Question

Find the inflection points and intervals in which the function $f\left(x\right)=\frac{x^2}{x^2+3}$ is concave up and concave down.

Answer 1

$f\left(x\right)=\frac{x^2}{x^2+3}$

$f^{\prime }\left(x\right)=\frac{2x\left(x^2+3\right)-2x^3}{{\left(x^2+3\right)}^2}$$=\frac{-6x}{{\left(x^2+3\right)}^2}$

$f^{\prime \prime }\left(x\right)=\frac{6x{\left(x^2+3\right)}^2-6x\times 2\left(x^2+3\right)\times 2x}{{\left(x^2+3\right)}^4}$

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

$=\frac{6x\left(x^2+3\right)\left(x-1\right)\left(x-3\right)}{{\left(x^2+3\right)}^4}$

For point of inflection

$f^{\prime \prime }\left(x\right)=0$

$\Rightarrow \frac{6x\left(x^2+3\right)\left(x-1\right)\left(x-3\right)}{{\left(x^2+3\right)}^4}=0$

$\Rightarrow 6x\left(x^2+3\right)\left(x-1\right)\left(x-3\right)=0$

$\Rightarrow x=0,1,3$

so, points of inflection are $x=0,1,3$

for concave upward

$f^{\prime \prime }\left(x\right)>0$

$\Rightarrow \frac{6x\left(x^2+3\right)\left(x-1\right)\left(x-3\right)}{{\left(x^2+3\right)}^4}>0$

$\Rightarrow 6x\left(x^2+3\right)\left(x-1\right)\left(x-3\right)>0$

$\Rightarrow x\in \left(0,1\right)\cup \left(3,\infty \right)$

For concave downward

$f^{\prime \prime }\left(x\right)<0$

$\Rightarrow \frac{6x\left(x^2+3\right)\left(x-1\right)\left(x-3\right)}{{\left(x^2+3\right)}^4}<0$

$\Rightarrow 6x\left(x^2+3\right)\left(x-1\right)\left(x-3\right)<0$

$\Rightarrow x\in \left(-\infty ,0\right)\cup \left(1,3\right)$