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AP Calculus AB : AP Calculus AB

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Example Questions

Example Question #12 : Second Derivatives

$\begin{align*}&\text{For the point }x=9\text{, find the concavity of the function}\\&y=8sin(20x)\end{align*}$

Possible Answers:

$\text{The function is concave upward.}$

$\text{The function is concave downward.}$

Correct answer:

$\text{The function is concave upward.}$

Explanation:

$\begin{align*}&\text{To find the concavity of a function, we'll wish to first find}\\&\text{its second derivative, so let's do it in steps. For the function}\\&\text{we're given,}\text{ we'll wish to use the following:}\\&d[sin(ax)]=acos(ax)dx\\&d[cos(ax)]=-asin(ax)dx\\&\text{Our first derivative is thus:}\\&160cos(20x)\\&\text{And our second is:}\\&-3200sin(20x)\\&\text{Now if we plug in our x-value we find}\\&-3200sin(20(9))=2563.69\\&\text{The function is concave upward.}\end{align*}$

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Example Question #13 : Second Derivatives

$\begin{align*}&\text{For the point }x=-2\text{, find the concavity of the function}\\&y=2cos(8x)\end{align*}$

Possible Answers:

$\text{The function is concave upward.}$

$\text{The function is concave downward.}$

Correct answer:

$\text{The function is concave upward.}$

Explanation:

$\begin{align*}&\text{To find the concavity of a function, we'll wish to first find}\\&\text{its second derivative, so let's do it in steps. For the function}\\&\text{we're given,}\text{ we'll wish to use the following:}\\&d[cos(ax)]=-asin(ax)dx\\&d[sin(ax)]=acos(ax)dx\\&\text{Our first derivative is thus:}\\&-16sin(8x)\\&\text{And our second is:}\\&-128cos(8x)\\&\text{Now if we plug in our x-value we find}\\&-128cos(8(-2))=122.58\\&\text{The function is concave upward.}\end{align*}$

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Example Question #14 : Second Derivatives

$\begin{align*}&\text{For the point }x=-10\text{, find the concavity of the function}\\&y=\frac{14}{x^{16}}\end{align*}$

Possible Answers:

$\text{The function is concave upward.}$

$\text{The function is concave downward.}$

Correct answer:

$\text{The function is concave upward.}$

Explanation:

$\begin{align*}&\text{To find the concavity of a function, we'll wish to first find}\\&\text{its second derivative, so let's do it in steps. For the function}\\&\text{we're given,}\text{ we'll wish to use the following:}\\&d[x^a]=ax^{a-1}dx\\&\text{Our first derivative is thus:}\\&-\frac{224}{x^{17}}\\&\text{And our second is:}\\&\frac{3808}{x^{18}}\\&\text{Now if we plug in our x-value we find}\\&\frac{3808}{(-10)^{18}}=3.81\cdot10^{-15}\\&\text{The function is concave upward.}\end{align*}$

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Example Question #304 : Derivatives

Tell whether f is concave up or concave down when x=5 . How do you know?

f(x)=14x^6-3x^5-137x^3+26x^2-12x-20

Possible Answers:

Concave down, because f''(5)< 0 .

Concave up, because .

Concave down, because .

Concave up, because f''(5)< 0 .

Correct answer:

Concave up, because .

Explanation:

Tell whether f is concave up or concave down when x=5 . How do you know?

f(x)=14x^6-3x^5-137x^3+26x^2-12x-20

To test for concavity, we need to find the second derivative.

In this case, we can use the power rule to do all our differentiation.

Power rule:

$\frac{d}{dx}x^n=(n)x^{n-1}$

We will use this on each term in order to find our first and then second derivative.

For each term, we will decrease the exponent by 1, and then multiply by the original exponent.

f(x)=14x^6-3x^5-137x^3+26x^2-12x-20

f'(x)=(6)14x^5-(5)3x^4-(3)137x^2+(2)26x-12

f'(x)=84x^5-15x^4-411x^2+52x-12

Do the same thing to f'(x) to get our final function.

f'(x)=84x^5-15x^4-411x^2+52x-12

f''(x)=(5)84x^4-(4)15x^3-(2)411x+52

f''(x)=420x^4-60x^3-822x+52

Now, to test for concavity, we will plug in our x value (5), and look at the sign of the result.

f''(5)=420(5)^4-60(5)^3-822(5)+52=262500-1500-4110+52=256942

SO, our second derivative is positive, very positive.

This means that f(x) is concave up at x=5.

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Example Question #305 : Derivatives

Determine the intervals on which the function is concave down:

$f=\frac{x^3}{2}+2x$

Possible Answers:

$(0, \infty)$

$(-\infty, \infty)$

$(-\infty, 0)$

The function is never concave down

Correct answer:

$(-\infty, 0)$

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:

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

f''=3x

The derivatives were 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:

3x=0

x=0

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

$(-\infty, 0), (0, \infty)$

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

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Example Question #15 : Relationship Between The Concavity Of ƒ And The Sign Of ƒ''

Determine the intervals on which the function is concave up:

f=-3x^2+x

Possible Answers:

$(-\infty, \infty)$

$(-6, \infty)$

$(0, \infty)$

The function is never concave up

$(-\infty, -6)$

Correct answer:

The function is never concave up

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'=-6x

f''=-6

The derivatives were found using the following rules:

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

The second derivative is a constant and is negative, which means that the function is concave down on its domain, never concave up.

