Devin's Blog

Theorem 2.7 The Product Rule
The product of two differentiable functions f and g is itself differentiable. Moreover, the derivative of fg is the first function times the derivative of the second, plus the second function times the derivative of the first.
d/dx [f(x)g(x)] = f(x)g'(x) + g(x)f'(x)

Theorem 2.8 The Quotient Rule
The quotient f/g of two differentiable functions f and g is itself differentiable at all values of x for which g(x) does not equal 0. Moreover, the derivative of f/g is given by the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.
d/dx

ex. y = 2x^2 cos x
2x^2-sinx + cosx(4x)
2x(-xsinx + 4cosx)

d/dx [tan x] = sec^2 x
d/dx [sec x] = sec x tan x
d/dx [cot x] = -csc^2 x
d/dx [csc x] = -csc x cot x

Holiday blog.

Alrightttt, on to the next holiday blog.
CHAIN RULE!
You use the chain rule when you have a function inside of a function, f(g(x)).
Chain rule is way different that the product and quotient rule, however it can be used in both of the rules.

Heres an example of just the chain rule by itself:

Ex. (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.

Example with product rule and chain rule:
Ex. (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

Example with quotient rule and chain rule:
Ex. ((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)^3

Holiday Blog.

Alrighttttt I already did one blog, so this is my second one and I guess i'm going to review some of chapter three. Which is when we started to learn max, mins, critical values and ect. So to start off extreme values are defined as maximums and minimums. Absolute Max is defined as the highest point on an interval
-Absolute max is the highest point on a given interval not the highest point on the graph. Absolute min is defined as the lowest point on an interval
-Absolute min is the lowest point of a given interval not the lowest point on the graph. Relative Max (also referred to as local max) is defined as any max NOT on an interval. Relative Min (also referred to a local min) is defined as any min NOT on an interval. Critical Numbers are defined as any max or min typically referred to as x=c; When trying to solve problems concerning a critical number, extreme value, max, min, or horizontal tangents take the derivative and set equal to zero.

Example:
Find the critical values of f(x)= 9(x^2-3)/x^3
9x^2-27/x^3
*Take the derivative
X^3(18x)-[(9x^2-27)(3x^2)]/ x^6
18x^4-[27x^4-81x^2]/x^6
18x^4-27x^4+81x^2/x^6
-9x^4+81x^2/ x^6
9x^2(-x^2+9)/X^6
9(-x^2+9)/x^4
*Set equal to zero and solve
-When there's a fraction only worry about the top of the fraction (only set the top=0).
9(-x^2+9)=0
-9x^2+81=0
-9x^2=-81
X^2=9
ANSWER:
X=+/- 3

Example:
Find the extrema of f(x)= 3x^4-4x^3 on the interval [-1,2]
4(3)x^3-3(4)x^2
12x^3-12x^2=0
12x^2(x-1)=0
X=0,1

To determine max or min on an interval plug into original and the biggest result will be the max and the smallest result will be the min. **Also plug in endpoints which are -1 and 2.
f(-1)= 3(-1)^4-4(-1)^3=7
f(0)=3(0)^4-4(0)^3=0
f(1)= 3(1)^4 – 4(1)^3= -1 ABSOLUTE MIN
f(2)= 3(2)^4-4(2)^3= 16 ABSOLUTE MAX

Blog #17

For this holiday blog, I'm going to review 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

I'm also going to review how to take the derivative of exponential functions.

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

Also, 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 – 2) = 3
e^2 = x – 3
x = e^2 + 3

Holiday Blog #3

Fort this blog I'm going to talk about how to use the First Derivative Test. To start off, you use it to find the relative/absolute extrema, max/mins, where the function is increasing/decreasing etc...In order to use it, you first take the derivative of the function you're given. Then you set the new equation equal to zero and solve for x to get your x-value(s). Then you set up intervals between negative infinity and infinity with those x-values. After that, you plug in numbers that would fall between those intervals (plugging them into the derivative) to determine if it is negative or positive at that specific x-value. If you get a negative number after plugging in, then that means the function is decreasing at that point. If you get a positive number after plugging in, then that means the function is increasing at that point. Thennnn, if you end up getting negative then positive after plugging in the two intervals, that means you have a min at that point. And if you get positive then negative that means you have a max at that point. Buttttttt, if you happen to get positive then positive, or negative then negative, that means there's nothing at that point.
Here's an example:

