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AP Calculus AB : Derivatives

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Example Questions

Example Question #1 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Determine the intervals on which f is increasing:

$f=\cos(5x), 0\leq x \leq \frac{2\pi}{5}$

Possible Answers:

$(\frac{\pi}{5}, \frac{2\pi}{5})$

The function is never increasing

$(0, 2\pi)$

$(\frac{\pi}{5}, \infty)$

Correct answer:

$(\frac{\pi}{5}, \frac{2\pi}{5})$

Explanation:

To determine the intervals on which the function is increasing, we must determine the intervals on which the function's first derivative is positive.

The first derivative of the function is equal to

$f'=-5\sin(5x)$

and was found using the following rules:

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

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

$-5\sin(5x)=0$

$c=0, \frac{\pi}{5}, \frac{2\pi}{5}$

Note that we stopped writing critical values based on the given interval in the problem statement.

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

$(0, \frac{\pi}{5}), (\frac{\pi}{5}, \frac{2\pi}{5})$

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

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is negative, while on the second interval, the first derivative is positive. Thus, the function is increasing on the second interval, $(\frac{\pi}{5}, \frac{2\pi}{5})$ .

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Example Question #6 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Determine the local minima of the following function:

f=-x^3+2

Possible Answers:

$x=\infty$

The function has no local minima

x=-3

x=0

Correct answer:

The function has no local minima

Explanation:

To determine the local minima of the function, we must determine the points at which the function's first derivative changes from negative to positive.

The first derivative of the function is

f'=-x^3+2

and was found using the following rules:

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

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

-3x^2=0

c=0

Using the critical value as an endpoint, we now create intervals over which to evaluate the sign of the first derivative:

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

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

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is negative, and on the second interval, the first derivative is also negative. The first derivative never changed in sign from negative to positive, thus the function has no local minima.

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Example Question #2 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Find the local maxima for the following function:

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

Possible Answers:

No local maxima exist

$x=\infty$

x=1

x=0

x=-1

Correct answer:

No local maxima exist

Explanation:

To determine the local maxima of the function, we must determine the points at which the function's first derivative changes from positive to negative.

The first derivative of the function is

f'=x^2+2x+1

and 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 critical values, at which the first derivative is equal to zero:

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

c=-1

Using the critical value as an endpoint, we now create intervals over which to evaluate the sign of the first derivative:

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

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

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is positive, and on the second interval, the first derivative is positive as well. There is no sign change of the first derivative, thus there are no local maxima (the function is always increasing).

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Example Question #11 : Derivative As A Function

Determine the local minima of the following function:

$f=\frac{(x-1)^4}{4}+27x$

Possible Answers:

x=-2

There are no local minima

x=0

x=2

Correct answer:

x=-2

Explanation:

To determine the local minima of the function, we must determine the points at which the function's first derivative changes from negative to positive.

The first derivative of the function is

f'=(x-1)^3+27

and 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 critical values, at which the first derivative is equal to zero:

(x-1)^3+27=0

(x-1)^3=-27

(x-1)=-3

c=-2

Using the critical value as an endpoint, we now create intervals over which to evaluate the sign of the first derivative:

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

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

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is negative, and on the second interval, the first derivative is positive. There is a sign change from negative to positive of the first derivative at x=-2 , thus a local minima exists here.

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Example Question #12 : Derivative As A Function

Determine the intervals on which the function is increasing:

f=2x^3+4x^2+6x

Possible Answers:

(-3, -1)

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

$(-\infty, \infty)$

$(-1, \infty)$

$(-\infty, -3)$

Correct answer:

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

Explanation:

To determine the intervals on which the function is increasing, we must determine the intervals on which the function's first derivative is positive.