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Example Question #16 : Relationship Between The Concavity Of ƒ And The Sign Of ƒ''

$f(x) = x^{3}+ 4x^{2}+ 8x- 17$

A relative minimum of the graph of can be located at (nearest hundredth):

Possible Answers:

x = -2.28

x= 0.77

x= -3.44

x=-0.39

The graph of has no relative minimum.

Correct answer:

The graph of has no relative minimum.

Explanation:

At a relative minimum (c,f(c)) of the graph f(x) , it will hold that f'(c) = 0 and f''(c) > 0 .

First, find f'(x) . Using the sum rule,

$f(x) = x^{3}+ 4x^{2}+ 8x- 17$

$f'(x) =\frac{\mathrm{d} }{\mathrm{d} x}( x^{3}+ 4x^{2}+ 8x- 17)$

$f'(x) =\frac{\mathrm{d} }{\mathrm{d} x}( x^{3} )+ \frac{\mathrm{d} }{\mathrm{d} x}( 4x^{2} )+\frac{\mathrm{d} }{\mathrm{d} x}( 8x )-\frac{\mathrm{d} }{\mathrm{d} x}( 17)$

Apply the Constant Multiple and Power Rules:

$f'(x) =3 x^{2} + 4 \cdot 2x +8x -0$

$f'(x) =3 x^{2} + 8x +8x$

Set this equal to 0:

$3 x^{2} + 8x +8x =0$

$x= \frac{-8 \pm \sqrt{8^{2}-4(3)(8)}}{2(3)} = \frac{-8 \pm \sqrt{-32}}{6}$

Both zeroes of are imaginary. Since has no real zeroes, it follows that the graph of has no relative extrema - maximum or minimum - on the coordinate plane.

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Example Question #308 : Derivatives

Determine the intervals on which the function is concave down:

$f=\frac{x^5}{20}+\frac{x^4}{12}+\frac{x^3}{6}+2x^2$

Possible Answers:

$(-\infty, -2)\cup(-1, 2)$

$(-2, -1)\cup(2, \infty)$

(-1, 2)

$(2, \infty)$

$(-\infty, -2)$

Correct answer:

$(-\infty, -2)\cup(-1, 2)$

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,

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

f''=x^3+x^2-4x+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^3+x^2-4x+4=0

(x^2-4)(x+1)=0

x=-2, -1, 2

Note that we factored by grouping to solve for x.

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

$(-\infty, -2), (-2, -1), (-1, 2), (2, \infty)$

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, on the second interval, the second derivative is positive, on the third interval, the second derivative is negative, and on the fourth interval, the second derivative is positive. The function is concave down on the intervals where the second derivative is negative, $(-\infty, -2)\cup(-1, 2)$ .

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Example Question #309 : Derivatives

Determine the intervals on which the function is concave up:

$f=\frac{x^4}{12}-\frac{5x^3}{6}-3x^2$

Possible Answers:

$(6, \infty)$

$(-\infty, -1)\cup(6, \infty)$

$(-\infty, -1)$

(-1, 6)

None of the other answers

Correct answer:

$(-\infty, -1)\cup(6, \infty)$

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,

$f'=\frac{x^3}{3}-\frac{5x^2}{2}-6x$

f''=x^2-5x-6

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^2-5x-6=0

(x+1)(x-6)=0

x=-1, 6

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

$(-\infty, -1), (-1, 6), (6, \infty)$

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. So, the function is concave up on the intervals $(-\infty, -1)\cup(6, \infty)$

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

Determine the intervals on which the function is concave down:

$f(x)=\sin(3x), 0 \leq x \leq \frac{2\pi}{3}$

Possible Answers:

$(0, \frac{\pi}{3})$

$(\frac{\pi}{3}, \frac{2\pi}{3})$

$(-\infty, \infty)$

None of the other answers

Correct answer:

$(0, \frac{\pi}{3})$

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,

$f'(x)=3\cos(3x)$

$f''(x)=-9\sin(3x)$

which was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$ ,

$\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$
Next, we must find the values at which the second derivative is equal to zero:

$-9\sin(3x)=0$

$x=0, \frac{\pi}{3}, \frac{2\pi}{3}$

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

$(0, \frac{\pi}{3}), (\frac{\pi}{3}, \frac{2\pi}{3})$

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

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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #12 : Second Derivatives

Example Question #13 : Second Derivatives

Example Question #14 : Second Derivatives

Example Question #304 : Derivatives

Example Question #305 : Derivatives

Example Question #15 : Relationship Between The Concavity Of ƒ And The Sign Of ƒ''

Example Question #16 : Relationship Between The Concavity Of ƒ And The Sign Of ƒ''

Example Question #308 : Derivatives

Example Question #309 : Derivatives

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

All AP Calculus AB Resources

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