# How do you find the exact relative maximum and minimum of the polynomial function of 4x^8 - 8x^3+18?

Mar 19, 2016

Only an absolute minimum at $\left(\sqrt[5]{\frac{3}{4}} , 13.7926682045768 \ldots \ldots\right)$

#### Explanation:

You will have relative maxima and minima in the values in which the derivate of the function is 0.

$f ' \left(x\right) = 32 {x}^{7} - 24 {x}^{2} = 8 {x}^{2} \left(4 {x}^{5} - 3\right)$

Assuming that we are dealing with real numbers, the zeros of the derivate will be:

$0 \mathmr{and} \sqrt[5]{\frac{3}{4}}$

Now we must calculate the second derivate to see what kind of extreme these values correspond:

$f ' \left(x\right) = 224 {x}^{6} - 48 x = 16 x \left(14 {x}^{5} - 3\right)$

$f ' ' \left(0\right) = 0$-> inflection point

$f ' ' \left(\sqrt[5]{\frac{3}{4}}\right) = 16 \sqrt[5]{\frac{3}{4}} \left(14 \times \left(\frac{3}{4}\right) - 3\right) = 120 \sqrt[5]{\frac{3}{4}} > 0$-> relative minimum

which occurs at

$f \left(\sqrt[5]{\frac{3}{4}}\right) = 13.7926682045768 \ldots \ldots$

No other maxima or minima exist, so this one is also an absolute minimum.