# How do you find the x coordinates of all points of inflection, final all discontinuities, and find the open intervals of concavity for y=x^4-3x^2?

Feb 26, 2018

$x$-coordinates of inflection points: $x = \pm \frac{1}{\sqrt{2}}$

No discontinuities, all polynomials are continuous.

#### Explanation:

This function has no discontinuities . A polynomial is always continuous.

To find intervals of concavity and determine inflection points, take the second derivative, set it equal to $0 ,$ and solve for $x$ :

$y ' = 4 {x}^{3} - \left(2\right) \left(3\right) x = 4 {x}^{3} - 6 x$

$y ' ' = \left(3\right) \left(4\right) {x}^{2} - 6 = 12 {x}^{2} - 6$

$12 {x}^{2} - 6 = 0$

$12 {x}^{2} = 6$

${x}^{2} = \frac{6}{12} = \frac{1}{2}$

$x = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}}$

Now, we want to break up the domain of $y$ around these $x$-values. The domain of $y$ is (-∞,∞). Breaking it up yields:

(-∞,-1/sqrt(2)),(-1/sqrt(2),1/sqrt(2)),(1/sqrt(2),∞)

Now, we want to determine whether $y ' '$ is positive or negative in each of these intervals. If $y ' ' > 0$ in an interval, then $y$ is concave up on that interval; if $y ' ' < 0$ on an interval, then $y$ is concave down on that interval. If $y ' '$ changes signs at a certain value of $x$, then there is an inflection point at that $x$-value.

(-∞,-1/sqrt(2)):

$y ' ' \left(- 1\right) = 12 {\left(- 1\right)}^{2} - 6 = 12 - 6 > 0$

$y$ is concave up on (-∞,-1/sqrt(2))

$\left(- \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}}\right)$:

$y ' ' \left(0\right) = 12 \left({0}^{2}\right) - 6 = - 6 < 0$

$y$ is concave down on $\left(- \frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}}\right)$ and has changed signs at $x = - \frac{1}{\sqrt{2}}$; thus, there is an inflection point at $x = - \frac{1}{\sqrt{2}}$.

(1/sqrt(2),∞):

$y ' ' \left(1\right) = 12 \left({1}^{2}\right) - 6 = 12 - 6 > 0$

$y$ is concave up on (1/sqrt(2),∞) and has changed signs at $x = \frac{1}{\sqrt{2}}$; thus, there is an inflection point at $x = \frac{1}{\sqrt{2}} .$