the 2nd derivative.
2. Set it equal to 0.
3. Solve for x.
4. Check differentiability.
5. Set up intervals.
6. Pick numbers on the intervals and plug them into the 2nd derivative.
7. If you get a positive number, it is concave up. If you get a negative number it is concave down.
8. If there is a change (like from positive to negative or vice versa), it is a point of inflection.

Ex. 1) Determine the open intervals on which the graph f(x) = -x^3 + 6x^2 – 9x – 1 is concave up or down.
f’(x) = -3x^2 + 12x – 9
f’’(x) = -6x + 12
-6x + 12 = 0
-6x = -12
x = 2
It is differentiable.
(-infinity, 2) u (2, infinity)
f’’(1) = positive #
f’’(3) = negative #
concave up: (-infinity, 2)
concave down: (2, infinity)
*Note that the question does not ask for the point of inflection, but if it did, you would say: x = 2 is a point of inflection.

Now, the 2nd Derivative Test. It is simply a shortcut for the 1st Derivative Test. Also, this test does not always work. If for step 7 you get a 0, the test fails. If that ever happens, you must use the 1st Derivative Test. Anyway, the steps are:
1. Take the 1st derivative.
2. Set it equal to 0.
3. Solve for x.
4. Check differentiability.
5. Take the 2nd derivative.
6. Plug in your critical points (your x-values) into the 2nd derivative.
7. If you get a positive number, it is a min. If you get a negative number, it is a max.

Ex. 2) Use the 2nd Derivative Test to find the relative extrema of f(x) = x^3 – 3x^2 + 3.
f’(x) = 3x^2 – 6x
3x^2 – 6x = 0
3x(x – 2) = 0
x = 0, 2
f’’(x) = 6x – 6
6(0) – 6 = negative #
6(2) – 6 = positive #
x = 0 is a max
x = 2 is a min

11/14/10

Well this week went by super fast! Only one more week until the holidays!!!!!!! :D Anyway, this week all we really did was take time to review for our Chapter 5 test which we took Thursday I think? And we didn't learn anything new except for Friday. Friday we learned about concavity and the Second Derivative Test. Not too many steps changed for that so it's not too bad. It's pretty easy so far. Sooooo, here are some examples:

Ex. 1) Determine the open intervals on which the graph is concave upward or concave downward.
a.) f(x)=3x^2-x^3
*Okay when the directions ask you specifically for concavity, all you have to do is find the second derivative of the function, set it equal to zero, solve for x, check for differentiability, and set up intervals (intervals just like in the 1st derivative test..except this time, when you get a positive number, that means it is concave up on that interval and when you get a negative number that means it is concave down on that interval)
*So first let's take the derivative of this function. You should get 6x-3x^2
*Now let's take the 2nd derivative...You should end up with 6-6x
*Now set 6-6x equal to zero and solve for x
*You should get x=1; Now set up your intervals like this:
(-infinity,1)u(1,infinity)
*Now plug values in those intervals into the 2nd derivative to find out if it is negative or positive
*For the first interval you can plug in 0, and you should get a positive number
*For the second interval you can plug in 2, and you should get a negative number
**Soooo, your answer is:
(-infinity,1) concave up
(1,infinity) concave down
b.) f(x)=(x^2+1)/(x^2-1)
*We're doing the same thing as problem a. so first, take the derivative of the function using the quotient rule
*For the 1st step of the first derivative you should get this: (x^2-1)(2x)-[(x^2+1)(2x)/(x^2-1)^2
*f'(x)=(-4x)/(x^2-1)^2
*Now for the 2nd derivative...For the 1st step of that you should get this:
(x^2-1)^2(-4)-[(-4x)(2(x^2-1)(2x)]/(x^2-1)^4
*Simplifying all of that (which takes quite a while) you should get this:
f''(x)=4(3x^2+1)/(x^2-1)^3
*Now, take the top of that fraction^ and set it equal to zero to find your x-values
*4(3x^2+1)=0
3x^2=-1
...**You get an i, so this does NOT work..
*Now you have to check for differentiability [which I didn't really do in problem a., because it is a polynomial and they are differentiable everywhere :)]
-So to check for differentiability here you take the bottom of the fraction (where there is a vertical asymptote) and set it equal to zero to find the x-values
*So you get x^2-1=0
x^2=1
x=+/-1
*Now you set up your intervals like this:
(-infinity,-1)u(-1,1)u(1,infinity)
*Now plug values in the intervals into the 2nd derivative to find out if it's positve or negative
*For the first interval you can plug in -2, and you should get a positive number
*For the second interval you can plug in 0, and you should get a negative number
*For the third interval you can plug in 2, and you should get a positive number
***So your answer is:
(-infinity,-1)u(1,infinity) concave up
(-1,1) concave down

