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Part 2 of 5

Completed: 9 of 12

My score: [tex]$7.33 / 12$[/tex] pts (61.11\%)

Perform a first derivative test on the function [tex]f(x) = 2x^3 + 3x^2 - 72x + 9 \; ;[-4,6][/tex].

a. Locate the critical points of the given function.
b. Use the First Derivative Test to locate the local maximum and minimum values.
c. Identify the absolute maximum and minimum values of the function on the given interval (when they exist).

a. Locate the critical points of the given function. Select the correct choice below and, if necessary, fill in the answer box within your choice.
A. The critical point(s) is/are at [tex]x = 3[/tex].
(Type an integer or a simplified fraction. Use a comma to separate answers as needed.)
B. There are no critical points.

b. Locate the local maximum value. Select the correct choice below and, if necessary, fill in the answer box within your choice.
A. There is a local maximum at [tex]x = \square[/tex]
(Type an integer or a simplified fraction.)
B. There is no local maximum.

Sagot :

GinnyAnsweridn
To solve the given problem, we need to perform several steps:

Step 1: Locate the critical points of the function [tex]\( f(x) = 2x^3 + 3x^2 - 72x + 9 \)[/tex].

To find the critical points, we need to calculate the first derivative of the function and set it to zero:

[tex]\[ f'(x) = \frac{d}{dx} (2x^3 + 3x^2 - 72x + 9) \][/tex]

Calculating the first derivative:

[tex]\[ f'(x) = 6x^2 + 6x - 72 \][/tex]

Next, we set the first derivative equal to zero to find the critical points:

[tex]\[ 6x^2 + 6x - 72 = 0 \][/tex]

Solving this quadratic equation:

[tex]\[ x^2 + x - 12 = 0 \][/tex]

Factoring the quadratic equation:

[tex]\[ (x + 4)(x - 3) = 0 \][/tex]

So, the critical points are:

[tex]\[ x = -4 \quad \text{and} \quad x = 3 \][/tex]

Therefore, the correct choice for part (a) is:
A. The critical point(s) is/are at [tex]\( x = -4, 3 \)[/tex].

Step 2: Use the First Derivative Test to locate the local maximum and minimum values.

To determine whether each critical point is a local maximum or minimum, we evaluate the sign of the first derivative around the critical points.

1. For [tex]\( x = -4 \)[/tex]:
- Pick a test point less than [tex]\(-4\)[/tex], say [tex]\( x = -5 \)[/tex]:
[tex]\[ f'(-5) = 6(-5)^2 + 6(-5) - 72 = 150 - 30 - 72 = 48 \][/tex] (positive)
- Pick a test point greater than [tex]\(-4\)[/tex], say [tex]\( x = -3 \)[/tex]:
[tex]\[ f'(-3) = 6(-3)^2 + 6(-3) - 72 = 54 - 18 - 72 = -36 \][/tex] (negative)

Since the first derivative changes from positive to negative, [tex]\( x = -4 \)[/tex] is a local maximum.

2. For [tex]\( x = 3 \)[/tex]:
- Pick a test point less than [tex]\( 3 \)[/tex], say [tex]\( x = 2 \)[/tex]:
[tex]\[ f'(2) = 6(2)^2 + 6(2) - 72 = 24 + 12 - 72 = -36 \][/tex] (negative)
- Pick a test point greater than [tex]\( 3 \)[/tex], say [tex]\( x = 4 \)[/tex]:
[tex]\[ f'(4) = 6(4)^2 + 6(4) - 72 = 96 + 24 - 72 = 48 \][/tex] (positive)

Since the first derivative changes from negative to positive, [tex]\( x = 3 \)[/tex] is a local minimum.

Therefore, the correct choice for part (b) is:
A. There is a local maximum at [tex]\( x = -4 \)[/tex].
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