# How do I find the critical points for the function f(x)=7x^4-6x^2+1?

May 5, 2015

The critical points for a continuous function occur at those points where the derivative is zero.

Given $f \left(x\right) = 7 {x}^{4} - 6 {x}^{2} + 1$

$f ' \left(x\right) = 28 {x}^{3} - 12 x$

If $f ' \left(x\right) = 0$
then
$28 {x}^{3} - 12 x g r a p h \left\{7 {x}^{4} - 6 {x}^{2} + 1 \left[- 2.982 , 3.178 , - 0.404 , 2.675\right]\right\} = x \left(28 {x}^{2} - 12\right) = 0$
and
either $x = 0$
or $\left(28 {x}^{2} - 12\right) = 0$

If $28 {x}^{2} - 12 = 0$
then $x = \pm \sqrt{\frac{12}{28}} = \pm \sqrt{\frac{3}{7}}$

The critical points occur at $x = 0 , x = - \sqrt{\frac{3}{7}} , \text{ and } x = \sqrt{\frac{3}{7}}$