Question

determine the location and value of the absolute extreme values of f on the given interval, if they exist.
$f(x)=x^{2}-5$ on $-3,4$
what is/are the absolute maximum/maxima of f on the given interval? select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
a. the absolute maximum/maxima is/are at $x=$
(use a comma to separate answers as needed.)
b. there is no absolute maximum of f on the given interval.

Explanation:

Step1: Find the derivative of the function

The function is \(f(x)=x^{2}-5\). Using the power rule \((x^{n})^\prime = nx^{n - 1}\), the derivative \(f^\prime(x)=2x\).

Step2: Find the critical points

Set \(f^\prime(x) = 0\), so \(2x=0\), which gives \(x = 0\). The critical point \(x = 0\) lies within the interval \([-3,4]\).

Step3: Evaluate the function at the critical point and endpoints

• At \(x=-3\): \(f(-3)=(-3)^{2}-5=9 - 5=4\).
• At \(x = 0\): \(f(0)=0^{2}-5=-5\).
• At \(x = 4\): \(f(4)=4^{2}-5=16 - 5 = 11\).

Answer:

A. The absolute maximum/maxima is/are at \(x = 4\)