# How do you find local maximum value of f using the first and second derivative tests: f(x) = x^5 - 5x + 5?

Sep 3, 2016

$f \left(- 1\right) = - 1 + 5 + 5 = 9 \text{ is local maximum}$.

#### Explanation:

It is known from Calculus that a fun. $f : \mathbb{R} \rightarrow \mathbb{R}$ has a local

maxima at x=c", then,":(1): f'(c)=0, &, :(2): f''(c)<0.

$f \left(x\right) = {x}^{5} - 5 x + 5 \Rightarrow f ' \left(x\right) = 5 {x}^{4} - 5 , \mathmr{and} , f ' ' \left(x\right) = 20 {x}^{3}$.

$\text{Now, } f ' \left(x\right) = 0 \Rightarrow 5 {x}^{4} - 5 = 0 \Rightarrow x = \pm 1$

Since, $f ' ' \left(- 1\right) = - 20 < 0 ,$, we find that,

$f \left(- 1\right) = - 1 + 5 + 5 = 9 \text{ is local maximum}$.