For what values of x is f(x)=(4x)/(x^2-1) concave or convex?

Jul 10, 2017

$f \left(x\right)$ is concave in the region $\left(- \infty , - 1\right) \cup \left(0 , 1\right)$ convex in the region $\left(- 1 , 0\right) \cup \left(0 , \infty\right)$.

Explanation:

A function $f \left(x\right)$, whicg is twice continuously differentiable on an interval , then the function ݂is

• convex if $f ' ' \left(x\right) > 0$ for all $x$ in the interval, and
• concave if $f ' ' \left(x\right) < 0$ for all $x$ in the interval.

Now as $f \left(x\right) = \frac{4 x}{{x}^{2} - 1}$

$f ' \left(x\right) = \frac{4 \left({x}^{2} - 1\right) - 4 x \cdot 2 x}{{x}^{2} - 1} ^ 2 = - \frac{1 + 4 {x}^{2}}{{x}^{2} - 1} ^ 2$

and $f ' ' \left(x\right) = - \frac{8 x {\left({x}^{2} - 1\right)}^{2} - \left(1 + 4 {x}^{2}\right) \left(4 x \left({x}^{2} - 1\right)\right)}{{x}^{2} - 1} ^ 4$

= $- \frac{8 {x}^{5} - 16 {x}^{3} + 8 x - \left(16 {x}^{5} + 4 {x}^{3} - 16 {x}^{3} - 4 x\right)}{{x}^{2} - 1} ^ 4$

= $- \frac{- 8 {x}^{5} - 4 {x}^{3} + 12 x}{{x}^{2} - 1} ^ 4 = \frac{4 x \left(2 {x}^{4} + {x}^{2} - 3\right)}{{x}^{2} - 1} ^ 4$

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

Observe that $f ' ' \left(x\right) < 0$ when $x \in \left(- \infty , - 1\right) \cup \left(0 , 1\right)$ and hence in this region f(x) is concave and

$f ' ' \left(x\right) > 0$ when $x \in \left(- 1 , 0\right) \cup \left(0 , \infty\right)$ and hence in this region f(x) is convex.

graph{(4x)/(x^2-1) [-10, 10, -5, 5]}