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AP Calculus AB : Derivatives

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Example Questions

Example Question #21 : Relationship Between The Concavity Of ƒ And The Sign Of ƒ''

Determine the intervals on which the function is concave up:

$f=\frac{x^5}{20}-4x^2$

Possible Answers:

$(-\infty, 2)$

(-2,2)

$(2, \infty)$

$(-\infty, -2)$

Correct answer:

$(2, \infty)$

Explanation:

To determine the intervals on which the function is concave up, we must determine the intervals on which the function's second derivative is positive.

First, we must find the second derivative of the function,

$f'=\frac{x^4}{4}-8x$

f''=x^3-8

which was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Next, we must find the values at which the second derivative is equal to zero:

x^3-8=0

x=2

Using this value, we now create intervals on which to evaluate the sign of the second derivative:

$(-\infty, 2), (2, \infty)$

Notice how at the bounds of the intervals, the second derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the second derivative function, we find that on the first interval, the second derivative is negative, and on the second interval, the second derivative is positive. Thus, the function is concave up on the second interval, $(2, \infty)$ .

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Example Question #22 : Second Derivatives

Determine the intervals on which the following function is concave up:

$f(x)= \frac{x^4}{12}+\frac{x^3}{2}-2x^2$

Possible Answers:

(-4, 1)

$(1, \infty)$

$(-\infty, -4)\cup(1, \infty)$

$(-\infty, -4)$

Correct answer:

$(-\infty, -4)\cup(1, \infty)$

Explanation:

To determine the intervals on which the function is concave up, we must determine the intervals on which the function's second derivative is positive.

First, we must find the second derivative of the function,

$f'(x)=\frac{x^3}{3}+\frac{3x^2}{2}-4x$

f''(x)=x^2+3x-4

which was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}, \frac{\mathrm{d} }{\mathrm{d} x}ax=a$

Next, we must find the values at which the second derivative is equal to zero

x^2+3x-4=0

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

x=1, -4

Using these values, we now create intervals on which to evaluate the sign of the second derivative:

$(-\infty, -4), (-4, 1), (1, \infty)$

Notice how at the bounds of the intervals, the second derivative is neither positive nor negative.

Evaluating the sign of the second derivative simply by plugging in any value on the given interval into the second derivative function, we find that on the first interval, the second derivative is positive, on the second interval, the second derivative is negative, and on the third interval, the second derivative is positive. Thus, the intervals on which the function is concave up are $(-\infty, -4)\cup(1, \infty)$ .

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Example Question #23 : Second Derivatives

Determine the intervals on which the function is concave down:

$f=\frac{2x^6}{30}-\frac{5x^4}{3}+9x^2$

Possible Answers:

$(-\infty, -3)\cup(-1, 1)\cup(1, \infty)$

$(-3, -1)\cup(1, 3)$

(-3, -1)

(1, 3)

Correct answer:

$(-3, -1)\cup(1, 3)$

Explanation:

To determine the intervals on which the function is concave down, we must determine the intervals on which the function's second derivative is negative.

First, we must find the second derivative of the function,

$f'=\frac{2x^5}{5}-\frac{20x^3}{3}+18x$

f''=2x^4-20x^2+18

which was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Next, we must find the values at which the second derivative is equal to zero:

2x^4-20x^2+18=0

(2x^2-18)(x^2-1)=0

$x=\pm1, \pm3$

Using these values, we now create intervals on which to evaluate the sign of the second derivative:

$(-\infty, -3), (-3, -1), (-1, 1), (1, 3), (3, \infty)$

Notice how at the bounds of the intervals, the second derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the second derivative function, we find that on the first interval, the second derivative is positive, on the second interval, the second derivative is negative, on the third interval, the second derivative is positive, on the fourth interval, the second derivative is negative, and on the fifth interval, the second derivative is positive. Thus, the function is concave down on $(-3, -1)\cup(1, 3)$ .

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Example Question #24 : Second Derivatives

Determine the intervals on which the function is concave up:

$f=\frac{x^4}{6}-25x^2$

Possible Answers:

$(-\infty, -5)$

(-5, 5)

$(-\infty, -5)\cup(5, \infty)$

$(5, \infty)$

Correct answer:

$(-\infty, -5)\cup(5, \infty)$

Explanation:

To determine the intervals on which the function is concave up, we must determine the intervals on which the function's second derivative is positive.

