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Calculus AB : Analytical Applications of Derivatives

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Example Questions

Example Question #1 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

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

Possible Answers:

$\frac{301}{6}$

$\frac{401}{6}$

$\frac{434}{6}$

$\frac{343}{6}$

$\frac{334}{6}$

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 #2 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

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

Possible Answers:

$3$

$5$

$2$

$1$

$4$

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 #3 : Analytical Applications Of Derivatives

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 .

Screen shot 2020 09 04 at 1.20.38 pm

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Example Question #1 : Analytical Applications Of Derivatives

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

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 #5 : Analytical Applications Of Derivatives

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 #2 : Analytical Applications Of Derivatives

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:

True

False

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 #5 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

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

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

Possible Answers:

f'(c) = -3

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

f'(c) =0

f'(c) = 3

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

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 #6 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

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:

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

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

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

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

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

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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Example Question #1 : Analytical Applications Of Derivatives

Find the mean value of the function f(x)=cos(x) over the interval $(0,\frac{\pi}{2})$ .

Possible Answers:

$c=sin^{-1}\frac{2}{\pi}$

$c=sin\frac{2}{\pi}$

$c=cos^{-1}\frac{2}{\pi}$

$c=cos\frac{2}{\pi}$

$c=sin^{-1}\frac{\pi}{2}$

Correct answer:

$c=sin^{-1}\frac{2}{\pi}$

Explanation:

To find the mean value of a function over some interval (a,b) , one mus use the formula: (f(b)-f(a))=(b-a)f'(c) .

Plugging in

$(f(\frac{\pi}{2})-f(0))=(\frac{\pi}{2})(-Sin(c))$

Simplifying

$-1=\frac{\pi}{2}(-Sin(c))$

$\frac{2}{\pi}=Sin(c)$

One must then use the inverse Sine function to find the value c:

$c=sin^{-1}\frac{2}{\pi}$

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Example Question #1 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

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

Possible Answers:

$\frac{301}{6}$

$\frac{401}{6}$

$\frac{434}{6}$

$\frac{343}{6}$

$\frac{334}{6}$

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 $\frac{343}{6}$ .

Report an Error

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Calculus AB : Analytical Applications of Derivatives

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #1 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

Example Question #2 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

Example Question #3 : Analytical Applications Of Derivatives

Example Question #1 : Analytical Applications Of Derivatives

Example Question #5 : Analytical Applications Of Derivatives

Example Question #2 : Analytical Applications Of Derivatives

Example Question #5 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

Example Question #6 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

Example Question #1 : Analytical Applications Of Derivatives

Example Question #1 : Use Mean Value Theorem, Extreme Value Theorem, And Critical Points

All Calculus AB Resources

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