Question

question for the function below, find the local extrema by using the first derivative test. ( h(x) = -\frac{7}{3}x^3 + 14x^2 + 147x + 2 ) select the correct answer below: there is a local minimum at ( x = 7 ). there is a local minimum at ( x = -3 ) and a local minimum at ( x = 7 ). there is a local maximum at ( x = -3 ) and a local minimum at ( x = 7 ). there is a local minimum at ( x = -3 ) and a local maximum at ( x = 7 ). there is a local maximum at ( x = -3 ).

Explanation:

Step1: Find the first derivative

Given \( h(x) = -\frac{7}{3}x^3 + 14x^2 + 147x + 2 \), the first derivative \( h'(x) \) is:
\( h'(x) = -7x^2 + 28x + 147 \)

Step2: Set derivative to zero and solve

Set \( h'(x) = 0 \):
\( -7x^2 + 28x + 147 = 0 \)
Divide by -7: \( x^2 - 4x - 21 = 0 \)
Factor: \( (x - 7)(x + 3) = 0 \)
Solutions: \( x = 7 \) or \( x = -3 \)

Step3: Test intervals for sign of \( h'(x) \)

• For \( x < -3 \) (e.g., \( x = -4 \)): \( h'(-4) = -7(16) + 28(-4) + 147 = -112 - 112 + 147 = -77 < 0 \) (function decreasing)
• For \( -3 < x < 7 \) (e.g., \( x = 0 \)): \( h'(0) = 0 - 0 + 147 = 147 > 0 \) (function increasing)
• For \( x > 7 \) (e.g., \( x = 8 \)): \( h'(8) = -7(64) + 28(8) + 147 = -448 + 224 + 147 = -77 < 0 \) (function decreasing)

Step4: Determine extrema type

• At \( x = -3 \): function changes from decreasing to increasing → local minimum.
• At \( x = 7 \): function changes from increasing to decreasing → local maximum.

Answer:

There is a local minimum at \( x = -3 \) and a local maximum at \( x = 7 \). (Corresponding option: "There is a local minimum at \( x = -3 \) and a local maximum at \( x = 7 \).")