Question

find the x-values of all points where the function has any relative extrema. find the value(s) of any relative extrema. f(x) = x³ + 3x² - 9x + 4 select the correct choice below and, if necessary, fill in any answer boxes within your choice. a. there are no relative minima. the function has a relative maximum of at x = . (use a comma to separate answers as needed.) b. the function has a relative maximum of at x = and a relative minimum of at x = . (use a comma to separate answers as needed.) c. there are no relative maxima. the function has a relative minimum of at x = . (use a comma to separate answers as needed.) d. there are no relative extrema.

Explanation:

Step1: Find the first derivative

To find relative extrema, we first find the derivative of \( f(x) = x^3 + 3x^2 - 9x + 4 \). Using the power rule, the derivative \( f'(x) \) is:
\( f'(x) = 3x^2 + 6x - 9 \)

Step2: Find critical points

Set \( f'(x) = 0 \) to find critical points:
\( 3x^2 + 6x - 9 = 0 \)
Divide both sides by 3:
\( x^2 + 2x - 3 = 0 \)
Factor the quadratic:
\( (x + 3)(x - 1) = 0 \)
So, the critical points are \( x = -3 \) and \( x = 1 \).

Step3: Find the second derivative

To determine if these critical points are maxima or minima, we find the second derivative \( f''(x) \):
\( f''(x) = 6x + 6 \)

Step4: Test critical points with the second derivative

• For \( x = -3 \):

\( f''(-3) = 6(-3) + 6 = -18 + 6 = -12 \)
Since \( f''(-3) < 0 \), the function has a relative maximum at \( x = -3 \).

• For \( x = 1 \):

\( f''(1) = 6(1) + 6 = 6 + 6 = 12 \)
Since \( f''(1) > 0 \), the function has a relative minimum at \( x = 1 \).

Step5: Find the values of the extrema

• For the relative maximum at \( x = -3 \):

\( f(-3) = (-3)^3 + 3(-3)^2 - 9(-3) + 4 = -27 + 27 + 27 + 4 = 31 \)

• For the relative minimum at \( x = 1 \):

\( f(1) = (1)^3 + 3(1)^2 - 9(1) + 4 = 1 + 3 - 9 + 4 = -1 \)

Answer:

B. The function has a relative maximum of \( 31 \) at \( x = -3 \) and a relative minimum of \( -1 \) at \( x = 1 \).