What are the global and local extrema of f(x)=8x^3-4x^2+6 ?

Oct 30, 2016

The local extrema are $\left(0 , 6\right)$ and $\left(\frac{1}{3} , \frac{158}{27}\right)$
and the global extrema are $\pm \infty$

Explanation:

We use $\left({x}^{n}\right) ' = n {x}^{n - 1}$
Let us find the first derivative

$f ' \left(x\right) = 24 {x}^{2} - 8 x$
For local extrema $f ' \left(x\right) = 0$
So $24 {x}^{2} - 8 x = 8 x \left(3 x - 1\right) = 0$

$x = 0$ and $x = \frac{1}{3}$

So let's do a chart of signs
$x$$\textcolor{w h i t e}{a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a}$$0$$\textcolor{w h i t e}{a a a a a}$$\frac{1}{3}$$\textcolor{w h i t e}{a a a a a}$$+ \infty$
$f ' \left(x\right)$$\textcolor{w h i t e}{a a a a a}$$+$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a a}$$+$
$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$\uparrow$$\textcolor{w h i t e}{a a a a a}$$\downarrow$$\textcolor{w h i t e}{a a a a a}$$\uparrow$

So at the point $\left(0 , 6\right)$ we have a local maximum
and at $\left(\frac{1}{3} , \frac{158}{27}\right)$
We have a point a point of inflexion $f ' ' \left(x\right) = 48 x - 8$
$48 x - 8 = 0$$\implies$$x = \frac{1}{6}$
limit$f \left(x\right) = - \infty$
$x \rightarrow - \infty$
limit$f \left(x\right) = + \infty$
$x \rightarrow + \infty$
graph{8x^3-4x^2+6 [-2.804, 3.19, 4.285, 7.28]}