The first derivative of the function is equal to

f'=2x^2+8x+6

and 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 critical values, at which the first derivative is equal to zero:

2x^2+8x+6=0

(2x+2)(x+3)=0

c=-3, -1

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

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

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

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

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Example Question #13 : Derivative As A Function

Determine the intervals on which the function is decreasing:

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

Possible Answers:

$(-\infty, -2-\sqrt{3}) \cup (-2+\sqrt{3}, \infty)$

The function is never decreasing

$(-2+\sqrt{3}, \infty)$

$(-2-\sqrt{3}, -2+\sqrt{3})$

$(-\infty, -2-\sqrt{3})$

Correct answer:

$(-2-\sqrt{3}, -2+\sqrt{3})$

Explanation:

To determine the intervals on which the function is decreasing, we must determine the intervals on which the function's first derivative is negative.

The first derivative of the function is equal to

f'(x)=x^2+4x+1

and 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 critical values, at which the first derivative is equal to zero:

x^2+4x+1=0

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

(x+2)^2=3

$x+2=\pm \sqrt{3}$

$c=-2\pm \sqrt{3}$

(Note that the method of completing the square was shown for solving for the critical values. One could use the quadratic formula as well.)

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

$(-\infty, -2-\sqrt{3}), (-2-\sqrt{3}, -2+\sqrt{3}),(-2+\sqrt{3}, \infty)$

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

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is positive, on the second interval, the first derivative is negative. and on the third interval, the first derivative is positive. Thus, the interval on which the first derivative is negative is the interval where the function is decreasing, $(-2-\sqrt{3}, -2+\sqrt{3})$ .

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Example Question #14 : Derivative As A Function

Find the intervals on which the function is decreasing:

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

Possible Answers:

(1, 2)

The function is never decreasing

$(-\infty, -2)$

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

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

Correct answer:

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

Explanation:

To determine the intervals on which the function is decreasing, we must determine the intervals on which the function's first derivative is negative.

The first derivative of the function is equal to

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

and 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 critical values, at which the first derivative is equal to zero:

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

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

c=-2, 1, 2

Note that factoring by grouping was used to find the critical values.

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

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

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

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is negative, on the second interval, the first derivative is positive, on the third interval, the first derivative is negative, and on the fourth interval, the first derivative is positive. The intervals on which the function is decreasing are the intervals on which the first derivative was negative, $(-\infty, -2)\cup(1, 2)$ .

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Example Question #11 : Derivative As A Function

Determine the intervals on which the following function is increasing:

$f=\frac{x^3}{3}-x$

Possible Answers:

$(-\infty, \infty)$

$(1, \infty)$

(-1, 1)

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

$(-\infty, -1)$

Correct answer:

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

Explanation:

To determine the intervals on which the function is increasing, we must determine the intervals on which the function's first derivative is positive.

The first derivative of the function is equal to

f'=x^2-1

and 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 critical values, at which the first derivative is equal to zero:

x^2-1=0

c=-1, 1

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

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

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

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is positive, on the second interval, the first derivative is negative, and on the third interval, the first derivative is positive. The function is therefore increasing on two intervals, $(-\infty, -1)\cup(1, \infty)$ , on which we just showed the first derivative was positive.

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Example Question #21 : Derivative As A Function

Let $f(x) = \frac{1}{7}x^{7} - \frac{6}{5} x^{5} + \frac{5}{3}x^{3} - 30x$ .

A relative minimum of the graph of can be located at:

Possible Answers:

$x= \sqrt[4]{5}$

$x =- \sqrt {6}$

$x=- \sqrt[4]{5}$

The graph of has no relative minimum.

