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

1 Answer
Feb 4, 2016

Answer:

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]}