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Calculus AB : Calculus AB

Study concepts, example questions & explanations for Calculus AB

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45 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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 #9 : Analytical Applications Of Derivatives

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

Possible Answers:

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

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

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

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

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

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 : Determine Increasing/Decreasing Intervals

What is a critical value?

Possible Answers:

The point at which the derivative is equal to zero

A local min or max

A point of inflection

All of the above

Correct answer:

All of the above

Explanation:

A point of inflection can be the point at which the derivative is equal to zero, a local minimum or maximum, and a point of inflection. A critical point will always be one that the tangent line to the original function is either completely horizontal or vertical. When the tangent line is completely horizontal, this will be a local maximum or minimum for the function. When the tangent line is completely vertical, then this will be a point of inflection. A point of inflection is a point in which there is a change of curvature in the graph of the function.

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Example Question #14 : Analytical Applications Of Derivatives

What is the first derivative test in regards to increasing/decreasing intervals?

Possible Answers:

You choose a number greater than the critical value. You plug this number into the derivative and if the solution is positive then the function f(x) is increasing, but if the solution is negative then the function f(x) is decreasing.

You choose a number less than the critical value. You plug this number into the derivative and if the solution is positive then the function f(x) is increasing, but if the solution is negative then the function f(x) is decreasing.

If the critical value is positive, the function f(x) is increasing. If the critical value is negative, the function f(x) is decreasing.

You choose a number less than, and a number greater than the critical value. You plug these numbers into the derivative and if the solutions are positive, then the function f(x) is increasing but if the solutions are negative then the function f(x) is decreasing.

Correct answer:

Explanation:

To find whether a function is decreasing or increasing along an interval, we look at the critical values and use what we call the first derivative test. Take the example f(x) = 2x^2+ 3x . The derivative would be f'(x) = 4x + 3 . To find the critical value we set the derivative equal to zero and solve for .

f'(x) = 4x + 3

0 = 4x + 3

-3 = 4x

$\frac{-3}{4} = x$

Now we have our critical point $\frac{-3}{4}$ . So we choose a number greater than this and plug it back into our derivative function.

f'(0) = 4(0) + 3

f'(0) = 3

The solution is positive. This means that the interval $(\frac{-3}{4}, \infty)$ on the graph of f(x)

f'(-1) = 4(-1) + 3

f'(-1) = -4 + 3 ,

f'(x) = -1

The solution is decreasing. This means that the interval $(-\infty, \frac{-3}{4})$ is decreasing on the graph of f(x) .

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Example Question #31 : Analytical Applications Of Derivatives

Find the critical value(s) for the function f(x) = 3x^3 - 2x .

Possible Answers:

$-\sqrt{\frac{2}{9}}$

$\pm \sqrt{\frac{2}{3}}$

$\sqrt{\frac{2}{9}}$

$\pm \sqrt{\frac{2}{9}}$

Correct answer:

$\pm \sqrt{\frac{2}{9}}$

Explanation:

We begin by finding the derivative of our function, f(x) .

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

f'(x) = 9x^2 -2

Now we set our derivative function equal to zero and solve for .

f'(x) = 9x^2 - 2

0 = 9x^2 - 2

2 = 9x^2

$\frac{2}{9} = x^2$

$\pm \sqrt{\frac{2}{9}} = x$

And we are left with two critical points $\pm \sqrt{\frac{2}{9}}$

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Example Question #3 : Determine Increasing/Decreasing Intervals

Find the critical point(s) of the function f(x) = 4x^2 + 8 and determine which interval(s) are decreasing or increasing.

Possible Answers:

Critical point x = 0 , increasing: $(0, \infty)$ , decreasing: $(-\infty, 0)$ .

Critical point x = -8 , increasing: $(-8, \infty)$ , decreasing: $(-\infty, -8)$ .

Critical point x = 8 , increasing: $(8, \infty)$ , decreasing: $(-\infty, 8)$ .

Critical point x = 0 , increasing: $(-\infty, 0))$ , decreasing: $(0, \infty)$ x = 0 .

Correct answer:

Critical point x = 0 , increasing: $(0, \infty)$ , decreasing: $(-\infty, 0)$ .

Explanation:

Critical point , increasing: , decreasing: .

Explanation: We first have to find the critical point(s). To do this we begin by finding the derivative.

f(x) = 4x^2 + 8

f'(x) = 8x

Now to find the critical point(s), we set the derivative equal to zero and solve for .

f'(x) = 8x

0 = 8x

0 =x

And so our critical point is . Now we will choose a number greater than to plug back into our derivative. We will arbitrarily choose the number .

f'(1) = 8(1)

f'(1) = 8 .

