Question

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]?

Answer 1

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