How do you find the absolute maximum and absolute minimum values of f on the given interval #f(x) = x^5 - x^3 - 1# and [-1, 1]?

1 Answer
May 14, 2016

Answer:

Absolute minimum #f(sqrt(3/5))=-1.1859# and absolute maximum #f(-sqrt(3/5)) = -0.8141#

Explanation:

Determining the critical points #(df)/(dx)=0# we obtain the set
#Z={-sqrt(3/5),0, 0, sqrt(3/5)}# Testing the second derivative for qualification #(d^2f)/(dx^2)(Z)# we obtain the values
#{-6sqrt(3/5),0,0,6sqrt(3/5)}# which indicates that #{-sqrt(3/5), sqrt(3/5)}# are local maximum and minimum respectively. Testing now the extremal values #f(-1)=-1, f(1) = -1# we conclude that the absolute maximum and minimum are located at #-{sqrt[3/5],sqrt[3/5]}#respectively