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AP Calculus AB : Derivatives

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Example Questions

Example Question #31 : Derivative As A Function

Determine the intervals on which the function is decreasing:

f=x^3-27x

Possible Answers:

(-3, 3)

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

$(-\infty, -3)$

$(3, \infty)$

Correct answer:

(-3, 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'=3x^2-27

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:

3x^2-27=0

$c=\pm3$

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

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

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

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Example Question #32 : Derivative As A Function

Find the intervals on which the following function is increasing:

$f=3\cos(4x), 0\leq x \leq \frac{\pi}{2}$

Possible Answers:

$(\frac{3\pi}{4}, \frac{5\pi}{4})$

$(0, \frac{\pi}{4})$

None of the other answers

$(\frac{\pi}{4}, \frac{\pi}{2})$

Correct answer:

$(\frac{\pi}{4}, \frac{\pi}{2})$

Explanation:

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

The first derivative of the function is equal to

$f'=-12\sin(4x)$

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:

$-12\sin(4x)=0$

$c=0, \frac{\pi}{4}, \frac{\pi}{2}$

Note that the interval stated in the beginning of the problem means the critical values must stay within this interval as well.

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

$(0, \frac{\pi}{4}), (\frac{\pi}{4}, \frac{\pi}{2})$

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, but on the second interval, the first derivative is positive; the function is therefore increasing on the second interval, $(\frac{\pi}{4}, \frac{\pi}{2})$ .

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Example Question #33 : Derivative As A Function

Determine the relative minima of the function:

f=(x+3)^2-2

Possible Answers:

The function has no relative minima

x=-3

x=3

x=0

Correct answer:

x=-3

Explanation:

To determine the values at which the function has relative minima we must determine the intervals on which the function's first derivative changes from negative to positive.

The first derivative of the function is equal to

f'=2(x+3)

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))= f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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

2(x+3)=0

c=-3

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

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

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 negative, and on the second interval, the first derivative is positive. Thus, a relative minimum exists at x=-3 .

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Example Question #31 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Find the relative maxima of the function:

$f=\sqrt{x+4}$

Possible Answers:

None of the other answers

x=0

x=-4

The function has no relative maxima

$x=\infty$

Correct answer:

The function has no relative maxima

Explanation:

To determine the values at which the function has a relative maximum, we must determine the intervals on which the function's first derivative changes from positive to negative.

The first derivative of the function is equal to

$f'=\frac{1}{2\sqrt{x+4}}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Next, we must find the critical values, at which the first derivative is equal to zero. The first derivative can never equal zero, so the function can have no relative maxima.

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Example Question #32 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Determine the relative maxima of the following function:

$f=\frac{x^5}{5}-\frac{13x^3}{3}+36x$

Possible Answers:

x=-2, 3

x=-3, 2

x=2

x=-3

Correct answer:

x=-3, 2

Explanation:

To determine the x-values at which the function has a relative maximum, we must determine the intervals on which the function's first derivative changes from positive to negative.

The first derivative of the function is equal to

f'=x^4-13x^2+36

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^4-13x^2+36=0

(x^2-9)(x^2-4)=0

$c=\pm 2, \pm3$

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

$(-\infty, -3), (-3, -2), (-2, 2), (2, 3), (3, \infty)$

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, on the third interval, the first derivative is positive, on the fourth interval, the first derivative is negative, and on the fifth interval, the first derivative is positive. Thus, the function has relative maxima at x=-3, 2 .

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Example Question #33 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Determine the intervals on which the function is increasing:

$f=\cos(2x), -\pi \leq x \leq \pi$

Possible Answers:

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

$(-\frac{\pi}{2}, 0)\cup(\frac{\pi}{2}, \pi)$

$(-\pi, -\frac{\pi}{2})\cup(0, \frac{\pi}{2})$

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

Correct answer:

$(-\pi, -\frac{\pi}{2})\cup(0, \frac{\pi}{2})$

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

$f'=-2\sin(2x)$

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:

$-2\sin(2x)=0$

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

Note the critical values are entirely within the given interval of x in the problem statement.

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

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

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, on the third interval, the first derivative is positive, and on the fourth interval, the first derivative is negative. Thus, the intervals on which the function is increasing are $(-\pi, -\frac{\pi}{2})\cup(0, \frac{\pi}{2})$ .

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Example Question #34 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Determine the intervals on which the function is decreasing:

$f(x)=\ln (x-5)$ , x> 5

Possible Answers:

The function is never decreasing

$(-\infty, \infty)$

$(-\infty, 5)$

$(5, \infty)$

Correct answer:

The function is never decreasing

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

$f'(x)=\frac{1}{x-5}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \ln(u)=\frac{1}{u}\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Next, we must find the critical values, at which the first derivative is equal to zero. The first derivative function can never equal zero, so the function is never decreasing.

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Example Question #35 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Determine the intervals on which the function is increasing:

$f=\frac{x^5}{5}-\frac{26}{3}x^3+25x$

Possible Answers:

$(-\infty, -5)\cup(-1,1)\cup(5, \infty)$

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

$(-\infty, -5)\cup(-1,1)$

$(-1,1)\cup(5, \infty)$

$(-5, -1)\cup(1, 5)$

Correct answer:

$(-\infty, -5)\cup(-1,1)\cup(5, \infty)$

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^4-26x+25

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^4-26x+25=0

(x^2-25)(x^2-1)=0

$c=\pm1, \pm5$

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

$(-\infty, -5), (-5, -1), (-1, 1), (1, 5), (5, \infty)$

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

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Example Question #36 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Find the intervals on which the function is decreasing:

$g=\frac{x^2}{2}-5x$

Possible Answers:

$(-\infty, \infty)$

$(-\infty, -5)$

$(5, \infty)$

$(-\infty, 5)$

Correct answer:

$(-\infty, 5)$

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

g'=x-5

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-5=0

c=5

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

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

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. We know, therefore, that the function is decreasing on the first interval, $(-\infty, 5)$ .

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Example Question #37 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Determine the relative minima of the function:

f=x^3-48x

Possible Answers:

x=-4, 4

x=4

x=-4

x=0

The function has no relative minima

Correct answer:

x=4

Explanation:

To determine the values at which the function has a relative minimum, we must determine the intervals on which the function's first derivative changes from negative to positive.

The first derivative of the function is equal to

3x^2-48

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:

3(x^2-16)=0

$c=\pm4$

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

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

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 x value at which the function has a relative minimum is x=4 , because here the function's first derivative changed from negative to positive.

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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 #31 : Derivative As A Function

Example Question #32 : Derivative As A Function

Example Question #33 : Derivative As A Function

Example Question #31 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Example Question #32 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Example Question #33 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Example Question #34 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Example Question #35 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Example Question #36 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Example Question #37 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

All AP Calculus AB Resources

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