First, we must find the second derivative of the function,

$f'=\frac{2x^3}{3}-50x$

f''=2x^2-50

which was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Next, we must find the values at which the second derivative is equal to zero:

2x^2-50=0

$x=\pm 5$

Using these values, we now create intervals on which to evaluate the sign of the second derivative:

$(-\infty, -5), (-5, 5), (5, \infty)$

Notice how at the bounds of the intervals, the second derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the second derivative function, we find that on the first interval, the second derivative is positive, on the second interval, the second derivative is negative, and on the third interval, the second derivative is positive. Therefore, the function is concave up on $(-\infty, -5)\cup(5, \infty)$ .

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Example Question #25 : Second Derivatives

Explain whether f(x) is concave up or concave down when x=1 , and choose the correct explanation for its behavior.

f(x)=16e^x+12x^6-48x+13

Possible Answers:

f(x) is concave up, because f''(1)<0

f(x) is concave down, because f''(1)<0

f(x) is concave down, because f''(1)>0

f(x) is concave up, because f''(1)>0

Correct answer:

f(x) is concave up, because f''(1)>0

Explanation:

Explain whether f(x) is concave up or concave down when x=1 , and choose the correct explanation for its behavior.

f(x)=16e^x+12x^6-48x+13

To determine concavity, we need to find the second derivative of our function. Then, we will plug in the given value of x and see what the sign is. If f" is positive at the point, then we have a concave up curve. If it is negative, then we have concave down. If f" is zero at the point, then we have a point of inflection.

To find our derivatives, we need to recall two rules.

$\frac{dy}{dx}e^x=e^x$

And

$\frac{dy}{dx}x^n=nx^{n-1}$

Using these two rules, we can find the derivative of f(x).

Our first term can be derived using our first rule. The derivative of e to the x is just e to the x.

This means that our first term will remain 16e to x.

For our other three terms, we follow the second rule. We will decrease each term's exponent by 1, and then multiply the coefficient by the old exponent.

$f'(x)=16e^x+(6)12x^{6-1}-(1)48x^{1-1}$

Notice that the 13 will drop out. It is a constant term, and as such when we multiply it by it's original exponent (0) it wil be reduced to zero as well.

Clean up the above to get:

$f'(x)=16e^x+72x^{5}-48$

Using those same to rules, repeat the process to get our second derivative.

$f''(x)=16e^x+(5)72x^{4}=16e^x+360x^4$

So with our second derivative, plug in 1 and find the sign.

f''(x)=16e^x+360x^4

$f''(1)=16e^1+360(1)^4=403.49\approx 403$

We get a positive result, so our function is concave up.

So, we can say that f(x) is concave up, because f''(1)>0

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Example Question #21 : Second Derivatives

Determine the intervals on which the function is concave up:

$f=\frac{x^7}{42}-\frac{x^5}{20}-\frac{x^3}{6}+\frac{x^2}{2}$

Possible Answers:

$(-\infty, -1)\cup(1, \infty)$

$(-\infty, -1)$

$(1,\infty)$

(-1,1)

Correct answer:

(-1,1)

Explanation:

To determine the intervals on which the function is concave up, we must determine the intervals on which the function's second derivative is positive.

First, we must find the second derivative of the function,

$f'=\frac{x^6}{6}-\frac{x^4}{4}-\frac{x^2}{2}+x$

f''=x^5-x^3-x^2+1

which was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Next, we must find the values at which the second derivative is equal to zero:

x^5-x^3-x^2+1=0

(x^3-1)(x^2-1)=0

x=-1, 1

Using these values, we now create intervals on which to evaluate the sign of the second derivative:

$(-\infty, -1), (-1, 1), (1, \infty)$

Notice how at the bounds of the intervals, the second derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the second derivative function, we find that on the first interval, the second derivative is negative, on the second interval, the second derivative is positive, and on the third interval, the second derivative is negative. So, the function is concave up on (-1,1) .

