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Calculus AB : Calculus AB

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Example Questions

Example Question #392 : Calculus Ab

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

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

Possible Answers:

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

$\displaystyle f'(c) =0$

$\displaystyle f'(c) = -3$

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

$\displaystyle f'(c) = 3$

Correct answer:

$\displaystyle f'(c) = -3$

Explanation:

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

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

Evaluating f(-2) $\displaystyle f(-2)$ and f(2) $\displaystyle f(2)$:

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

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

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

The expression for f'(c) $\displaystyle f'(c)$ is equal to

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

the correct choice.

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Example Question #393 : Calculus Ab

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

The values of $\displaystyle f$ for five different values of $\displaystyle x$ are as follows:

$\displaystyle f(0) = 3$

$\displaystyle f(1) = 4$

$\displaystyle f(2) = 4$

$\displaystyle f(3) = 2$

$\displaystyle f(4) = -1$

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

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

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

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Example Question #394 : Calculus Ab

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

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

Plugging in

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

Simplifying

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

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

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

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

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Example Question #1 : Determine Increasing/Decreasing Intervals

What is a critical value?

Possible Answers:

A local min or max

The point at which the derivative is equal to zero

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 #13 : Analytical Applications Of Derivatives

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

Possible Answers:

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) $\displaystyle f(x)$ is increasing, but if the solution is negative then the function f(x) $\displaystyle f(x)$ is decreasing.

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) $\displaystyle f(x)$ is increasing, but if the solution is negative then the function f(x) $\displaystyle f(x)$ is decreasing.

If the critical value is positive, the function f(x) $\displaystyle f(x)$ is increasing. If the critical value is negative, the function f(x) $\displaystyle f(x)$ 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) $\displaystyle f(x)$ is increasing but if the solutions are negative then the function f(x) $\displaystyle f(x)$ 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 $\displaystyle f(x) = 2x^2+ 3x$. The derivative would be $\displaystyle f'(x) = 4x + 3$. To find the critical value we set the derivative equal to zero and solve for $\displaystyle x$.

$\displaystyle f'(x) = 4x + 3$

$\displaystyle 0 = 4x + 3$

-3 = 4x $\displaystyle -3 = 4x$

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

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

$\displaystyle f'(0) = 4(0) + 3$

$\displaystyle f'(0) = 3$

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

$\displaystyle f'(-1) = 4(-1) + 3$

$\displaystyle f'(-1) = -4 + 3$,

$\displaystyle f'(x) = -1$

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

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Example Question #31 : Analytical Applications Of Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

$\displaystyle f(x) = 3x^3 - 2x$

$\displaystyle f'(x) = 9x^2 -2$

Now we set our derivative function equal to zero and solve for $\displaystyle x$.

$\displaystyle f'(x) = 9x^2 - 2$

$\displaystyle 0 = 9x^2 - 2$

$\displaystyle 2 = 9x^2$

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

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

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

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Example Question #32 : Analytical Applications Of Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

Critical point , increasing: , decreasing: .

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

$\displaystyle f(x) = 4x^2 + 8$

$\displaystyle f'(x) = 8x$

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

$\displaystyle f'(x) = 8x$

0 = 8x $\displaystyle 0 = 8x$

0 =x $\displaystyle 0 =x$

And so our critical point is $\displaystyle 0$. Now we will choose a number greater than $\displaystyle 0$ to plug back into our derivative. We will arbitrarily choose the number $\displaystyle 1$.

$\displaystyle f'(1) = 8(1)$

$\displaystyle f'(1) = 8$.

The solution is positive, so we know that on the interval $(0, \infty)$ $\displaystyle (0, \infty)$ our function f(x) $\displaystyle f(x)$ is increasing. Now we must choose a number less than to plug back into our derivative. We will arbitrarily choose $\displaystyle -1$.

$\displaystyle f'(-1) = 8(-1)$

$\displaystyle f'(-1) = -8$

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

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Example Question #33 : Analytical Applications Of Derivatives

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

Possible Answers:

(-1,0) $\displaystyle (-1,0)$

(0,4) $\displaystyle (0,4)$

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

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

Correct answer:

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

Explanation:

We begin by finding the derivative of the function.

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

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

Now we must find the critical points of the function

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

$\displaystyle 0 = 2x^3 - 6x^2 -8x$

$\displaystyle 0 = 2x(x^2 - 3x - 4)$

$\displaystyle 0 = 2x(x - 4)(x+1)$

0=2x $\displaystyle 0=2x$ 0=x-4 $\displaystyle 0=x-4$ 0=x+1 $\displaystyle 0=x+1$

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

So we need to look at the intervals $(-\infty, -1)$ $\displaystyle (-\infty, -1)$, (-1, 0) $\displaystyle (-1, 0)$, (0, 4) $\displaystyle (0, 4)$, and $(4, \infty)$ $\displaystyle (4, \infty)$. We will begin with the interval $(-\infty, -1)$ $\displaystyle (-\infty, -1)$. We will arbitrarily choose $\displaystyle -2$ to plug back into our derivative function.

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

$\displaystyle f'(-2) = 2(-2)^3 - 6(-2)^2 - 8(-2)$

$\displaystyle f'(-2) = -16 -24 + 16$

$\displaystyle f'(-2) = -24$

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

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

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

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

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

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

$\displaystyle f'(1) = 1 - 6 - 8$

$\displaystyle f'(1) = -13$

The solution is negative, so the interval (0, 4) $\displaystyle (0, 4)$ is decreasing. Now we will look at the interval $(4, \infty)$ $\displaystyle (4, \infty)$. We will choose the number $\displaystyle 5$ to plug back into our derivative function.

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

$\displaystyle f'(5) = 250-125-40$

$\displaystyle f'(5) =85$

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

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

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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:

False

True

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 #35 : Analytical Applications Of Derivatives

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

Possible Answers:

True

False

Correct answer:

True

Explanation:

Begin by finding the derivative.

$\displaystyle f(x) = x^2 - 1$

$\displaystyle f'(x) = 2x$

Now find the critical point of the function.

$\displaystyle f'(x) = 2x$

0 = 2x $\displaystyle 0 = 2x$

0 = x $\displaystyle 0 = x$

Now we will plug a number that is greater than $\displaystyle 0$ into the derivative function. We will arbitrarily choose the number $\displaystyle 1$.

$\displaystyle f'(1) = 2(1)$

$\displaystyle f'(1) = 1$

The solution is positive, so the interval $(0, \infty)$ $\displaystyle (0, \infty)$ 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 #392 : Calculus Ab

Example Question #393 : Calculus Ab

Example Question #394 : Calculus Ab

Example Question #1 : Determine Increasing/Decreasing Intervals

Example Question #13 : Analytical Applications Of Derivatives

Example Question #31 : Analytical Applications Of Derivatives

Example Question #32 : Analytical Applications Of Derivatives

Example Question #33 : Analytical Applications Of Derivatives

Example Question #2 : Determine Increasing/Decreasing Intervals

Example Question #35 : Analytical Applications Of Derivatives

All Calculus AB Resources

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