**I understood what we learned this week so I don't have any questions...yet.

Devin's Reflection

Definition of the Natural Exponential Function

The inverse function of the natural logarithmic function f(x) = ln x is called the natural exponential function and is denoted by f^-1(x) = e^x, that is, y = e^x if and only if x = lny.

The inverse relationship between the natural logarithmic function and the natural exponential function can be summarized as follows.

Ln(e^x) = x and e^lnx = x

Ex. 9 = e^x-3

Ln3 = ln(e^x-3)

Ln9 = x-3

X = 3 + ln9

Theorem 5.10 Operations with exponential functions

Let a and b be any real numbers.

1. E^a e^b = e^(a+b)

2. E^a / e^b = e^(a-b)

Properties of the Natural Exponential Function

1. The domain of f(x) = e^x is (-oo,00), and the range is (0, oo)/

2. The function f(x) = e^x is continuous, increasing, and one to one on its entire domain.

3. The graph of f(x) = e^x is concave upward on its entire domain.

4. Lim e^x = 0 and lim e^x = oo

Taylor blog #12

The majority of this week we were reviewing and studying for the chapter five test. However, friday we did learn something new.
We learned the second derivative test and the shortcut to the first derivative test.

Some key words to recognize when to use the second derivative test are:
"Use the second derivative test"
"which direction is the concavity"
"determine the intervals on which the graph is concave up or down"
"determine the points of inflection"


an easy way to try to remember the steps for the second derivative test is to remember that you will be doing the same steps as the first derivative test except with a second derivative

STEPS TO USING THE SECOND DERIVATIVE TEST:

Step #1:
take the first derivative set equal to zero and solve. ((keep in mind that this step will not always be necessary but can always be performed anyway. This step is not necessary when one looking for concavity or points of inflection.))

Step #2:

take the second derivative set = 0 and solve

Step #3:
Check differentiability

Step #4:
set up intervals

Step #5:
pick numbers on the interval and plug them into the SECOND derivative

Step #6:
if POSITIVE: the function is concave up on that interval
if NEGATIVE: the function is concave down on that interval

Step #7:
If there is a change it is a point of inflection ((This step only needs to be done when the problem is asking for the points of inflection specifically))

Example #1:

Determine the open intervals on which the graph is concave up or concave down:
F(x)= 6/ x^2+3

Step #1:
take the first derivative set equal to zero and solve. (( But this step is not necessary because the problem is looking for concavity.))

Step #2:

take the second derivative set = 0 and solve

(x^2+3)(0)-[(6)(2x)]/ (x^2+3)^2

F'(x)= -12x/ (x^2+3)^2

(x^2+3)^2(-12)-[(-12x)(2(x^2+3)(2x))]/ (x^2+3)^2

(x^2+3)(-12)+48x^2

-12x^2-36+48x^2

F"(x)= -36x^2-36=0

-36x^2=36

x^2= 1

x= +/- 1


Step #3:
Check differentiability

DIFFERENTIABLE?
yes.