Ex 1) Determine the open intervals on which the function is increasing or decreasing
y = x^3/9 - 3x
*Okay first you need to take the derivative of this equation. To make it easier, I'm going to find a common denominator and then use the quotient rule to derive it
*So once you get a common denominator you should now have this equation: (x^3-27x)/9
*Now for the 1st step of the quotient rule you should have
(9)(3x^2-27)-[(x^3-27x)(0)]/(81)
*Simplifying that you get (27x^2-243)/81
*And to simplify that further, you can take out a 27 in the numerator leaving you with (x^2-9) in the numerator and 3 in the denominator
*So now you set the top of the fraction equal to zero (because the bottom doesn't matter)
x^2-9=0
x=3,-3
*Now you set up your intervals
(-infinity,-3)u(-3,3)u(3,infinity)
*Now plug in values for each interval...For the first interval you can plug in -4 (into the derivative) and you should get a positive number; then for the second interval you can plug in 0 and you should get a negative number; then for the last interval you can plug in 4 and you should get a postive number again
*So that means the function is increasing on (-infinity,-3)u(3,infinity)
And it's decreasing on (-3,3)
*And that also means you have a max at x=-3 and a min at x=3

Holiday Blog #2

Okay for this blog I'm going to review implicit differentiation-and all that means is to take the derivative of an equation with respect to other variables (like y) besides x. So that would make it dy/dx, dt/dx, etc...
So here are some examples:

Ex 1.) Find dy/dx......x^2+y^2=36
*For these you're going to take the derivative all the way across the equation so for the derivative of 36 you'd get zero
*So once you take the derivative you should get:
2x+2ydy/dx=0
*Now you have to move the dy/dx to the other side of the equal sign so you can solve for it
*So now you have:
2ydy/dx = -2x
*Now divide by 2y and should should get this:
dy/dx = -2x/2y which simplifies to -x/y

Ex. 2) Find dy/dx.....x^8+9x+2xy-y^5=4
*First take the derivative of each term all the way across the equation and you should get:
8x^7+9+2(x)(dy/dx)+(y)(1)-5y^4dy/dx = 0
*Now simplify that and you should have:
8x^7+9+2xdy/dx+2y-5y^4dy/dx = 0
*Now take the dy/dx's and move them to one side of the equal sign and the other terms to the other side
*So you should now have:
2xdy/dx-5ydy/dx = -8x^7-2y-9
*Now divide -8x^7-2y-9 by 2x-5y^4
*And your answer is (-8x^7-2y-9)/(2x-5y^4)

Holiday Blog #1

For this blog I'm going to review what we previously learned in Chapter 5..taking derivatives of logs, natural logs, e, exponents and whatnot.
*So first let's start with natural logs. The rule for taking the derivative of those is to first write the inside as a fraction over 1 and then multiply it by its derivative. So it would be like this:
1/# x d/dx of #
So here are some examples like that:
1.) ln(5x+4)
=1/5x+4 x 5
=5/5x+4

2.) ln(5x^2+2x+4)
=1/(5x^2+2x+4) x 10x+2
=(10x+2)/(5x^2+2x+4)

3.) ln(x^2-16)^1/2
=1/(x^2-16)^1/2 x 1/2(x^2-16)^-1/2 x 2x
=x/(x^2-16)

*Now let's go over how to derive exponential functions dealing with e. The rule says that all you do to take the derivative is first recopy the function they give you, then multiply it by the exponent's derivative
So here are some examples:
1.) f(x)=(4e)^-3x^2
-First instead of recopying the whole thing, you leave the 4 out in front just like you would do when deriving any other equation. Then you recopy the e^-3x^2
-So you should have 4(e^-3x^2
-Now multiply that by the derivative of e^-3x^2 which is -6x
-So now you have 4(e^-3x^2 x -6x)
= -24xe^-3x^2

