# What are the critical points of f(x)=7x^4-6x^2+1 ?

Jun 10, 2018

$\left(0.655 , - 0.286\right)$
$\left(- 0.655 , 4.863\right)$
$\left(0 , 1\right)$

#### Explanation:

$f \left(x\right) = y$

$y = 7 {x}^{4} - 6 {x}^{2} + 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 28 {x}^{3} - 12 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$28 {x}^{3} - 12 x = 0$

$7 {x}^{3} - 3 x = 0$

$x \left(7 {x}^{2} - 3\right) = 0$

$7 {x}^{2} - 3 = 0$

$7 {x}^{2} = 3$

${x}^{2} = \frac{3}{7}$

$x = \pm \sqrt{\frac{3}{7}}$

$x = 0.655$ and $x = - 0.655$ and $x = 0$

So the critical points of this equation is:

$y = 7 {\left(0.655\right)}^{4} - 6 {\left(0.655\right)}^{2} + 1 = - 0.286$
$y = 7 {\left(- 0.655\right)}^{4} - 6 {\left(- 0.655\right)}^{2} + 1 = 4.863$
$y = 7 {\left(0\right)}^{4} - 6 {\left(0\right)}^{2} + 1 = 1$

Critical points:

$\left(0.655 , - 0.286\right)$
$\left(- 0.655 , 4.863\right)$
$\left(0 , 1\right)$