Step #4:
set up intervals

(-infinity,-1)U(-1,1)U(1, infinity)


Step #5:
pick numbers on the interval and plug them into the SECOND derivative

F"(-2)= 36(-2)^2-36/ (-2^2+3)^2 = POSITIVE

F"(0)= 36(0)^2-36/ (0^2+3)^2 = NEGATIVE

F"(2)= 36(2)^2-36/ (2^2+3)^2= POSITIVE

Step #6:
if POSITIVE: the function is concave up on that interval

Therefore the function is concave up on the intervals (-infinity, -1)U(1,Infinity)

if NEGATIVE: the function is concave down on that interval

Therefore the function is concave down on the interval (-1,1)

Step #7:
If there is a change it is a point of inflection.

This step only needs to be done when the problem is asking for the points of inflection specifically therefore this step is not necessary for this problem because it is not asking for the points of inflection.




THE SHORTCUT TO THE FIRST DERIVATIVE TEST

Steps to the shortcut for the first derivative test:

Step #1:

take the derivative, set equal to zero, and solve

Step #2:

Check to see if differentiable

Step #3:

Plug in critical points into the second derivative

Step #4

If POSITIVE there is a min
If NEGATIVE there is a max
If 0 the test is false

(((NOTICE THAT FOR THE SHORTCUT WHETHER THE PLUG IN IS POSITIVE OR NEGATIVE HAS A REVERSED MEANING)))


Example:

Step #1:

take the derivative, set equal to zero, and solve


-15x^4+15x^2=0

15x^2(-x^2+1)

x= 0, 1,-1


Step #2:

Check to see if differentiable

differentiable?
yes.


Step #3:

Plug in critical points into the second derivative

F"(X)=60x^3 + 30x

-60(0)^3+30(0)=0
-60(-1)^3+30(-1)=POSITIVE
-60(1)^3+30(1)=NEGATIVE

Step #4

If POSITIVE there is a min

Therefore, at x= -1 there is a min


If NEGATIVE there is a max

Therefore, at x= 1 there is a max


If 0 the test is false

Therefore, at x=0 the test fails

Week 4 Blog Prompt

How can you find the domain of a graph? How do you know if a functions is increasing or decreasing? Give an example of each.

Devin Blog Prompt Stuff

uhhhh I do not see the blog prompt, so I just making it known that I came on here to do my prompt. So I hope you feel better Mrs. Robinson. Get Well Soon :-)

Taylor Rodriguez Concerning the blog prompt

The blog prompt hasnt been posted yet and it is due today so im letting all know that i checked for it and i will complete it if it is ever posted

Taylor Rodriguez blog #11

This week we skipped to chapter five
Chapter five is dealing with derivatives of logs
and
exponents that are variables.

Chapter 5 section one-
covers taking the derivative of a natural log.
For completing this task there is a simple formula to follow.
this formula is
D/dx ln(u)= 1/u X (du/dx)
This formula is read as the derivative of natural log u is found by putting one over what is in parenthesis multiplied by the derivative of what had been in parenthesis.
Therefore you would plug into the formula
Simplify
and
solve to get your result.

DO NOT FORGET THAT YOUR DERIVATIVE RULES SUCH AS THE CHAIN RULE THE PRODUCT RULE AND THE QUOTIENT RULE STILL APPLY.


Some examples of chapter five section one include:
'
d/dx ln(2x) = 1/2x (2)
d/dx ln(2x)= 2/2x
Therefore ln(2x)= 1/x

d/dc ln (x^2+1)= 1/x^2+1 (2x)
Therefore ln(x^2+1)= 2x/x^2+1

d/dx x ln (x) = x(1/x (1)) + lnx (1)
d/dx x ln (x)= x(1/x)+ln x
d/dx x ln (x)= x/x+lnx
Therefore d/dx x ln (x)= 1 + lnx

d/dx (ln(x))^3= 3(ln x)^2 (1/x) (1)
d/dx (ln(x))^3= 3(ln x)^2 (1/x)
Therefore d/dx (ln(x))^3= 3(ln x)^2/ x


Chapter five section two


covers taking the derivative of an ecponential function.
For completing this task there is also a simple formula to follow.
the formula for solving these problems is
D/dx e^u = e^u X (du/dx)
This formula is read as the derivative of the exponential function e is found by recopying what the problem is asking to take the derivative of and multiplying it by the derivative of the exponent.
Therefore when trying to solve these problem too you would plug into the formula
Simplify
and
solve to get your result.
AGAIN!
DO NOT FORGET THAT YOUR DERIVATIVE RULES SUCH AS THE CHAIN RULE THE PRODUCT RULE AND THE QUOTIENT RULE STILL APPLY.

