Solved on Feb 13, 2024

Find critical values and relative extrema of $g(x) = -x^3 + 12x - 19$ . Select A if the critical value(s) exist, and fill in the answer box. Select B if the function has no critical values.

STEP 1

Assumptions
1. The function given is $g(x) = -x^3 + 12x - 19$ .
2. Critical values occur where the derivative of the function is zero or undefined.
3. Relative extrema occur at critical values where the function changes concavity or slope.

STEP 2

To find the critical values, we need to find the first derivative of the function $g(x)$ .
$g'(x) = \frac{d}{dx}(-x^3 + 12x - 19)$

STEP 3

Differentiate the function with respect to $x$ .
$g'(x) = \frac{d}{dx}(-x^3) + \frac{d}{dx}(12x) - \frac{d}{dx}(19)$

STEP 4

Apply the power rule and constant rule for differentiation.
$g'(x) = -3x^2 + 12$

STEP 5

Set the derivative equal to zero to find critical points.
$-3x^2 + 12 = 0$

STEP 6

Solve for $x$ to find the critical values.
$-3x^2 = -12$

STEP 7

Divide both sides by -3 to simplify the equation.
$x^2 = 4$

STEP 8

Take the square root of both sides.
$x = \pm 2$

STEP 9

We have found the critical values. They are $x = -2$ and $x = 2$ .

STEP 10

To determine if these critical values are relative extrema, we need to analyze the second derivative or use the first derivative test.

STEP 11

Let's find the second derivative of $g(x)$ .
$g''(x) = \frac{d}{dx}(-3x^2 + 12)$

STEP 12

Differentiate the first derivative with respect to $x$ .
$g''(x) = -6x$

STEP 13

Evaluate the second derivative at the critical values.
For $x = -2$ : $g''(-2) = -6(-2) = 12$
For $x = 2$ : $g''(2) = -6(2) = -12$

STEP 14

Since $g''(-2) > 0$ , there is a relative minimum at $x = -2$ .

STEP 15

Since $g''(2) < 0$ , there is a relative maximum at $x = 2$ .

STEP 16

To find the relative extrema values, we need to evaluate $g(x)$ at the critical points.
For $x = -2$ : $g(-2) = -(-2)^3 + 12(-2) - 19$

STEP 17

Calculate $g(-2)$ .
$g(-2) = -(-8) - 24 - 19$

STEP 18

Simplify the expression.
$g(-2) = 8 - 24 - 19$

STEP 19

Calculate the value of $g(-2)$ .
$g(-2) = -35$

STEP 20

For $x = 2$ : $g(2) = -(2)^3 + 12(2) - 19$

STEP 21

Calculate $g(2)$ .
$g(2) = -8 + 24 - 19$

STEP 22

Simplify the expression.
$g(2) = 16 - 19$

STEP 23

Calculate the value of $g(2)$ .
$g(2) = -3$

STEP 24

Now we can state the critical values and relative extrema.
a) The critical values of the function are $x = -2$ and $x = 2$ .
b) The relative minimum occurs at $x = -2$ with $g(-2) = -35$ , and the relative maximum occurs at $x = 2$ with $g(2) = -3$ .

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Find critical values and relative extrema of g(x)=−x3+12x−19g(x) = -x^3 + 12x - 19g(x)=−x3+12x−19. Select A if the critical value(s) exist, and fill in the answer box. Select B if the function has no critical values.

STEP 1

Assumptions1. The function given is g(x)=−x3+12x−19 g(x) = -x^3 + 12x - 19 g(x)=−x3+12x−19.2. Critical values occur where the derivative of the function is zero or undefined.3. Relative extrema occur at critical values where the function changes concavity or slope.

STEP 2

To find the critical values, we need to find the first derivative of the function g(x) g(x) g(x).g′(x)=ddx(−x3+12x−19) g'(x) = \frac{d}{dx}(-x^3 + 12x - 19) g′(x)=dxd​(−x3+12x−19)

STEP 3

Differentiate the function with respect to x x x.g′(x)=ddx(−x3)+ddx(12x)−ddx(19) g'(x) = \frac{d}{dx}(-x^3) + \frac{d}{dx}(12x) - \frac{d}{dx}(19) g′(x)=dxd​(−x3)+dxd​(12x)−dxd​(19)

STEP 4

Apply the power rule and constant rule for differentiation.g′(x)=−3x2+12 g'(x) = -3x^2 + 12 g′(x)=−3x2+12

STEP 5

Set the derivative equal to zero to find critical points.−3x2+12=0 -3x^2 + 12 = 0 −3x2+12=0

STEP 6

Solve for x x x to find the critical values.−3x2=−12 -3x^2 = -12 −3x2=−12

STEP 7

Divide both sides by -3 to simplify the equation.x2=4 x^2 = 4 x2=4

STEP 8

Take the square root of both sides.x=±2 x = \pm 2 x=±2

STEP 9

We have found the critical values. They are x=−2 x = -2 x=−2 and x=2 x = 2 x=2.

