# 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]?

Absolute minimum $f \left(\sqrt{\frac{3}{5}}\right) = - 1.1859$ and absolute maximum $f \left(- \sqrt{\frac{3}{5}}\right) = - 0.8141$
Determining the critical points $\frac{\mathrm{df}}{\mathrm{dx}} = 0$ we obtain the set
$Z = \left\{- \sqrt{\frac{3}{5}} , 0 , 0 , \sqrt{\frac{3}{5}}\right\}$ Testing the second derivative for qualification $\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} \left(Z\right)$ we obtain the values
$\left\{- 6 \sqrt{\frac{3}{5}} , 0 , 0 , 6 \sqrt{\frac{3}{5}}\right\}$ which indicates that $\left\{- \sqrt{\frac{3}{5}} , \sqrt{\frac{3}{5}}\right\}$ are local maximum and minimum respectively. Testing now the extremal values $f \left(- 1\right) = - 1 , f \left(1\right) = - 1$ we conclude that the absolute maximum and minimum are located at $- \left\{\sqrt{\frac{3}{5}} , \sqrt{\frac{3}{5}}\right\}$respectively