Some examples of these type problems include

d/dx e^2x-1 = e^2x-1 (2)
Therefore d/dx e^2x-1 = 2e^2x-1

d/dx e^-3/x= e^-3/x (-3x^-1)
d/dx e^-3/x= e^-3/x (3x^-2)
d/dx e^-3/x= e^-3/x (3/x^2)
Therefore d/dx e^-3/x= 3e^-3/x / x^2




Chapter five section four
covered taking the derivative for exponents other than e
For completing this task there are a few simple formulas to follow.
one of the formula for solving these problems is
D/dx A^u = A^u(ln (a))(du/dx)
This formula is read as the derivative of a number raised to a variable as a power is found by multiplying what the problem is asking to take the derivative of by ln of the number being raised to the variable and then multiplying all of that by the derivative of the exponent.
Therefore when trying to solve these problem too you would plug into the formula
Simplify
and
solve to get your result.
Some examples of these type problems include

d/dx 2^x= 2^xln2 (1)
Therefore the derivative of 2^x is 2^x ln 2

d/dx 2^3x= 2^3x ln 2 (3)
Threfore the derivative of d/dx 2^3x is 3 (2^3xln2)

d/dx 7 ^x^2= 7^x^2ln7 (2x)
Therefore the derivative of d/dx 7 ^x^2 is 2x(7^x^2ln7)


The other formula for solving these problems is
d/dx logaU= 1/U(ln(a)) X (du/dx)
This formula is read as the derivative of a log is found by placing what the problem is asking for in asking for the derivative of the log under one then multiplying it by the derivative of what the problem is asking for in asking for the derivative of the log.
Therefore when trying to solve these problem too you would plug into the formula
Simplify
and
solve to get your result.
Some examples of these type problems include

d/dx log cosx = 1/cosx(ln10) (-sin x)
d/dx log cosx = -sinx/ cosx(ln10)
Therefore the derivative of log cos x= -tanx/ ln10

d/dx Log5 x^3+4x= 1/x^3+4x(ln5) (3x^2+4)
Therefore d/dx Log5 x^3+4x= 3x^2+4/ (x^3+4x)(ln5)


AGAIN!
Please!
DO NOT FORGET THAT YOUR DERIVATIVE RULES SUCH AS THE CHAIN RULE THE PRODUCT RULE AND THE QUOTIENT RULE STILL APPLY.

11/7/2010

I dislike logssssssssss anyways. This week went by super fast. We learned basic logs well we actually relearned basic logs in some ways. We now know how to derive natural logs and logs with “e” in them. The two formulas that basically sum all of it up.

derivative of a natural log:

d/dx ln u = (1/u)(du/dx)

Exponential functions:

number raised to a variable

d/dx e^u = e^u x du/dx

These two different formulas are super basic so you dont even need examples because I know all of you know basic algebra.

Blog of Blogginess

Intro of darkness, then wordiness, then spacious-ness

lol, stole that from TobyTurner

well, last week, we learned some of the most basic stuff in math, logs

unfortunately, we learned the derivatives for them too

d/dx log(x^2 + 1) = 1/(x^2 + 1) * 2x
2x/(x^2 + 1)

and then there's the basic, basic, BASIC, log rules

11/7/10

So we did some chapter 5, and... i liked it! hahahaha it's some pretty simplistic stuff when you keep up with it. what we are doing includes the derivatives of natural logs

d/dx ln(x)=1/x x d/dx

1 over inside multiplied by derivative of the inside

d/dx lnx^2= 1/x^2 x 2x= 2x/x^2=2/x

exponents:

raising numbers to a variable, recopy then multiply by the derivative of the exponent

x^3x= x^3x x 3 = 3x^3x

ln e^x=7
ln e cancels out

x=7

pretty basic stuff, remember your log properties and formulas and you do well!