2.) f(x)=x^4e^x
-Okay first thing you should realize is that this is a product rule. The same rules apply even though there's an e involved..and to save time, the derivative of e^x is e^x
-So using the product rule you get:
(x^4)(e^x)+(e^x)(4x^3)
=x^4e^x + 4x^3e^x
-Then you can simplify that^ further by taking out an x^3 and an e^x
-So your final answer is:
x^3e^x(x+4)

*Last, let's go over how to derive logs. You take the inside of the function, put it as a fraction over 1 times ln(of whatever base) times the derivative of the inside..I'm not sure if I explained that right haha, but anyway here's an example of that:

1.) log (base 3) (6x^2+5x+2)
=1/(6x^2=5x+2)ln3 x 12x+5
= (12x+5)/(6x^2+5x+2)ln3

Blog #16

This week, we just reviewed the previous chapters and worked on our huge study guide for the upcoming exam. So I'll just review the following:
1) Rolle’s Theorem, 2) the Mean Value Theorem, and 3) the First Derivative Test. First of all, you need to know when to use all of them.
1) Rolle’s Theorem tells you if there’s AT LEAST one max or min, but it doesn’t tell you how many or what they are.
2) The Mean Value Theorem tells you that the slope of a secant line equals the slope of a tangent line between a and b.
3) The First Derivative Test tells you were on the graph is there a max or min by telling you were the graph is decreasing or increasing. It also finds the max or min.

Next, you need to know when to use them.
But, in order to use Rolle’s Theorem or the Mean Value Theorem, you first need to make sure your graph is continuous and differentiable.
1) If a problem asks “Find all values of c in the given interval such f’(c) = 0”, use Rolle’s Theorem.
*c = x
**Rolle’s Theorem generally uses x-intercepts.
2) If a problem asks “Find all values of c in the open interval [a, b] such that
f’(c) = f(b) – f(a) / b – a”, use the Mean Value Theorem.
***This theorem always uses the formula above.
3) If a problem asks “Find the relative extrema of…” or “Find the open intervals on which the graph is increasing or decreasing”, use the First Derivative Test.

Ex. 1) Let f(x) = x^4 – 2x^2. Find all values of c in the interval [-2, 2] such f’(c) = 0.
1. Is it continuous? Yes
2. Is it differentiable? Yes
The next step for Rolle’s Theorem is to check if the y-values are the same by plugging in the x-values from the interval into the original equation.
3. f(-2) = (-2)^4 – 2(-2)^2 = 8
f(2) = (2)^4 – 2(2)^2 = 8
Are the y-values the same? Yes
Next, take the derivative and set it equal to 0, then solve for x.
4. f’(x) = 4x^3 – 4x
4x^3 – 4x = 0
4x(x^2 – 1) = 0
x = 0, 1, -1
Therefore, c = 0, 1, -1
Finally, justify your answer.
By Rolle’s theorem, the function is continuous, differentiable, and f(-2) = f(2), therefore there is at least one max or min on [-2, 2].

Ex. 2) Given f(x) = 5 – (4/x), find all values of c in the open interval [1, 4] such that f’(c) = f(4) – f(1) / 4 – 1.
1. Is it continuous? Yes
2. Is it differentiable? Yes
The next step of the Mean Value Theorem is to find the slope by plugging into the formula.
3. 5 – (4/4) = 4
5 – (4/1) = 1
4 – 1 = 3
4 – 1 / 3 = 1
Next, take the derivative and set it equal to the slope you just found, then solve for x.
4. f’(x) = 4/x^2
4/x^2 = 1
x = 2, -2
Therefore, c = 2, -2
By the Mean Value Theorem, the function is continuous and differentiable on the interval, therefore there is some value (c = 2) where the derivative equals the slope between 1 and 4.

Devin's Blog

Rolle’s Theorem states that “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.

