Question

find the absolute extremum, if any, for the following function.

$f(x)=3x^{4}-5$

select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

a. the absolute minimum is -5 at $x = 0$.
b. there is no absolute minimum.

select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.

a. the absolute maximum is at $x=$
b. there is no absolute maximum.

Explanation:

Step1: Find the derivative of the function

The function is \( f(x)=3x^{4}-5 \). Using the power rule \( (x^{n})^\prime = nx^{n - 1} \), the derivative \( f^\prime(x)=3\times4x^{3}=12x^{3} \).

Step2: Find the critical points

Set \( f^\prime(x) = 0 \), so \( 12x^{3}=0 \). Solving for \( x \), we get \( x = 0 \).

Step3: Analyze the second - derivative (or use the behavior of the function)

The second - derivative \( f^{\prime\prime}(x)=36x^{2} \). When \( x = 0 \), \( f^{\prime\prime}(0)=0 \). Another way: as \( x
ightarrow\pm\infty \), \( y = f(x)=3x^{4}-5
ightarrow+\infty \) since the leading term \( 3x^{4}\) (where the degree \( n = 4\) is even and the coefficient \(a = 3>0\)) dominates. And \( f(0)=3\times0^{4}-5=-5 \).

Answer:

A. The absolute minimum is \(-5\) at \(x = 0\); B. There is no absolute maximum.