The solution is positive, so we know that on the interval $(0, \infty)$ our function f(x) is increasing. Now we must choose a number less than to plug back into our derivative. We will arbitrarily choose .

f'(-1) = 8(-1)

f'(-1) = -8

The solution is negative. So we know that on the interval $(-\infty, 0)$ our function f(x) is decreasing.

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Example Question #4 : Determine Increasing/Decreasing Intervals

Which intervals is the function $f(x) = \frac{1}{2}x^4 - 3x^3 +4x^2$ increasing on?

Possible Answers:

$(-\infty, -1), (0,4)$

(-1,0)

(0,4)

$(-1, 0), (4, \infty)$

Correct answer:

$(-1, 0), (4, \infty)$

Explanation:

We begin by finding the derivative of the function.

$f(x) = \frac{1}{2}x^4 - 2x^3 -4x^2$

f'(x) = 2x^3 - 6x^2 -8x

Now we must find the critical points of the function

f'(x) = 2x^3 - 6x^2 -8x

0 = 2x^3 - 6x^2 -8x

0 = 2x(x^2 - 3x - 4)

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

0=2x 0=x-4 0=x+1

$\Rightarrow$ 0=x 4=x -1=x

So we need to look at the intervals $(-\infty, -1)$ , (-1, 0) , (0, 4) , and $(4, \infty)$ . We will begin with the interval $(-\infty, -1)$ . We will arbitrarily choose to plug back into our derivative function.

f'(x) = 2x^3 - 6x^2 -8x

f'(-2) = 2(-2)^3 - 6(-2)^2 - 8(-2)

f'(-2) = -16 -24 + 16

f'(-2) = -24

The solution is negative, so the interval $(-\infty, -1)$ is decreasing. Now we will look at the interval (-1, 0) . We need to choose a number between and . We will choose $-\frac{1}{2}$ .

$f'(-\frac{1}{2}) = 2(-\frac{1}{2})^3 - 6(-\frac{1}{2})^2 - 8(-\frac{1}{2})$

$f'(-\frac{1}{2}) = -\frac{1}{4} - \frac{6}{4} + \frac{16}{4}$

$f'(-\frac{1}{2}) = \frac{9}{4}$

The solution is negative. So from the interval (-1, 0) , the function is increasing. Now we will look at the interval (0, 4) . We will choose the number to plug back into our equation.

f'(1) = 2(1)^3 - 6(1)^2 - 8(1)

f'(1) = 1 - 6 - 8

f'(1) = -13

The solution is negative, so the interval (0, 4) is decreasing. Now we will look at the interval $(4, \infty)$ . We will choose the number to plug back into our derivative function.

f'(5) = 2(5)^3 - 6(5)^2 - 8(5)

f'(5) = 250-125-40

f'(5) =85

The solution is positive. So the interval $(4, \infty)$ is increasing.

The increasing intervals are (-1, 0) and $(4, \infty)$

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Example Question #2 : Determine Increasing/Decreasing Intervals

True or False: If a function has a critical point, then it must be increasing on one interval and decreasing on the other. It cannot be completely increasing or completely decreasing.

Possible Answers:

True

False

Correct answer:

False

Explanation:

Think back to the types of critical values that we can have. We can have a maximum or a minimum; if we have one of these then the intervals will change from increasing to decreasing or vice versa. Or we will have an inflection point. An inflection point is a change in curvature of the function. So if the critical point is an inflection point, then the function can be completely increasing or completely decreasing.

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Example Question #6 : Determine Increasing/Decreasing Intervals

True or False: The function f(x) = x^2 - 1 is increasing on the interval $(0, \infty)$ .

Possible Answers:

False

True

Correct answer:

True

Explanation:

Begin by finding the derivative.

f(x) = x^2 - 1

f'(x) = 2x

Now find the critical point of the function.

f'(x) = 2x

0 = 2x

0 = x

Now we will plug a number that is greater than into the derivative function. We will arbitrarily choose the number .

f'(1) = 2(1)

f'(1) = 1

The solution is positive, so the interval $(0, \infty)$ is increasing.

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Calculus AB : Calculus AB

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

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 #9 : Analytical Applications Of Derivatives

Example Question #1 : Determine Increasing/Decreasing Intervals

Example Question #14 : Analytical Applications Of Derivatives

Example Question #31 : Analytical Applications Of Derivatives

Example Question #3 : Determine Increasing/Decreasing Intervals

Example Question #4 : Determine Increasing/Decreasing Intervals

Example Question #2 : Determine Increasing/Decreasing Intervals

Example Question #6 : Determine Increasing/Decreasing Intervals

All Calculus AB Resources

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