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AP Calculus AB : Derivative as a function

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Example Questions

Example Question #41 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Determine the intervals on which the function is increasing:

$f=\frac{x^2}{x-2}, x\neq 2$

Possible Answers:

$(-\infty, -4)$

$(-4, -2)\cup(-2, 0)$

$(0, \infty)$

$(-\infty, -4)\cup(0, \infty)$

Correct answer:

$(-\infty, -4)\cup(0, \infty)$

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

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

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

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

$\frac{x^2+4x}{(x+2)^2}=0$

x(x+4)=0

c=0, -4

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

$(-\infty, -4), (-4, -2), (-2, 0), (0, \infty)$

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative, nor is it defined at x=-2 .

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 and third intervals, the first derivative is negative, and on the fourth interval, the first derivative is positive. Thus, the function is increasing on $(-\infty, -4)\cup(0, \infty)$ .

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Example Question #42 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

f'(x)=x^2-3x-36

g'(x)=x^3-7x^2+45

h'(x)=3x^2+4x-130

i'(x)=-x^3+9x^2-3x-24

Which function has a positive slope at x=5?

Possible Answers:

f(x)

i(x)

h(x)

g(x)

None of the other answers.

Correct answer:

i(x)

Explanation:

i(x) is the only function listed with a positive slope at x=5 . The sign of the slope at of a function at a point is determined by the value first derivative of the function at the corresponding point (i.e., f'(x) , g'(x) , h'(x) , and i'(x) ).

i'(5)=-(5)^3+9(5)^2-3(5)-24

=-125+225-15-24=61

61 is positive, and so, i(x) has a positive slope at x=5 . The other derivative functions listed have non-positive outputs at x=5 .

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Example Question #51 : Derivative As A Function

Find the area of the region enclosed by the parabola $y=12-x^2$ and $y=-x$ .

Possible Answers:

$\frac{334}{6}$

$\frac{301}{6}$

$\frac{401}{6}$

$\frac{343}{6}$

$\frac{434}{6}$

Correct answer:

$\frac{343}{6}$

Explanation:

The two curves intersect in between x=-3 and x=4 , which can be found by solving the quadratic equation $12-x^2=-x$ .

To solve for the area between curves, y=f(x) and y=g(x) , we use the formula $A=\int^b_a(f(x)-g(x))dx$

For our problem:

$A=\int^4_{-3}(12-x^2-(-x))dx$

$A=12x-\frac{x^3}{3}+\frac{x^2}{2}$ evaluated from to which yields $\frac{343}{6}$ .

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Example Question #52 : Derivative As A Function

Find the area under the curve $f(x)=\frac{1}{\sqrt{x+2}}$ between $2\leq x\leq 7$ .

Possible Answers:

$3$

$4$

$5$

$1$

$2$

Correct answer:

$2$

Explanation:

To find the area under the curve, we need to integrate. In this case, it is a definite integral.

$\int_{2}^{7}\frac{1}{\sqrt{x+2}}dx=2\sqrt{x+2}\Big|_2^7=2$

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Example Question #671 : Ap Calculus Ab

Find the area bounded by $y=2, y=x, y=\frac{1}{9}x^2, x=3$

Possible Answers:

2.5

1.5

3.5

Correct answer:

Explanation:

The easiest way to look at this is to plot the graphs. The shaded area is the actual area that we want to compute. We can first find area bounded by x=3 and y=2 in the first quadrant and subtract the excessive areas. The area of that rectangle box is 6. The area under the curve $\frac{1}{9}x^{2}$ is $\int_{0}^{3}\frac{1}{9}x^2=1$ .

The area of the triangle above the curve y=x is 2. Therefore, the area bounded is 6-1-2=3 .

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Example Question #51 : Derivative As A Function

$\int _0^\pi sin(x) dx$

Possible Answers:

Correct answer:

Explanation:

$\int sin(x) dx = -cos(x) + C$ $\int _0^\pi sin(x) dx = -cos(\pi) - -cos(0)= 1-(-1) = 2$

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Example Question #4 : Mean Value Theorem

Consider the region bounded by the functions

$f(x) = -x^{2} + 2x$ and

$g(x) = -sin(\frac{\pi x}{2})$

between x = 0 and x = 2 . What is the area of this region?

