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find the absolute extremum, if any, for the following function. $f(x)=8…

Question

find the absolute extremum, if any, for the following function.
$f(x)=8x^{3}-1$
select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
a. the absolute minimum is at $x=$
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

The derivative of \(f(x)=8x^{3}-1\) is \(f^{\prime}(x)=24x^{2}\).

Step2: Find critical points

Set \(f^{\prime}(x) = 0\), so \(24x^{2}=0\), which gives \(x = 0\).

Step3: Analyze the behavior of the function

As \(x\to-\infty\), \(y = f(x)=8x^{3}-1\to-\infty\). As \(x\to+\infty\), \(y = f(x)=8x^{3}-1\to+\infty\).

Answer:

B. There is no absolute maximum.