Blog 11

Alright, so this week wasn’t too bad, we’ve been able to do a lot of things to get our grade up which is really helpful :). We also started learning chapter five, which is pretty simple but can get hard just like everything else we learn in calculus. It still deals with derivatives but it goes back to simple concepts, like exponents and logs. *So you will have to review all of those rules before trying to work the problems. You want to make them as simple as possible before taking the derivative of a problem.

Section 5.1: d/dx ln u=1/u(du)

Example: d/dx= ln (2x)= 1/2x(2)=1/x
d/dx= ln(x^2+1)=1/x^2+1(2x)=2x/x^2+1
-For these problems you just put the number behind ln in the denominator, and leave it as it is. Then multiply it by the derivative of that number. They also have problems like this that involve chain rule, and product rule.

Here are some examples using log rules:
Expand:
ln xy/z=ln(x)(1)+(y)(1)-ln z
-Using the log rules when something is multiplied it means add, when something is divided it means subtract. This one also uses the product rule.
Condense:
ln(x-2)-ln(x+2)=ln(x-2)/(x+2)

The exponential functions are pretty much the same as the log ones.
d/dx=e^u=e^u(du/dx)
-Recopy the problem, then times it by the derivative of u.

Examples:
E^2x-1=e^2x-1(2)=2e^2x-1
E^-3/x=e^-3/x(3/x^2)=3e^-3/x/x^2
Rules to remember for exponential functions:
ln e^x=4, x=4
e^lnx^2=9, x=+/- 3
e^x=5, x=ln 5

----------------------------------------
Something that's still giving me trouble is using more than one rule together in these problems. I get confused when the chain rule and product rule get mixed, then when they throw in trig functions I just get even more confused. I think I just need more practice on this.

Blog 11/7/2010--Stephen

This week was a pretty good one. It went by pretty fast and what we learned is actually easy and I can speed through the problems lol. This week we learned how to take the derivatives of natural logs and logs, and we also reviewed some of the basic properties of logs and natural logs. I will describe how to take some derivatives of natural logs and logs, and I will also give many examples.

derivative of a natural log:

d/dx ln u = (1/u)(du/dx)

Example:

d/dx ln (2x) = (1/2x)(2) = (1/x)

d/dx ln (x^2 + 1) = ((1)/(x^2+1)) x (2x) = (2x)/(x^2 + 1)

Exponential functions:

number raised to a variable

d/dx e^u = e^u x du/dx

Example:

d/dx e^(2x-1) = e^(2x-1) x 2

2e^(2x-1)

d/dx e^(-3/x) = e^(-3/x) x (3/x^2)

(3e^(-3/x))/(x^2) = ((3)/((x^2)(e^(3/x)))

ln e^x = 4

x = 4

e^ln(x^2) = 9

x^2 = 9

x = plus or minus 3

e^x = 5

ln e^x = ln 5

x = ln 5

ln(x-2) = 3

e^3 = x-2

x = e^3 + 2

Alaina's make up blog for week 3

That week, we talked about Rolle’s Theorem, the first derivative test, and the mean value theorem. Each helps to find an absolute or relative extrema on a closed interval.
I’m only going to talk about Rolle’s Theorem because it takes a while to explain and my examples are long.

FIRST THERE ARE SOME RULES YOU NEED TO KNOW:
-To find the extrema of a continuous function f on a closed interval [a,b]:
1. find the critical numbers of f in [a,b]
2. evaluate f at each critical number in [a,b]
3. evaluate f at each endpoint in [a,b]
4. the least of these values is the absolute min, the greatest is the absolute max
5. anywhere the function is not not differentiable is a critical number

KEY WORDS:
Extreme values: maximums and minimums
Absolute max: highest point on an interval
Absolute min: lowest point on an interval
Relative (local) max: any max not on the interval
Relative (local) min: any min not on the interval
Critical numbers: any max or min; typically referred to as x=c

** Directions: to find a critical number, extreme value, max, min or horizontal tangent, set derivative = 0.

Rolle’s Theorem:
-Let f be continuous on the closed interval [a,b] and differentiable on the open interval (a,b). If f(a)=f(b) then there is at least one number c in (a,b) such that f’(c)=0. It basically tell you that you have at least one max or min on the interval.
*f’(x)=0 is short for there is at least one max or min on the interval.
**You must test for continuity and differentiability before using Rolle’s Theorem. You must check if the y-values match. (plug in and check).