Ex. Find the two x-intercepts of f(x) = x^2+5x+4

Set the equation equal to zero

F(x) = x^2+5x+4 = 0

Then factor the equation

F(x) = (x+4)(x+1) = 0

So, f(4) = f(1) = 0, and from Rolle’s Theorem you know that there exists at least one c in the interval (-1,-4) such that f’(c) = 0.

Take the derivative of the original equation

F(x) = x^2+5x+4 = 0

F’(x) = (2)x^2-1+5(1)x^1-1+4^0 = 0

F’(x) = 2x^1+5 = 0

And determine that f’(x) = 0 when x = -5/2, Note that this x-value lies in the open interval (-1,-5).

that blog with the letters and the numbers and the things

yea so we didn't learn anything new this week, and just did study guides all week, which is BLASPHEMY, but at least we didn't have to cram new stuff in for the exam

anyway, I'mma just do derivatives

f(x) = 2x^2 + 5x
f'(x) = 4x + 5

it's that basic

blog #examsohgoodie!

well i dont think we learned anything new this week but i was able to jesus up a few answers on a lovely study guide that had no magnitude what-so-ever when compared to the amount of sarcasm i am using at this moment.

so heres how you take a derivative :)

f (x) = 2x^3

yous multiplies the coefficient by the exponent and then subtract one from the exponent.

so you then get: f '(x) = 6x^2

0102/21/21 rof golb s'rettebdel nehpets

This week we didn't learn anything new in Calculus. All we did was review packets for the exam the whole week. JOYYY TO THE WORLD. Anyway, I guess I'll give a few examples of problems from the ginormous packets we received, and what might possibly be on the exam. >.>

1. Find the slope m of the line tangent to the graph of the function g(x) = 5 - x^2 at the point (2,1).

To do this, you must find the derivative, then plug in your x-value.

Derivative = -2x

Plug in: -2 (2) = -4

The slope is -4

2. Find the slope of the graph of the function at the given value.

f(x) = 2 (4x + 6)^2 when x = 5

Find the derivative and then plug in.

Derivative:

2 * (2 (4x + 6)) * 4

16(4x + 6)

16(4(5) + 6)

16(20 + 6)

16 (26) = 416

416 is the slope of the graph.

3. Find the derivative of the function f(x) = 4x^5 sin x + 5x^9 cos x

(4x^5)(cos x) + (20x^4)(sin x) + (5x^9)(-sin x) + (45x^8)(cos x)

sin x (20x^4 - 5x^9) + cos x (45x^8 + 4x^5)

So, the derivative equals

f ' (x) = (20x^4 - 5x^9)sin x + (4x^5 + 45x^8) cos x

Blog 12/12/10

Since we didn't learn anything new this week I guess i'll just review something simple that I remember. Soooo... differentiability. There are four ways to determine whether a function is differentiable. If it is differentiable it pretty much means it isn't continuous at some point in the graph. *Using these steps to figure out if a function is differentiable or not you can: look at the graph, factor out the problem, or simply just by looking at the problem. So here are the four ways to determine if a function is differentiable:

1) If there is a corner
Corners occur when absolute values and/or piecewises are found in the function that you are trying to take the derivative.
An example of a function with a corner 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
**When graphing this function you will see what a corner is(it will be the point where the graph makes a v).

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
**Vertical asymptotes can also be graphed

3) If it's not continuous
If a function is not continuous, then it's not differentiable. Discontinuities can occur if theres an vertical asymptote, removable, or jump.

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
**When graphing: if there is a point in there graph where there is a dip then it goes back straight, then it would be a cusp.

12/12/10

Alright well this week all we did was work on our study guides in class...And yet, I'm still not done with them haha. That's pretty sad..anyway I guess I'll just give a few random examples like some we'll probably see on the exam. Soooo, here we go..

