# How do you find points of inflection and determine the intervals of concavity given y=2x^4-4x^2+1?

Apr 22, 2018

inflection points $\left(\frac{\sqrt{3}}{3} , - \frac{1}{9}\right) , \left(- \frac{\sqrt{3}}{3} , - \frac{1}{9}\right)$
concave up $\left(- \infty , - \frac{\sqrt{3}}{3}\right) , \left(\frac{\sqrt{3}}{3} , \infty\right)$
concave down $\left(- \frac{\sqrt{3}}{3} , \frac{\sqrt{3}}{3}\right)$

#### Explanation:

Given: $y = 2 {x}^{4} - 4 {x}^{2} + 1$

Infection points can be found by setting $y ' ' = 0$; but first you need to know if you have minimums and/or maximums. You can use the first derivative test or the second derivative test. The second derivative test is easier if the second derivative is easy to find. However, it has a restriction. If y''(critical" "value)" = 0 you will then need to use the first derivative test.

Let's use the second derivative test:

Find the first derivative: $y ' = 8 {x}^{3} - 8 x = 0$

Factor: $8 x \left({x}^{2} - 1\right) = 8 x \left(x - 1\right) \left(x + 1\right) = 0$

critical values: $x = 0 , x = \pm 1$

Find the second derivative: $y ' ' = 24 {x}^{2} - 8$

Find the relative minimums and maximums:

$y ' ' \left(0\right) < 0$ relative maximum at $x = 0$

$y ' ' \left(- 1\right) > 0$ relative minimum at $x = - 1$

$y ' ' \left(1\right) > 0$ relative minimum at $x = 1$

Find the points of inflection:

$y ' ' = 24 {x}^{2} - 8 = 0$

Factor: $8 \left(3 {x}^{2} - 1\right) = 0$

$3 {x}^{2} = 1$

${x}^{2} = \frac{1}{3}$

$x = \pm \sqrt{\frac{1}{3}} = \pm \frac{1}{\sqrt{3}} = \pm \frac{\sqrt{3}}{3}$

$y \left(\frac{\sqrt{3}}{3}\right) = 2 {\left(\frac{\sqrt{3}}{3}\right)}^{4} - 4 {\left(\frac{\sqrt{3}}{3}\right)}^{2} + 1 = - \frac{1}{9}$

$y \left(- \frac{\sqrt{3}}{3}\right) = 2 {\left(\frac{\sqrt{3}}{3}\right)}^{4} - 4 {\left(\frac{\sqrt{3}}{3}\right)}^{2} + 1 = - \frac{1}{9}$

inflection points $\left(\frac{\sqrt{3}}{3} , - \frac{1}{9}\right) , \left(- \frac{\sqrt{3}}{3} , - \frac{1}{9}\right)$

Concave down is when $y '$ is decreasing. A relative maximum would be concave down to the point of inflection.

Concave up is when $y '$ is increasing. A relative minimum would be concave up to the point of inflection.

Intervals of concavity:

$\left(- \infty , - \frac{\sqrt{3}}{3}\right) , \left(- \frac{\sqrt{3}}{3.} \frac{\sqrt{3}}{3}\right) , \left(\frac{\sqrt{3}}{3} , \infty\right)$

remember from work above, that there is a relative maximum at $x = 0$ and two relative minimums at $x = \pm 1$

concave up $\left(- \infty , - \frac{\sqrt{3}}{3}\right) , \left(\frac{\sqrt{3}}{3} , \infty\right)$
concave down $\left(- \frac{\sqrt{3}}{3} , \frac{\sqrt{3}}{3}\right)$