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
Friday, December 31, 2010
Wednesday, December 22, 2010
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
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
-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
Sunday, December 19, 2010
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
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
Thursday, December 16, 2010
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
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
Wednesday, December 15, 2010
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)
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
*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
Sunday, December 12, 2010
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.
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
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.
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)
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**
Monday, December 6, 2010
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.
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?
Sunday, December 5, 2010
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?
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 axb , where a is your constant and b is your exponent, your formula is (a/b+1)(xb+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
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 axb , where a is your constant and b is your exponent, your formula is (a/b+1)(xb+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
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.
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
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
Saturday, December 4, 2010
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..
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.
Wednesday, December 1, 2010
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.
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