Question

If \( f(x) = (x^2 - 16)^2 \), where \( x \) is real, \( f(x) \) has

• A. Two maxima and one minimum
• B. One maxima and 2 minima
• C. Three minima
• D. Three maxima

Hint:

To analyze the maxima and minima of a function, find its derivative and set it equal to zero to determine critical points.

Answer 1

The Correct Option is B

We are given the function \( f(x) = (x^2 - 16)^2 \), and we are tasked with determining the number and type of its critical points (maxima and minima).

Step 1: Take the first derivative of the function:

\[ f'(x) = 2(x^2 - 16) \cdot 2x = 4x(x^2 - 16) \]

Step 2: Find the critical points by setting \( f'(x) = 0 \):

\[ 4x(x^2 - 16) = 0 \]

This gives the critical points:

• \( x = 0 \)
• \( x = \pm 4 \)

Step 3: Determine the nature of the critical points using the second derivative test.

The second derivative is:

\[ f''(x) = 4(x^2 - 16) + 4x(2x) = 12x^2 - 64 \]

Step 4: Evaluate \( f''(x) \) at the critical points:

• At \( x = 0 \), \( f''(0) = -64 \), which is negative, indicating a maximum.
• At \( x = 4 \) and \( x = -4 \), \( f''(4) = f''(-4) = 64 \), which is positive, indicating minima at both points.

Thus, the function has one maximum at \( x = 0 \) and two minima at \( x = 4 \) and \( x = -4 \), which corresponds to option (B).