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)=x^{4}-32x^{2}+256
select the correct choice below and, if necessary, fill in any answer boxes within your choice.
a. there are no relative maxima. the function has a relative minimum of □ at x = □.
(use a comma to separate answers as needed.)
b. there are no relative minima. the function has a relative maximum of □ at x = □.
(use a comma to separate answers as needed.)
c. the function has a relative maximum of □ at x = □ and 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

Given \( f(x)=x^{4}-32x^{2}+256 \).
Using the power rule \( (x^{n})^\prime = nx^{n - 1} \), we have \( f^\prime(x)=4x^{3}-64x=4x(x^{2}-16)=4x(x - 4)(x + 4) \).

Step2: Find the critical points

Set \( f^\prime(x)=0 \), so \( 4x(x - 4)(x + 4)=0 \).
Solving for \( x \), we get \( x=-4,0,4 \).

Step3: Find the second derivative

Differentiate \( f^\prime(x)=4x^{3}-64x \) using the power rule. \( f^{\prime\prime}(x)=12x^{2}-64 \).

Step4: Use the second - derivative test

• For \( x = - 4 \): \( f^{\prime\prime}(-4)=12\times(-4)^{2}-64=192 - 64 = 128>0 \). So \( x=-4 \) is a relative minimum.
• For \( x = 0 \): \( f^{\prime\prime}(0)=12\times0^{2}-64=-64<0 \). So \( x = 0 \) is a relative maximum.
• For \( x = 4 \): \( f^{\prime\prime}(4)=12\times4^{2}-64=192 - 64 = 128>0 \). So \( x = 4 \) is a relative minimum.

Answer:

B. There are no relative minima. The function has a relative maximum of \( 256 \) at \( x = 0 \)