How do you find the relative extrema of f(x)=3x^5-5x^3?

Jul 10, 2018

Explanation:

The function is

$f \left(x\right) = 3 {x}^{5} - 5 {x}^{3}$

Calculate the first derivative

$f ' \left(x\right) = 15 {x}^{4} - 15 {x}^{2} = 15 {x}^{2} \left({x}^{2} - 1\right)$

The critical points are when $f ' \left(x\right) = 0$

That is,

$15 {x}^{2} \left({x}^{2} - 1\right) = 0$

The solutions to this equation are

$\left\{\begin{matrix}x = 0 \\ x = - 1 \\ x = 1\end{matrix}\right.$

Let's build a variation chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 1$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a a}$$1$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$f ' \left(x\right)$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a}$↗$\textcolor{w h i t e}{a a a a}$↘$\textcolor{w h i t e}{a a a a}$↘$\textcolor{w h i t e}{a a a a}$↗

When $x = - 1$, there is a relative maximum at $\left(- 1 , 2\right)$

When $x = 1$, there is a relative minimum at $\left(1 , - 2\right)$

When $x = 0$, there is an inflection point at $\left(0 , 0\right)$

graph{3x^5-5x^3 [-10, 10, -5, 5]}