Ex. 1) Find the slope of the graph of the function at the given value.
f(x)=2x^2+(3/x^2) when x=3
*First you should notice that there's two things going on in this problem--a term and a fraction. So what you can do is either take the derivative of each one separately or find a common denominator and turn it into one fraction...I'm going to find a common denominator and then use the quotient rule
*So to find your common denominator you multiply 2x^2 by x^2 and now your new fraction is 2x^4+3/x^2
*Now using the quotient rule you get:
(x^2)(8x^3)-[(2x^4+3)(2x)]/(x^4)
=4x^5-6x/x^4
^From there you can take out an x in the numerator, so you'll be able to cancel an x on the denominator also
*So now you have 4x^4-6/x^3
*Now all you do is plug 3 into that equation^
*So you get 318/27....which reduces to 106/9

Ex. 2) Determine all values of x (if any) at which the graph of the function has a horizontal tangent.
f(x)=x^3+3x^2+4
*To find a horizontal tangent, all you have to do is take the derivative, set it equal to zero and then solve for x
*So for the derivative you get 3x^2+6x
*Set it equal to zero:
3x^2+6x=0
3x(x+2)=0
x=0 and x=-2

Ex. 3) Suppose the position function for a free-falling object on a certain planet is given by s(t)=-16t^2+v0t+s0. A silver coin is dropped from the top of a building that is 1380 feet tall. Determine the average velocity of the coin over the time interval [3,4].
*First off, since it's asking for the average velocity, that means you're going to have to find the slope. So you won't be taking the derivative at all in this problem
*So to find the slope first you need to find the y-values for 3 and 4. To do that you plug 3 into the original equation -16t^2 and then plug in 4 after
*So when you plug in 3 you should get -144
*And when you plug in 4 you should get -256
*Now using the points you just found--(3,-144) & (4,-256)-- you have to use the slope formula which is (y2-y1)/(x2-x1)
*So once you plug in your points you should end up with:
-256+144/4-3
=-112 ft/sec

**I don't understand a lot of the problems in our studyguide packets...(packets 2 and 3)

Holiday Blog Prompt 3

Freebie: What are you asking for this Christmas? :) Merry Christmas and Happy New Year!

Holiday Blog Prompt 2

How many things can a derivative be used to find? Give at least 3 example problems to illustrate your point.

Holiday Blog Prompt 1

How do you do optimization? Give a unique example and site it. **Everyone should have a different example**

Devin's Blog

Derivative is a slope of a tangent line, slope of a curve at a point, or a rate of change. If an equations states something about a rate, a velocity, a speed, instantaneous velocity, or an acceleration then you must find a derivative. When using the constant rule, you must take the derivative of the equation and then set the derivative equal to zero. When using the power rule, every time you take a derivative, you must lose a power. For composite functions you must use the chain rule. When using the chain rule you work from outside to the inside. You must take the derivative of the outside and then recopy the inside of the equation and continue with the problem. You can also use the chain rule inside of using a product or quotient rule.
Ex. y’ 6x^4
4(6)x^4-1
=24x^3
Ex. 5x^2+11x
2(5)x^2-1+1(11)x^1-1
=10x+11
Ex. x^5+3x
5(1)x^5-1+1(3)x^1-1
=5x^4+3

I just posted it on here because Mr. Waller's printer was working on my nerves and was acting dumb.

Week 7 Prompt

What have you personally accomplished in math this semester? Do you feel more confident? What will you do differently next semester?

the blog of absence

yea........the only days we learned anything new this week were apparently days that I was at the conference
so yea.................how about them antiderivatives that look pretty scary?

alaina's blog, 5 Dec, 2010

This week, we talked about Antiderivatives and Integrals.
First of all, an antiderivative is a general solution, while an integral is more specific where you must solve for c. Second, an antiderivative is just the derivative backwards. Your given is f'(x) or f''(x) and you are asked to find f(x).
Third, how to find an antiderivative: If your given ax^b , where a is your constant and b is your exponent, your formula is (a/b+1)(x^b+1).
*When solving for the antiderivative, always add “+ c” in place of a constant.
**When solving for an integral, you will be given something like f(1) = 2, in that case, plug in 1 for x into your antiderivative and set it equal to 2 and that’s your constant.
***When given the second derivative, just solve twice.
****When given an integral, it is always used with the integral symbol in front (“S”) and dx afterwards.
***Shortcut: If you have a fractional exponent, add one and multiply by the reciprocal.
**Trig functions: You just do the opposite of the derivative.

