# How do you find the maximum, minimum, and inflection points for h(x) = 7x^5 - 12x^3 + x?

Apr 3, 2015

Maximum and minimum:
Find critical numbers for $h$:

$h ' \left(x\right) = 35 {x}^{4} - 36 {x}^{2} + 1$ which exists for all $x$ and is $0$ at,

$35 {x}^{4} - 36 {x}^{2} + 1 = 0$

$\left(35 {x}^{2} - 1\right) \left({x}^{2} - 1\right) = 0$

Which has 4 solutions $x = \pm \frac{1}{\sqrt{35}}$, $x = \pm 1$.

Now, test each critical number using either the first or second derivative tests for local extreme values.

Inflection points:
$h ' ' \left(x\right) = 140 {x}^{3} - 72 x = 4 x \left(35 {x}^{2} - 18\right)$

So concavity could change at $x = 0$ and at $x = \pm \sqrt{\frac{18}{35}}$

A quick check of the sign (or an observation the these are all single zeros (i.e. not even multiplicity)) shows that the sign of $h ' '$ and hance the concavity does indeed change at each of these values of $x$.

The inflection points are: $\left(0 , 0\right)$, $\left(\sqrt{\frac{18}{35}} , h \left(\sqrt{\frac{18}{35}}\right)\right)$ and
$\left(- \sqrt{\frac{18}{35}} , h \left(- \sqrt{\frac{18}{35}}\right)\right)$

(If you are required to find the $y$ values, save yourself some trouble by noticing that $h$ is an odd function, so you only need to find $h \left(\sqrt{\frac{18}{35}}\right)$ and then $h \left(- \sqrt{\frac{18}{35}}\right) = - h \left(\sqrt{\frac{18}{35}}\right)$ (which you just found).