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Sagot :
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].
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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