Ex. 1) S dx = x + c

Ex. 2) S (x+2) dx = (1/1+1)(x^1+1) = x^2/2 + 2x + c

Ex. 3) S (3x^4 – 5x^2 + x) dx = 3x^5/5 – 5x^3/3 + x^2/2 + c

Ex. 4) S sinx dx = -cosx

bloggeroonie

2.MORE.WEEKS. EH.UH.OH.MY.GOD.

We started 4.1 this week, and we learned Antiderivatives, which is pretty much just following simple short steps. It the problem doesn’t ask for the Antiderivative, it will ask for the general solution. The formula for finding the antiderivative: ax^#; a/#+1 x^#+1 **To make this easier you have to know all of the derivatives for trig (like cos=-sin, sin=cos.. ect.). Here’s an example of what the problem and instructions would look like: Find the general solution to y’=2. Y=2x+c for this problem you have to work the derivative backwards. So you know if the derivative is 2, you know the general solution would be 2x. **Be aware that you have to but +c after every problem or it WILL be wrong.

taylor blog # 15

This week we were only able to tackle on major concept broken into two concepts.
These two concepts were Antiderivatives and integrals

ANTIDERIVATIVES

antiderivatives are different from an integral only because antiderivatives are classified as a general solution whereas integrals are classified as particular solutions.

Basically to find antiderivatives you apply the steps of taking a derivative backwards
or
you can memorize the formula for antiderivatives

THE FORMULA FOR FINDING AN ANTIDERIVATIVE:

Given Problem: a X ^#

A/#+1 X ^#+1
and because this is a general solution we have to account for the fact that there may or may not be a constant to account for
to do this we always add "+ c" to the end of the solved antiderivative.


EXAMPLE:

Given: 3x^2

3/2+1 X^2+1
3/3X^3
X^3

Therefore the antiderivative would be
X^3+C

For Problems with polynomials you solve piece by piece just like you would do to take the derivative of a polynomial

Example:
Given: (x+2)

1/1+1 X ^ 1+1 AND 2x
Therefore the antiderivative would be
1/2 X^2 + 2x + c



INTEGRALS


As stated earlier
integrals are different from an antiderivatives only because antiderivatives are classified as a general solution whereas integrals are classified as particular solutions.

the only component of an Integral which makes it a Particular solution is the fact that you solve for c

You still use the formula but once you solve for the antiderivative you plug in the given f(#)=0 in for X and set = 0 and solve for C and complete the solution by rewriting the antiderivative with c= # plugged in for c

Example: 1/x^2 and F'(1)=0

1/-2+1 X ^ -2+1
1/-1 X ^-1
-X^-1 +c
-1/X^1+c

plug in

-1/1 + c =0
-1 + c=0
Therefore c=1

Complete the solution

-1/x+1

Blog 15

So were almost off for breakkkkkk :D and all we have left is a couple of quizzes this week, which hopefully won’t be too bad! We started 4.1 this week, and so far it is pretty easy.. surprisingly. We learned Antiderivatives, which is pretty much just following simple short steps. It the problem doesn’t ask for the Antiderivative, it will ask for the general solution. The formula for finding the antiderivative: ax^#; a/#+1 x^#+1 **To make this easier you have to know all of the derivatives for trig (like cos=-sin, sin=cos.. ect.). Here’s an example of what the problem and instructions would look like: Find the general solution to y’=2. Y=2x+c for this problem you have to work the derivative backwards. So you know if the derivative is 2, you know the general solution would be 2x. **Be aware that you have to but +c after every problem or it WILL be wrong.

Example: Find the antiderivative of 3x^1
*Put it into the formula
3/1+1=3/2x^2=3/2x^2 + c

Alright, now for something I don’t really understand. I guess what I don’t really understand is using the antiderivative in word problems. Like with velocity, and acceleration. I know velocity is the antiderivative, then acceleration is the antiderivative twice, but I don’t get how to work the entire problem. We had some like this for homework for 4.1.