Ex1: Find the 2 x-intercepts of f(x)=x^2-3x+2 and show that f’(x)=0 at some point between the 2 x-intercepts.
a)continuous? Yes. All polynomials are.
b)Differentiable? Yes. Everywhere.
c)x^2-3x+2
(x-2)(x-1)
s=1,2 =>interval [1,2]
d)f(1)=(1)^2-3(1)+2=0
f(2)=(2)^2-3(2)+2=0
They have the same y-value
By Rolle’s Theorem, the function is continuous, differentiable and f(1)=f(2) thus there must be at least one max or min (f’(x)=0) on the interval [1,2].

Ex2: Let f(x)=x^4-2x^2. Find all values of c in the interval (-2,2) such that f’(c)=0.
a)continuous? Yes.
b)Differentiable? Yes.
c)f(-2)=(-2)^4-2(-2)^2=8
f(2)=(2)^4-2(2)^2
d)find all values:
4x^3-4x=0
4x(x^2-1)=0
4x=0, x^2-1=0
x=0, x=-1, 1
By Rolle’s Theorem, the function is continuous, differentiable, and c=0.

blog#golb

okay so this week wasn't too bad. we learned about how to find the derivative of a natural log or log as well as a logarithm and number raised to an exponent.

in order to find the derivative of a natural log you have to multiply 1/u by du where u is the inside of the natural log and du is the derivative of the inside of the natural log.

for example: ln(7x)

d/dxln(7x) = 1/7x multiplied by 7 = x

and when finding the derivative of e^u you take e^u multiplied by the derivative of u.

for example: d/dx e^ x^2 - 1

e^x^2 - 1 multiplied by 2x = 2xe^x^2-1

Devin's Reflection

Definition of the Natural Logarithmic Function

-the natural logarithmic function is defined by: ln x = f1^x 1/t dt, x > 0.

the domain of the natural logarithmic function is the set of all positive real numbers.

Theorem 5.1 Properties of the Natural Logarithmic Function

The natural logarithmic function has the following properties.

1. The domain is (0, oo) and the range is (-oo,oo).

2. The function is continuous, increasing, and one to one.

3. The graph is concave downward.

Domain of f(x) = ln x^2

First derivative f’ = ln 2x

Second derivative f’’ = ln 2

Using the definition of the natural logarithmic function, you can prove several important properties involving operations with natural logarithms. If you are already familiar with logarithms, you will recognize that these properties are characteristic of all logarithms.

Theorem 5.2 Logarithmic Properties

If a and b are positive numbers and n is rational, then the following properties are true.

1. Ln(1) = 0

2. Ln(ab) = ln a + ln b

3. Ln(a^n) = n ln a

4. Ln(a/b) = ln a – ln b

11/7/10

Welllllllllllllllll, this week we skipped over to Chapter 5 and learned how to derive natural logs, logs, exponential functions, etc...Thank god we finally had a break from the more difficult things. This week was such a breeze for me; and I understood EVERYTHING! Soooo, with that, here are some examples :)

Ex. 1) Find the derivative of the function:
a.) lnx^2
*the rules for taking the derivative of a natural log say that you have to make it 1/the inside x the derivative
*So you should get 1/x^2(2x)
=2x/x^2..which simplifies to give you 2/x
b.) ln(2x^2+1)
=1/2x^2+1(4x)
=4x/(2x^2+1) ..and that does not simplify so that's your derivative
c.) ln(x+1)^2
*seeing that the function is square should give you the hint that you need to use the chain rule for this...but not yet
*First you should get 1/(x+1)^2
*Now you have to multiply that^ by the derivative of the denominator which is (using the chain rule) 2(x+1)
*So you should get 2(x+1)/(x+1)^2
*the (x+1)'s can cancel and you're left with 2/(x+1)
d.) lnx/x^2
*this is in fact a quotient rule...so here we go
*Using the quotient rule you should end up with this as your first step:
(x^2)(1/x)-[(lnx)(2x) / (x^4) ...keep in mind that the derivative of lnx is 1/x
*simplifying that^ you should get (x-2xlnx)/x^4

