Answer
concave up $x$ < $-1$, $x$ > $1$
concave down $-1$ < $x$ < $1$
inflection point is at $x$ = $-1$, $x$ = $1$
Work Step by Step
let
$f(x)$ = $\frac{1}{x^{2}+3}$
$f'(x)$ = $\frac{-2x}{(x^{2}+3)^{2}}$
$f''(x)$ = $ \frac{(x^{2}+3)^{2}(-2)-(-2x)(2)(x^{2}+3)(2x)}{(x^{2}+3)^{4}}$ = $\frac{6x^{2}-6}{(x^{2}+3)^{3}}$
concave up
$\frac{6x^{2}-6}{(x^{2}+3)^{3}}$ > $0$
$x^{2}-1$ > $0$
$(x-1)(x+1)$ > $0$
$x$ < $-1$, $x$ > $1$
concave down
$\frac{6x^{2}-6}{(x^{2}+3)^{3}}$ < $0$
$x^{2}-1$ < $0$
$-1$ < $x$ < $1$
inflection point is at
$x$ = $-1$, $x$ = $1$