Blog #15

Antiderivatives/Integrals:

First of all, an antiderivative is a general solution, while an integral is more specific where you must solve for c.
Second, an antiderivative is just the derivative backwards. You will be given f’(x) and asked to find f(x).
Third, how to find an antiderivative: If your given ax^b (a being a constant and b being an exponent), your formula is (a/b+1)(x^b+1).
*Note: When solving for the antiderivative, always add “+ c” in place of a constant. When solving for an integral, you will be given something like “f(1) = 2”, in that case, plug in 1 for x into your antiderivative and set it equal to 2 and that’s your constant.
**Note: When given the second derivative, just solve twice.
***Note: When given an integral, it is always used with the integral symbol in front (“S”) and dx afterwards.
****Note: Shortcut: If you have a fractional exponent, add one and multiply by the reciprocal.
*****Note: Trig functions: You just do the opposite of the derivative.

Ex. 1) S dx = x + c
Ex. 2) S (x+2) dx = (1/1+1)(x^1+1) = x^2/2 + 2x + c
Ex. 3) S (3x^4 – 5x^2 + x) dx = 3x^5/5 – 5x^3/3 + x^2/2 + c
Ex. 4) S sinx dx = -cosx
Ex. 5) S x^1/2 dx = (1/2) + 1 = 3/2 = 2x^(3/2)/3 + c
Ex. 6) S 3x^2 dx = x^3 + c

Ex. 7) Find the integral of f’(x) = 1/x^2 that satisfies the initial f(1) = 0.
S x^-2 dx = (1/-1)(x^-1) = -1/x + c
-1/1 + c = 0
-1 + c = 0
c = 1
-1/x + 1

12/4/10

Okay so this week we finished up Chapter 3 and took a test on it. Then on Thursday we started learning antiderivatives, which is like integrating except you're just giving a general solution. Integrating from what I understand is when you are actually finding c, a constant. It's basically finding out what the original function is before you took the derivative of it. So, you're given the derivative and you have to find the original function.
Here are a few examples:
(**By the way, I'm just going to use a "S" for the integral notation because that's the closest thing I can think of to use..)

Ex. 1) S (x+7)dx
*Okay to find the antiderivative of this function you're going to use the formula (a/#+1)x^#+1..this formula goes for all polynomials..(where a is the number in front of the variable and # is the exponent)
*So just like taking a derivative you're going to take the antiderivative of each term separately. So let's start with x. The number in front is understood to be 1, and that goes over the number 2 because when you add 1 to the exponent you get 2. So you should have 1/2x^2
*N0w let's take the antiderivative of 7...and you don't have to use the formula for this because you should know that the derivative of a "#x" gives you that number. So that means the antiderivative of 7 is 7x
*So your answer is 1/2x^2+7x...buttttt, there's something you have to add to that! *Since an antiderivative is a general solution, you don't know for sure if there was a constant included in the original equation. So you just add +c to the equation you found to complete your answer
*So your final answer is 1/2x^2+7x+c

Ex. 2) S (x+1)(3x-2) dx
*First of all, (something I forgot to mention earlier), there is NO product rule, quotient rule, or chain rule in antiderivatives/integrating..So before you take the antiderivative you have to somehow get rid of what's being multiplied or dividing by combining it possibly or breaking up a fraction (if that's what you have..)
*In this case you would have a product rule IF you were taking the derivative of it..but since you're not you first have to foil it out before you can find its antiderivative
*So foiling it out you get 3x^2+x-2
*So your new problem is S (3x^2+x-2)
*Once you use the formula for each separate term you should get this:
3/3x^3+1/2x^2-2x+c
*Simplifying that you get x^3+1/2x^2-2x+c

**Um I'm still having trouble sketching graphs out of thin air. I'm really not sure where to start and at this point I think I'm making up my own ways of doing it..