Ex. 2) Find an equation of the tangent line to the graph of f at the given point:
f(x)=3x^2-lnx (1,3)
*First take the derivative of the function
*You should get: 6x - (1/x)
*You have the point (1,3)..to find the equation of the tangent line, you have to plug that x-value into your derivative to get your slope
*So you get 6(1) -1/1
=5
*Now you plug everything into the point-slope formula and you should get this:
y-3=5(x-1)

Ex. 3) Find the derivative of the following:
a.) f(x)=e^2x
*In order to find the derivative of anything with e, the rule says that you have to recopy the function and multiply by the derivative (derivative of e's exponent)
*So you should get e^2x(2)
*And that can be written as 2e^2x
b.) 3e^(1-x^2)
*First, like a regular derivative, you would leave the number out in front and then you would recopy the function..then multiply by its derivative
*So you should have:
3(e^1-x^2 x -2x) ....x means times
*And simplifying that you should get -6xe^(1-x^2)
c.) x^2e^-x
*looking at this problem you should notice that two things are being multiplied..sooo, that means you'll have to use the product rule
*So you should get this as your first step: (x^2)(e^-x x -1) + (e^-x)(2x)
*Simplifying that you get -x^2e^-x + 2xe^-x
d.) 7^(2x-1)
*hold upppppp, there's no e in this one so you can't take the derivative the same way you have been for the other ones..there's a separte rule for this!
*it says that you recopy the function, multiply it by "ln#", and also multiply it by the functions derivative. that probably didn't make sense, but here's what I mean..
*You should get 7^(2x-1) x ln7 x 2
*Well you can't really simplify anything I don't think so you can just rewrite it as:
2(ln7)7^(2x-1)

**Like I said before, this week was easyyyyy and I had no problems!

Blog #11

This week, we worked on Ch.5. It’s very simple, just derivatives of natural logs (ln).

Derivatives of ln:
d/dx ln u = (1/u)(du/dx)
*don’t forget ln properties, product rule, quotient rule, and chain rule

Ex. 1) d/dx ln 3x = (1/3x)(3) = 1/x

Ex. 2) d/dx ln (x^2 + 2) = (1/(x^2 + 2))(2x) = 2x/(x^2 + 2)

Ex. 3) d/dx xlnx = (x)(1/x) + (lnx)(1) = 1 + lnx

Ex. 4) d/dx (ln x)^4 = (4(ln x)^3)(1/x)(1) = 4(ln x)^3 / x

Ex. 5) d/dx ln square root (x + 1) = ln(x + 1)^1/2 = ½ ln(x + 1) = (1/2)(1/(x + 1))(1) = 1 / 2(x + 1)

Ex. 6) d/dx ln (2x/3x^2) = ln (2x) – ln (3x^2) = (1/2x)(2) – (1/(3x^2))(6x) = (1/x) – (2/x) = - 1/x

We also learned how to take the derivative exponential functions with, which are numbers raised to variables.

Derivates of exponential functions:
d/dx e^u = (e^u)(du/dx)

Ex. 7) e^2x = (e^2x)(2) = 2e^2x

Ex. 8) e^(-5/x) = (e^(-5/x))(5x^-2) = 5e^(-5/x) / x^2

Another thing we learned is how to solve for x in ln equations.

Ex. 9) ln e^x = 6
x = 6

Ex. 10) e^(lnx^2) = 9
x^2 = 9
x = 3, -3

Ex. 11) e^x = 3
ln e^x = ln 3
x = ln 3

Ex. 12) ln (x – 3) = 2
e^2 = x – 3
x = e^2 + 3

MakeUp Blogs

Week 1 Prompt:
What concept do you feel like you mastered this nine weeks? What concept did you struggle with the most? Why? What can you change this nine weeks in your study habits, etc to improve your grade?