Possible Answers:

$\frac{4}{3}$

$\frac{3+\pi }{4}$

$\frac{3\pi }{4}$

$\frac{4}{3} +\frac{4}{\pi }$

$\frac{\pi }{4}$

Correct answer:

$\frac{4}{3} +\frac{4}{\pi }$

Explanation:

The area of this region is given by the following integral:

$Area = \int_{0}^{2} f(x) - g(x) = \int_{0}^{2} -x^{2} + 2x - (-sin(\frac{\pi x}{2}))$ or

$Area = \int_{0}^{2} -x^{2} + 2x + sin(\frac{\pi x}{2})$

Taking the antiderivative gives

$-\frac{x^{3}}{3} + x^{2} -\frac{2}{\pi }cos(\frac{\pi x}{2})$ , evaluated from x = 0 to x = 2 .

$F(2) = \frac{-8}{3} + 4 + \frac{2 }{\pi }$ , and

$F(0) = -\frac{2}{\pi }$ .

Thus, the area is given by:

$F(2) - F(0) =- \frac{8}{3} + 4 + \frac{2}{\pi } - (-\frac{2}{\pi }) =\frac{4}{3} +\frac{4}{\pi }$

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Example Question #52 : Derivative As A Function

Let $f(x) = x^{4}+ 6x - 17$ .

True or false: As a consequence of Rolle's Theorem, has a zero on the interval (0, 2) .

Possible Answers:

False

True

Correct answer:

False

Explanation:

By Rolle's Theorem, if is continuous on [a,b] and differentiable on (a,b) , and f(a)= f(b) , then there must be $c \in \left ( a,b \right )$ such that f'(c) = 0 . Nothing in the statement of this theorem addresses the location of the zeroes of the function itself. Therefore, the statement is false.

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Example Question #1 : Mean Value Theorem

$f(x) = x^{3} - 7x + 16$

As a consequence of the Mean Value Theorem, there must be a value $c \in (-2, 2 )$ such that:

Possible Answers:

$f'(c) = -\frac{1}{3}$

f'(c) = 3

$f'(c) = \frac{1}{3}$

f'(c) =0

f'(c) = -3

Correct answer:

f'(c) = -3

Explanation:

By the Mean Value Theorem (MVT), if a function is continuous and differentiable on [a,b] , then there exists at least one value $c \in (a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$ . , a polynomial, is continuous and differentiable everywhere; setting a= -2, b= 2 , it follows from the MVT that there is $c \in (-2, 2 )$ such that

$f'(c) = \frac{f(2)-f(-2)}{2-(-2)}$

Evaluating f(-2) and f(2) :

$f(x) = x^{3} - 7x + 16$

$f(2) = 2^{3} - 7 (2) + 16 = 8 -14+16 = 10$

$f(-2) = (-2) ^{3} - 7 (-2) + 16 = -8 +14+16 = 22$

The expression for f'(c) is equal to

$f'(c) = \frac{10-22}{2-(-2)} = \frac{-12}{4} = -3$ ,

the correct choice.

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Example Question #55 : Derivative As A Function

is continuous and differentiable on $\mathbb{R}$ .

The values of for five different values of are as follows:

f(0) = 3

f(1) = 4

f(2) = 4

f(3) = 2

f(4) = -1

Which of the following is a consequence of Rolle's Theorem?

Possible Answers:

f(x) cannot have a zero on the interval (0,3 ) ,

f(x) must have a zero on the interval (3,4) ,

There cannot be $c \in (2,3)$ such that f'(x) = 0 .

There must be $c \in (1,2)$ such that f'(x) = 0 .

None of the statements in the other choices follows from Rolle's Theorem.

Correct answer:

There must be $c \in (1,2)$ such that f'(x) = 0 .

Explanation:

By Rolle's Theorem, if is continuous on [a,b] and differentiable on (a,b) , and f(a)= f(b) , then there must be $c \in \left ( a,b \right )$ such that f'(c) = 0 .

is given to be continuous. Also, if we set a=1, b=2 , we note that f(1) = f(2)=0 . This sets up the conditions for Rolle's Theorem to apply. As a consequence, there must be $c \in (1,2)$ such that f'(c) = 0 .

Incidentally, it does follow from the given information that f(x) must have a zero on the interval (3,4) , but this is due to the Intermediate Value Theorem, not Rolle's Theorem.

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AP Calculus AB : Derivative as a function

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #41 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Example Question #42 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Example Question #51 : Derivative As A Function

Example Question #52 : Derivative As A Function

Example Question #671 : Ap Calculus Ab

Example Question #51 : Derivative As A Function

Example Question #4 : Mean Value Theorem

Example Question #52 : Derivative As A Function

Example Question #1 : Mean Value Theorem

Example Question #55 : Derivative As A Function

All AP Calculus AB Resources

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