Question

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

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

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

a. the absolute minimum is ( square ) at ( x=square ).

b. there is no absolute minimum.

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: Use the second - derivative test

Find the second - derivative \(f^{\prime\prime}(x)\). Differentiating \(f^\prime(x)=12x^{3}\) with respect to \(x\) using the power rule, \(f^{\prime\prime}(x)=36x^{2}\).
Evaluate \(f^{\prime\prime}(x)\) at \(x = 0\): \(f^{\prime\prime}(0)=36\times0^{2}=0\). We can also use the behavior of the function.
As \(x
ightarrow\pm\infty\), \(y = f(x)=3x^{4}-5
ightarrow+\infty\) (since the leading term \(3x^{4}\) dominates).
Substitute \(x = 0\) into the function \(f(x)\): \(f(0)=3\times0^{4}-5=- 5\).

Answer:

A. The absolute minimum is \(-5\) at \(x = 0\).