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
Cesareo R.
May 14, 2016

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

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