Question

The interval in which the function $\frac{x^2}{x^2+3}$ is concave up

• A. $(1,\infty )$
• B. $(-\infty ,-1)\cup (1,\infty )$
• C. $(0,\infty )$
• D. $(-1,1)$

Answer 1

The correct option is D $(-1,1)$

Given, $f\left(x\right)=\frac{x^2}{x^2+3},x\in R$
$\Rightarrow f^{\prime }\left(x\right)=\frac{6x}{(x^2+3{)}^2}$
$\Rightarrow f^{\prime \prime }\left(x\right)=\frac{18(x^2+3)(1-x^2)}{(x^2+3{)}^4}$
For concave up, $f^{\prime \prime }\left(x\right)>0$
$\Rightarrow 1-x^2>0$
$\Rightarrow x^2-1<0$
$\Rightarrow x\in (-1,1)$