What intervals is f(x) = (5x^2)/(x^2 + 4) concave up/down?

May 28, 2015

In order to investigate concavity, we shall look at the sign of the second derivative.

$f \left(x\right) = \frac{5 {x}^{2}}{{x}^{2} + 4}$

$f ' \left(x\right) = \frac{- 40 x}{{x}^{2} + 4} ^ 2$

Finding $f ' ' \left(x\right)$:

$f ' ' \left(x\right) = \frac{- 40 {\left({x}^{2} + 4\right)}^{2} - \left(- 40 x\right) 2 \left({x}^{2} + 4\right) \left(2 x\right)}{{x}^{2} + 4} ^ 4$

$= \frac{\left({x}^{2} + 4\right) \left[- 40 \left({x}^{2} + 4\right) + 160 {x}^{2}\right]}{{x}^{2} + 4} ^ 4$

$= \frac{- 40 \left({x}^{2} + 4\right) + 160 {x}^{2}}{{x}^{2} + 4} ^ 3$

So:

$f ' ' \left(x\right) = \frac{120 {x}^{2} - 160}{{x}^{2} + 4} ^ 3$

In general, a function can change sign by either crossing the $x$ axis (being equal to $0$), or by being discontinuous (teleporting across the $x$ axis).

In this case $f ' ' \left(x\right)$ is never discontinuous for real $x$, and

$f ' ' \left(x\right) = 0$ when $3 {x}^{2} - 4 = 0$, so $x = \pm \frac{2}{\sqrt{3}}$

Cut the number line into intervals and test each interval:

$\left(- \infty , - \frac{2}{\sqrt{3}}\right)$ $f ' ' \left(x\right)$ is positive, so the graph of $f$ is concave up

$\left(- \frac{2}{\sqrt{3}} , \frac{2}{\sqrt{3}}\right)$ $f ' ' \left(x\right)$ is negative, so the graph of $f$ is concave down

$\left(\frac{2}{\sqrt{3}} , \infty\right)$ $f ' ' \left(x\right)$ is positive, so the graph of $f$ is concave up