How do you determine where the graph of the given function is increasing, decreasing, concave up, and concave down for h(x) = (x^2) / (x^2+1)?

May 28, 2015

First we need to evaluate the domain and the first and second derivatives:
${D}_{h} = \mathbb{R}$
$h ' \left(x\right) = \frac{2 x}{{x}^{2} + 1} ^ 2$
$h ' ' \left(x\right) = \frac{- 2 \left(3 {x}^{2} - 1\right)}{{x}^{2} + 1} ^ 3 = \frac{- 6 \left(x + \frac{\sqrt{3}}{3}\right) \left(x - \frac{\sqrt{3}}{3}\right)}{{x}^{2} + 1} ^ 3$

Now, $h$ is increasing when $h ' > 0$ and decreasing when $h ' < 0$. Notice that ${\forall}_{x \in \mathbb{R}} \setminus {x}^{2} + 1 > 0$ so as long and we're interested only in the sing of the derivative we can ommit the denominator.
$2 x > 0 \iff x > 0$ - function $h$ increases when x in (0;+oo)
$2 x < 0 \iff x < 0$ - function $h$ decreases when x in (-oo;0)

Function $h$ is concave up when $h ' ' > 0$ and concave down when $h ' ' < 0$.
concave up: -6(x+sqrt(3)/3)(x-sqrt(3)/3)>0 iff x in (-sqrt(3)/3;sqrt(3)/3)
concave down: $- 6 \left(x + \frac{\sqrt{3}}{3}\right) \left(x - \frac{\sqrt{3}}{3}\right) < 0 \iff$
iff x in (-oo;-sqrt(3)/3) cup (sqrt(3)/3;+oo)