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Example Question #1 : Optimization, Both Global And Local Extrema

Given the function $\inline y=sin(2x)+x^2z+z^3$ ,

find $\frac{\mathrm{d} y}{\mathrm{d} x}$ at $(x,z)=(\frac{\pi}{2},3)$ .

Possible Answers:

$2+3\pi$

$-2-3\pi$

$-3\pi$

$2-3\pi$

$-2+3\pi$

Correct answer:

$-2+3\pi$

Explanation:

We are differentiating with respect to . So we can treat as a constant.

Therefore, the derivative becomes $y'=2cos(2x)+2xz$ .

By plugging in $\frac{\pi }{2}$ and $z=3$ , we get $-2+3\pi$ .

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Example Question #1 : Optimization, Both Global And Local Extrema

Find the maximum value of y = x^3 - x^2 + 3 on the interval [0, 2] .

Possible Answers:

Correct answer:

Explanation:

The max's and min's a function on a closed interval can occur either at local extrema, or the endpoints. Local extrema occur when the derivative is 0. First, taking the derivative,

y' = 3x^2 - 2x = x(3x -2)

We see that the extrema will occur at x = 0 and $x =\frac{2}{3}$ . Here, we can either check to see which of these extrema are max's using the first or second derivative test, or we can just plug them into our function -- if one of them was actually a min, it just won't be our answer.

If you choose to go the first route, you have y'' = 3x-2 + 3x = 6x-2

y''(0) = -2

$y''(\frac{2}{3}) = 2$

So we know that the extrema at 0 is a local max, and at 2/3 is a local min.

Testing our max and two endpoints, we have

f(0) = 3

f(2) = 7

So our maximum value is 7.

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Example Question #3 : Optimization, Both Global And Local Extrema

It's Mother's Day and you want to make a wonderful picture for your mom. You know that you will put a nice ribbon on the border of your rectangular picture, and that you will put double ribbons on the left and right sides. You have 80 inches of ribbon. What dimensions should your picture have if you want it to have the most area possible without running out of ribbon?

Possible Answers:

$20\ in\times20\ in$

$10\ in\times10\ in$

$15\ in\times15\ in$

$10\ in\times20\ in$

$40\ in\times40\ in$

Correct answer:

$10\ in\times20\ in$

Explanation:

Let the dimensions of the picture be x and y.

Area=xy

Our constraint is

80=2x+4y

Solve for x

x=2y-40

Let us substitute that value of x back into our area equation

$Area=(2y-40)\cdot y=2y^2-40y$

In order to maximize this equation, we need to find its derivative and set it equal to zero

Area'=4y-40

0=4y-40

0=y-10

10=y

Now that we have found our first dimension, we need to find the second

80=2x+4y

80=2x+4(10)

80=2x+40

40=2x

20=x

Then our dimension are 10in and 20in

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Example Question #2 : Optimization, Both Global And Local Extrema

Locate any points of inflection for the function:

f(x)=xe^x-x-1

Possible Answers:

$x=ln(\frac{1}{2})$

x=2

x=-2

x=ln(2)

$x=\frac{1}{2}$

Correct answer:

x=-2

Explanation:

Points of inflection are found when the second derivative of a function is equal to zero. For our function

f(x)=xe^x-x-1

The first derivative is:

f'(x)=e^x+xe^x-1

By the product rule.

Our second derivative then is:

f'(x)=2e^x+xe^x

Setting this equal to zero we obtain:

2e^x+xe^x=0

e^x(2+x)=0

Since we know e^x can never be equal to zero, we are only concerned with 2+x=0 , so x=-2

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AP Calculus AB : Derivatives

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #21 : Relationship Between The Concavity Of ƒ And The Sign Of ƒ''

Example Question #22 : Second Derivatives

Example Question #23 : Second Derivatives

Example Question #24 : Second Derivatives

Example Question #25 : Second Derivatives

Example Question #21 : Second Derivatives

Example Question #1 : Optimization, Both Global And Local Extrema

Example Question #1 : Optimization, Both Global And Local Extrema

Example Question #3 : Optimization, Both Global And Local Extrema

Example Question #2 : Optimization, Both Global And Local Extrema

All AP Calculus AB Resources

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