Question

What are the extrema of `f(x)=x^3-2x+5` on #[-2,2]?

Answer 1

Minimum: `f(-2)=1`
Maximum: `f(+2)=9`

Explanation:

Steps:

1. Evaluate the endpoints of the given Domain

   `f(-2)=(-2)^3-2(-2)+5 = -8+4+5=color(red)(1)`
   `f(+2)=2^3-2(2)+5 =8-4+5 = color(red)(9)`

2. Evaluate the function at any critical points within the Domain.

   To do this find the point(s) within the Domain where `f'(x)=0`
   `f'(x)=3x^2-2=0`
   `rarrx^2=2/3`
   `rarr x=sqrt(2/3)" or "x=-sqrt(2/3)`
   `f(sqrt(2/3))~~color(red)(3.9)` (and, no, I didn't figure this out by hand)
   `f(-sqrt(2/3))~color(red)(~6.1)`

Minimum of `{color(red)(1, 9, 3.9, 6.1)} =1` at `x=-2`
Maximum of `{color(red)(1,9,3.9,6.1)}=9` at `x=+2`

Here is the graph for verification purposes:
graph{x^3-2x+5 [-6.084, 6.4, 1.095, 7.335]}