$x = \sqrt {6}$

Correct answer:

$x = \sqrt {6}$

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,

$f'(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{7}x^{7} - \frac{6}{5} x^{5} + \frac{5}{3}x^{3} - 30x \right )$

$f'(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{7}x^{7} \right )- \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{6}{5} x^{5} \right )+ \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{5}{3}x^{3} \right )- \frac{\mathrm{d} }{\mathrm{d} x}\left ( 30x \right )$

Differentiate the individual terms:

$f'(x) = \frac{1}{7} \cdot 7 x^{6} - \frac{6}{5} \cdot 5 x^{4} + \frac{5}{3} \cdot 3 x^{2} - 30x$

$f'(x) = x^{6} - 6 x^{4} +5 x^{2} - 30$

Set this equal to 0:

$x^{6} - 6 x^{4} +5 x^{2} - 30 = 0$

$(x^{6} - 6 x^{4} )+(5 x^{2} - 30) = 0$

$x^{4} (x^{2} - 6 )+ 5 (x^{2} - 6 ) = 0$

$(x^{4}+ 5 ) (x^{2} - 6 ) = 0$

Either:

$x^{4}+5 = 0$ , in which case, $x^{4} = -5$ ; this equation has no real solutions.

$x^{2}- 6=0$ has two real solutions, $-\sqrt {6}$ and $\sqrt {6}$ .

Now take the second derivative, again using the sum rule:

$f'(x) = x^{6} - 6 x^{4} +5 x^{2} - 30$

$f''(x) = \frac{\mathrm{d} }{\mathrm{d} x}(x^{6} - 6 x^{4} +5 x^{2} - 30)$

$f''(x) = \frac{\mathrm{d} }{\mathrm{d} x}(x^{6} )- \frac{\mathrm{d} }{\mathrm{d} x} ( 6 x^{4} )+\frac{\mathrm{d} }{\mathrm{d} x}( 5 x^{2} )- \frac{\mathrm{d} }{\mathrm{d} x}( 30)$

$f''(x) =6 x^{5} - 6 ( 4 x^{3} )+ 5( 2x )- 0$

$f''(x) =6 x^{5} -2 4 x^{3} +10x$

Substitute $-\sqrt {6}$ for :

$f''(-\sqrt {6}) =6 (-\sqrt {6} )^{5} -2 4 (-\sqrt {6})^{3} +10(-\sqrt {6})$

$f''(-\sqrt {6}) =- 216\sqrt {6}+144 \sqrt {6} -10\sqrt {6}$

$f''(-\sqrt {6}) =-82\sqrt {6} < 0$

Therefore, has a relative maximum at $x =- \sqrt {6}$ .

Now. substitute $\sqrt {6}$ for :

$f''( \sqrt {6}) =6 ( \sqrt {6} )^{5} -2 4 ( \sqrt {6})^{3} +10( \sqrt {6})$

$f''(-\sqrt {6}) = 216\sqrt {6}-144 \sqrt {6} +10\sqrt {6}$

$f''(-\sqrt {6}) =82\sqrt {6} >0$

Therefore, has its only relative minimum at $x = \sqrt {6}$ .

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Example Question #21 : Derivative As A Function

Find the intervals on which the function is decreasing:

$f=\frac{x^3}{3}+4x^2-20x$

Possible Answers:

(-10, 2)

$(-\infty, 10)\cup(2,\infty)$

$(2, \infty)$

$(-\infty, 10)$

Correct answer:

(-10, 2)

Explanation:

To determine the intervals on which the function is decreasing, we must determine the intervals on which the function's first derivative is negative.

The first derivative of the function is equal to

f'=x^2+8x-20

and 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 critical values, at which the first derivative is equal to zero:

x^2-8x-20=0

(x-2)(x+10)=0

c=-10, 2

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

$(-\infty, -10), (-10, 2), (2, \infty)$

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

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is positive, on the second interval, the first derivative is negative, and on the third interval, the first derivative is positive. Thus, we know that the function is decreasing on the second interval, (-10, 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 #1 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Example Question #6 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Example Question #2 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Example Question #11 : Derivative As A Function

Example Question #12 : Derivative As A Function

Example Question #13 : Derivative As A Function

Example Question #14 : Derivative As A Function

Example Question #11 : Derivative As A Function

Example Question #21 : Derivative As A Function

Example Question #21 : Derivative As A Function

All AP Calculus AB Resources

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