What have I mastered from the first nine weeks? Well, i feel that I have basically mastered most of the derivative rules. I understand how to do all of the complex chain, product, and quotient rules along with the definition of a derivative.

What did I struggle with? I really didn't have problems with much, one thing I still haven't learned is infinity limits, but that's because i don't feel like studying the different outcomes.

What can I change? I really need to start doing my homework and taking this class more seriously. Due to my failures to keep up with my studies I'm doing a ton of homework tonight including these make up blogs. So, I will start doing homework, blogs, and studying for tests to keep my grade in a good place.

Week 2 Prompt:
What are the steps to a related rate problem? What are the key words you look for and what do they mean? Give an example of a related rate problem and solve it.

Well, the steps are pretty simple. First off scan the problem and identify all given info. Next you should determine an equation to use for the problem. Next you implicitly differentiate your equation with respect to time. Finally you plug in your given info, and solve for whatever is needed to be solved for.

Key words to look for:
in respect to
rate of change
per

Ex. A 25ft ladder is leaning against a house. The base of the ladder is 7 feet away from the wall and is being pulled away from the wall at 2ft per second. How fast is the ladder moving down the wall?

First, Identify what is given. The height of the ladder, y, is not given but can be figured is 24ft. dy/dt is what we are looking for. The length of the ladder, c, is 25ft. dc/dt is 0 because it is not moving. The distance between the wall and the base, x, is 7ft. dx/dt is given and is 2ft/s.

The equation we will use is the Pythagorean theorem: x^2+y^2=c^2

Differentiate: 2x(dx/dt)+2y(dy/dt)=2c(dc/dt)

Plug in and solve for dy/dt:
dy/dt=7/6 feet per second.

Week 3 Prompt
What are the rules for simplifying exponents? What are the rules for simplifying logs?

This is probably one of my favorite things in math. Logs! There are many properties for logs and exponents. Exponent rules are pretty simple.

An exponent just means you multiply that number by itself the exponents number of times. More simply put: x^y is x times x, y times.
Some simple rules to remember when dealing with exponents:

The opposite of the exponent is the root.
x^-1=1/x^1
(x^2)(x^2)=x^4
(x^3)^2=x^6
(x^2)/(x^1)=x^1

Log rules:
The base of a log is 10 if not stated otherwise.
The base of a natural log is e.
loga+logb=log(ab)
loga-logb=log(a/b)
lne^a=a

There are more rules, but these are the rules that I feel are the most important as far as logs and exponents go.

Alaina's Blog, 4 Nov. 2010

This week was fairly simple. It was easier than most weeks and the topics themselves were easier than most that we've covered. This week, we started taking the derivative of natural log and of some number, we first learned with e raised to an unknown exponent.

First, I’ll talk about taking the derivative of natural log.
When taking the derivative of natural log, drop the ln and change the variable to a fraction allowing the variable to be the denominator and then multiplying the fraction by the derivative of what is inside of natural log. There may also be a product rule, quotient rule, or chain rule within the natural log.

Ex1: ln x = 1/x
Ex2: ln (8x)^2= (1/(8x^2))(16x)= 16x/8x^2
Ex3: (ln 2x)^2= 2(ln 2x)(2)= 2(1/2x)(2)= 4(1/2x)=2/x

Next, I will talk about taking the derivative of e to an unknown exponent n. I would just like to justify that e is in fact a number. It has a numerical value; therefore, you are taking the derivative of a number raised to an unknown exponent n. First, you copy e to n exponent and then you multiply it by the derivative of the exponent. There may also be a product rule, quotient rule, or chain rule within this type of function.

Ex 1: e^x= e^x (1)= e^x
Ex 2: e^(5x)= e^(5x) (5)= 5e^(5x)
Ex 3: e^(x^2)= e^(x^2)(2x)= 2xe^(x^2)

Week 3 Prompt

What are the rules for simplifying exponents? What are the rules for simplifying logs?