Question

find any critical numbers for f and then use the second derivative test to decide whether the critical number(s) lead to relative maxima or relative minima. if f(c)=0 or f(c) does not exist for a critical number c, then the second derivative test gives no information. in this case, use the first derivative test instead.

f(x) = 8x³ - 2x² + 9

what is/are the critical number(s)? select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a. the critical number(s) is/are x =
(simplify your answer. use a comma to separate answers as needed.)
b. there are no critical numbers.

Explanation:

Step1: Find the first derivative

To find critical numbers, we first find the derivative of \( f(x) = 8x^3 - 2x^2 + 9 \). Using the power rule \( \frac{d}{dx}(x^n)=nx^{n - 1} \), we have:
\( f'(x)=\frac{d}{dx}(8x^3)-\frac{d}{dx}(2x^2)+\frac{d}{dx}(9)=24x^2 - 4x \)

Step2: Solve \( f'(x) = 0 \)

Set the first derivative equal to zero and solve for \( x \):
\( 24x^2-4x = 0 \)
Factor out \( 4x \):
\( 4x(6x - 1)=0 \)
Using the zero - product property (if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \)):

• If \( 4x=0 \), then \( x = 0 \)
• If \( 6x - 1=0 \), then \( 6x=1 \), and \( x=\frac{1}{6} \)

Answer:

A. The critical number(s) is/are \( x = 0,\frac{1}{6} \)