# What are the critical points of  f(x)=4x^3+8x^2+6x-2?

Feb 25, 2016

There is no critical point for the given function

#### Explanation:

$f ' \left(x\right) = 12 {x}^{2} + 8 x + 6$

for a critical point $f ' \left(x\right) = 0$
however the discriminant of $12 {x}^{2} + 8 x + 6$
$\textcolor{w h i t e}{\text{XXX}} {b}^{2} - 4 a c = {8}^{2} - 4 \left(12\right) \left(6\right) < 0$
so there are no Real values of $x$ for which the function has a zero slope (i.e. there are no critical points).

A graph of the function looks like:
graph{4x^3+8x^2+6x-2 [-2.87, 2.605, -4.936, -2.202]}