# How do you determine the intervals where f(x)=x^4-x^2 is concave up or down?

Oct 1, 2017

Concave up: $\left(- \infty , - \frac{1}{\sqrt{6}}\right) , \left(\frac{1}{\sqrt{6}} , \infty\right)$
Concave down: $\left(- \frac{1}{\sqrt{6}} , \frac{1}{\sqrt{6}}\right)$

#### Explanation:

To determine concavity of a function, you must find the second derivative $f ' ' \left(x\right)$, determine its critical points (places where $f ' ' \left(x\right) = 0$ or is not defined), and then examine the sign of $f ' ' \left(x\right)$ in all intervals defined by those critical points.

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

From examination $f ' ' \left(x\right)$ is a quadratic function, and therefore there are no places where it is undefined. Thus, our only critical points will come from the zeroes of $f ' ' \left(x\right)$:

$12 {x}^{2} - 2 = 0$
$12 {x}^{2} = 2$
${x}^{2} = \frac{1}{6} \therefore x = \pm \frac{1}{\sqrt{6}} \approx \pm 0.408$

Having two zeroes means we have 3 intervals to examine:

$\left(- \infty , - \frac{1}{\sqrt{6}}\right) , \left(- \frac{1}{\sqrt{6}} , \frac{1}{\sqrt{6}}\right) , \left(\frac{1}{\sqrt{6}} , \infty\right)$

Choose a single $x$ value inside of each interval and evaluate $f ' ' \left(x\right)$ at that value. If the result is positive, the function $f \left(x\right)$ is concave up in that interval; if the result is negative, the function is concave down. For simplicity, choose "easy" values of $x$ to evaluate:

$f ' ' \left(- 1\right) = 12 {\left(- 1\right)}^{2} - 2 = 12 - 2 = 10 > 0 \therefore \text{concave up}$

$f ' ' \left(0\right) = 12 {\left(0\right)}^{2} - 2 = 0 - 2 = - 2 < 0 \therefore \text{concave down}$

$f ' ' \left(1\right) = 12 {\left(1\right)}^{2} - 2 = 12 - 2 = 10 > 0 \therefore \text{concave up}$

graph{x^4-x^2 [-1.923, 1.923, -0.963, 0.959]}