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Calculus AB : Analytical Applications of Derivatives

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Example Question #416 : Calculus Ab

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

Possible Answers:

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

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

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

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

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 #417 : Calculus Ab

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 #418 : Calculus Ab

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

Possible Answers:

(0,4)

(-1,0)

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

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

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 #31 : Analytical Applications Of Derivatives

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 #420 : Calculus Ab

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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Example Question #2 : Determine Increasing/Decreasing Intervals

We know that a function has a critical point at x=2 and the function is increasing on the interval $(-\infty, 2)$ and decreasing on the interval $(2, \infty)$ . Is the critical point x=2 a local maximum?

Possible Answers:

Yes

There is not enough information

Correct answer:

Yes

Explanation:

When a critical point’s interval to the left of the point is increasing and the interval to the right of the point is decreasing, this means that the critical point is at least a local maximum. It could be a global maximum but we do not have enough information to determine this so we will say it is definitely a local maximum.

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Example Question #412 : Calculus Ab

For the function x^3 - 2x , which intervals are increasing and which intervals are decreasing?

Possible Answers:

Increasing: $(-\infty, - \sqrt{\frac{3}{2}})$ , Decreasing: $(\sqrt{\frac{3}{2}}, \infty)$ , $(-\sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}})$

Decreasing: $(-\infty, - \sqrt{\frac{3}{2}})$ , Increasing: $(\sqrt{\frac{3}{2}}, \infty)$ , $(-\sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}})$

Increasing: $(-\infty, - \sqrt{\frac{3}{2}})$ , $(\sqrt{\frac{3}{2}}, \infty)$ , Decreasing: $(-\sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}})$

Decreasing: $(-\infty, - \sqrt{\frac{3}{2}})$ , $(\sqrt{\frac{3}{2}}, \infty)$ , Increasing: $(-\sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}})$

Correct answer:

Increasing: $(-\infty, - \sqrt{\frac{3}{2}})$ , $(\sqrt{\frac{3}{2}}, \infty)$ , Decreasing: $(-\sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}})$

Explanation:

We begin by finding the critical point(s).

f(x) = x^3 - 2x

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

0 = 3x^2 - 2

2 = 3x^2

$\frac{3}{2} = x^2$

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

First we will consider the point $\sqrt{\frac{3}{2}}$ . We will choose the point for the point less than the critical point and the point for the point less than the critical point.

f'(0) = 3(0)^2-2

f'(0) = -2

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

f'(2) = 10

So the function is decreasing on the interval $(-\sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}})$ and increasing on the interval $(\sqrt{\frac{3}{2}}, \infty)$ .

Now we will consider the point $-\sqrt{\frac{3}{2}}$ . We already know that the function is increasing on the interval $(-\sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}})$ , so we only need to consider the interval $(-\infty, -\sqrt{\frac{3}{2}})$ . We will choose the point .

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

f'(-2) = 10

And so the function is also increasing on the interval $(-\infty, - \sqrt{\frac{3}{2}})$ .

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Example Question #413 : Calculus Ab

The function $f(x) = \frac{-1}{x-3}$ is:

Possible Answers:

Decreasing on the interval $(\infty, 3)4$ and increasing in the interval $(3,\infty)$

Decreasing on both intervals

Increasing on both intervals

Correct answer:

Increasing on both intervals

Explanation:

If we take the derivative of the function we have $f'(x) = \frac{1}{(x-3)^2}$ . The critical value of this function is x = 3 . So if we plug into the derivative function we are left with $f'(0) = \frac{1}{9}$ . If we plug in to the derivative function we are left with f'(4) = 1 . So both intervals give us positive solutions and so both intervals are increasing. This makes our critical point a point of inflection.

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Example Question #1 : Determine Local/Global Extrema, Optimization, And Concavity

Given the function $y=\sin \left ( 2x \right )+x^{2}z+z$ , find $\frac{dy}{dx}$ at $(x,z)=(\frac{\pi}{2},3)$ .

Possible Answers:

$-2+3\pi$

$-3\pi$

$2+3\pi$

$-2-3\pi$

$2-3\pi$

Correct answer:

$-2+3\pi$

Explanation:

We are differentiating with respect to . So we can treat as a constant.

Therefore, the derivative becomes $y'=2cos(2x)+2xz$ .

By plugging in $\frac{\pi }{2}$ and $z=3$ , we get $-2+3\pi$ .

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Example Question #2 : Determine Local/Global Extrema, Optimization, And Concavity

Find the maximum value of y = x^3 - x^2 + 3 on the interval [0, 2] .

Possible Answers:

Correct answer:

Explanation:

The max's and min's a function on a closed interval can occur either at local extrema, or the endpoints. Local extrema occur when the derivative is 0. First, taking the derivative,

y' = 3x^2 - 2x = x(3x -2)

We see that the extrema will occur at x = 0 and $x =\frac{2}{3}$ . Here, we can either check to see which of these extrema are max's using the first or second derivative test, or we can just plug them into our function -- if one of them was actually a min, it just won't be our answer.

If you choose to go the first route, you have y'' = 3x-2 + 3x = 6x-2

y''(0) = -2

$y''(\frac{2}{3}) = 2$

So we know that the extrema at 0 is a local max, and at 2/3 is a local min.

Testing our max and two endpoints, we have

f(0) = 3

f(2) = 7

So our maximum value is 7.

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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 #416 : Calculus Ab

Example Question #417 : Calculus Ab

Example Question #418 : Calculus Ab

Example Question #31 : Analytical Applications Of Derivatives

Example Question #420 : Calculus Ab

Example Question #2 : Determine Increasing/Decreasing Intervals

Example Question #412 : Calculus Ab

Example Question #413 : Calculus Ab

Example Question #1 : Determine Local/Global Extrema, Optimization, And Concavity

Example Question #2 : Determine Local/Global Extrema, Optimization, And Concavity

All Calculus AB Resources

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