0102 ,5 rebmeced ,0102/5/21 no topsgolbredrum

This week went by pretty quickly, and I'm glad there's only one week left of school until Christmas holidays, for me at least :P. So this week, we learned the joys of antiderivatives and integrals. And I'm not being sarcastic this time. So anyway, I'll discuss how to do some antiderivatives and give plenty of examples of how to do their funcztions.

An antiderivative is a general solution.

f ' (x) = f(x) where the antiderivative is f(x)

You always put "plus c" at the end of a solution

The shortcut to the antiderivative is as follows:

If f '(x) =

a x^#

then the antiderivative or f (x) =

((a)/(#+1)) (x^(#+1))

Examples

Find the antiderivative of 3x.

3/2 x^2 + c

the integral of 1/x^3 dx =

x^-3 dx

1/-2 x^-2

-1/2x^2 + c

the integral of x^1/2 dx =

x^1/2 dx

1/1/3/2 x^3/2 = 2/3 x^3/2 + c

If you have a fractional exponent, multiply by the reciprocal to get your coefficient.

Devin's Blog

Guidelines for solving applied minimum and maximum problems

1. Identify all given quantities and all quantities to be determined. If possible, make a sketch.

2. Write a primary equation for the quantity that is to be maximized or minimized. (A review of several useful formulas from geometry is presented inside the back cover.)

3. Reduce the primary equation to one having a single independent variable. This may involve the use of secondary equations relating the independent variables of the primary equation.

4. Determine the feasible domain of the primary equation. That is, determine the values for which the stated problem makes sense.

5. Determine the desired maximum or minimum value by the calculus techniques.

Ex. A rectangular page is to contain 20 square inches of print. The margins at the top and bottom of the page are to be 1 inch, and the margins on the left and right are to b 1. What should the dimensions of the page be so that the least amount of paper is used?

A = (x+2)(x+2)

20 = xy

A = (x+2)(20/x+2)

Devin's Blog

When you are sketching the graph of a function, either by hand or with a graphing utility, remember that normally you cannot show the entire graph. The decision as to which part of the graph you choose to show is often crucial.

Guidelines for analyzing the graph of a function

1. Determine the domain and range of the function

2. Determine the intercepts, asymptotes, and symmetry of the graph

3. Locate the x-values for which f’x and f’’x either are zero or do not exist. Use the results to determine relative extrema and points of inflection

To do curve sketching you must use the following skills and techniques

1. X-intercepts and y-intercepts

2. Symmetry

3. Domain and range

4. Continuity

5. Vertical asymptotes

6. Differentiability

7. Relative extrema

8. Concavity

9. Points of inflection

10. Horizontal asymptotes

11. Infinite limits at infinity

When curve sketching you must first take the derivative, then you must take the second derivative. Then you must find the x-intercepts, and then find the y-intercept. Then find the vertical asymptotes. Then find the horizontal asymptotes. Then you use your derivative to find the critical numbers. Then use your critical numbers to find your points of inflection. Then you must find your domain and symmetry. And then you must test your intervals.

Devin's Blog

When you are sketching the graph of a function, either by hand or with a graphing utility, remember that normally you cannot show the entire graph. The decision as to which part of the graph you choose to show is often crucial.

Guidelines for analyzing the graph of a function

1. Determine the domain and range of the function

2. Determine the intercepts, asymptotes, and symmetry of the graph

3. Locate the x-values for which f’x and f’’x either are zero or do not exist. Use the results to determine relative extrema and points of inflection

To do curve sketching you must use the following skills and techniques

1. X-intercepts and y-intercepts

2. Symmetry

3. Domain and range

4. Continuity

5. Vertical asymptotes

6. Differentiability

7. Relative extrema

8. Concavity

9. Points of inflection

10. Horizontal asymptotes

11. Infinite limits at infinity

When curve sketching you must first take the derivative, then you must take the second derivative. Then you must find the x-intercepts, and then find the y-intercept. Then find the vertical asymptotes. Then find the horizontal asymptotes. Then you use your derivative to find the critical numbers. Then use your critical numbers to find your points of inflection. Then you must find your domain and symmetry. And then you must test your intervals.

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).