STEP 10

To determine if these critical values are relative extrema, we need to analyze the second derivative or use the first derivative test.

STEP 11

Let's find the second derivative of g(x) g(x) g(x).g′′(x)=ddx(−3x2+12) g''(x) = \frac{d}{dx}(-3x^2 + 12) g′′(x)=dxd​(−3x2+12)

STEP 12

Differentiate the first derivative with respect to x x x.g′′(x)=−6x g''(x) = -6x g′′(x)=−6x

STEP 13

Evaluate the second derivative at the critical values.For x=−2 x = -2 x=−2: g′′(−2)=−6(−2)=12 g''(-2) = -6(-2) = 12 g′′(−2)=−6(−2)=12For x=2 x = 2 x=2: g′′(2)=−6(2)=−12 g''(2) = -6(2) = -12 g′′(2)=−6(2)=−12

STEP 14

Since g′′(−2)>0 g''(-2) > 0 g′′(−2)>0, there is a relative minimum at x=−2 x = -2 x=−2.

STEP 15

Since g′′(2)<0 g''(2) < 0 g′′(2)<0, there is a relative maximum at x=2 x = 2 x=2.

STEP 16

To find the relative extrema values, we need to evaluate g(x) g(x) g(x) at the critical points.For x=−2 x = -2 x=−2: g(−2)=−(−2)3+12(−2)−19 g(-2) = -(-2)^3 + 12(-2) - 19 g(−2)=−(−2)3+12(−2)−19

STEP 17

Calculate g(−2) g(-2) g(−2).g(−2)=−(−8)−24−19 g(-2) = -(-8) - 24 - 19 g(−2)=−(−8)−24−19

STEP 18

Simplify the expression.g(−2)=8−24−19 g(-2) = 8 - 24 - 19 g(−2)=8−24−19

STEP 19

Calculate the value of g(−2) g(-2) g(−2).g(−2)=−35 g(-2) = -35 g(−2)=−35

STEP 20

For x=2 x = 2 x=2: g(2)=−(2)3+12(2)−19 g(2) = -(2)^3 + 12(2) - 19 g(2)=−(2)3+12(2)−19

STEP 21

Calculate g(2) g(2) g(2).g(2)=−8+24−19 g(2) = -8 + 24 - 19 g(2)=−8+24−19

STEP 22

Simplify the expression.g(2)=16−19 g(2) = 16 - 19 g(2)=16−19

STEP 23

Calculate the value of g(2) g(2) g(2).g(2)=−3 g(2) = -3 g(2)=−3

STEP 24

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Find critical values and relative extrema of $g(x) = -x^3 + 12x - 19$ . Select A if the critical value(s) exist, and fill in the answer box. Select B if the function has no critical values.

Assumptions
1. The function given is $g(x) = -x^3 + 12x - 19$ .
2. Critical values occur where the derivative of the function is zero or undefined.
3. Relative extrema occur at critical values where the function changes concavity or slope.

To find the critical values, we need to find the first derivative of the function $g(x)$ .
$g'(x) = \frac{d}{dx}(-x^3 + 12x - 19)$

Differentiate the function with respect to $x$ .
$g'(x) = \frac{d}{dx}(-x^3) + \frac{d}{dx}(12x) - \frac{d}{dx}(19)$

Apply the power rule and constant rule for differentiation.
$g'(x) = -3x^2 + 12$

Set the derivative equal to zero to find critical points.
$-3x^2 + 12 = 0$

Solve for $x$ to find the critical values.
$-3x^2 = -12$

Divide both sides by -3 to simplify the equation.
$x^2 = 4$

Take the square root of both sides.
$x = \pm 2$

We have found the critical values. They are $x = -2$ and $x = 2$ .

Let's find the second derivative of $g(x)$ .
$g''(x) = \frac{d}{dx}(-3x^2 + 12)$

Differentiate the first derivative with respect to $x$ .
$g''(x) = -6x$

Evaluate the second derivative at the critical values.
For $x = -2$ : $g''(-2) = -6(-2) = 12$
For $x = 2$ : $g''(2) = -6(2) = -12$

Since $g''(-2) > 0$ , there is a relative minimum at $x = -2$ .

Since $g''(2) < 0$ , there is a relative maximum at $x = 2$ .

To find the relative extrema values, we need to evaluate $g(x)$ at the critical points.
For $x = -2$ : $g(-2) = -(-2)^3 + 12(-2) - 19$

Calculate $g(-2)$ .
$g(-2) = -(-8) - 24 - 19$

Simplify the expression.
$g(-2) = 8 - 24 - 19$

Calculate the value of $g(-2)$ .
$g(-2) = -35$

For $x = 2$ : $g(2) = -(2)^3 + 12(2) - 19$

Calculate $g(2)$ .
$g(2) = -8 + 24 - 19$

Simplify the expression.
$g(2) = 16 - 19$

Calculate the value of $g(2)$ .
$g(2) = -3$