Answer to Question #273991 in Calculus for Josa Kyme Palma

Question

Determine the open intervals on which the graph of f(x)= (x²)/(x²+1) is concave upward or concave downward

Answer 1

$f(x)=\dfrac{x^2}{x^2+1}$

Domain: $(-\infin, \infin)$

$f'(x)=\dfrac{2x(x^2+1)-2x(x^2)}{(x^2+1)^2}=\dfrac{2x}{(x^2+1)^2}$

$f''(x)=\dfrac{2((x^2+1)^2-2x(2x)(x^2+1))}{(x^2+1)^4}$

$=\dfrac{2(x^2+1-4x^2)}{(x^2+1)^3}=\dfrac{2(1-3x^2)}{(x^2+1)^3}$

Find the point(s) of inflection

$f''(x)=0=>\dfrac{2(1-3x^2)}{(x^2+1)^3}=0$

$1-3x^2=0$

$x_1=-\dfrac{\sqrt{3}}{3}, x_2=\dfrac{\sqrt{3}}{3}$

If $x<-\dfrac{\sqrt{3}}{3}, f''(x)<0, f(x)$ is concave downward.

If $-\dfrac{\sqrt{3}}{3}<x<\dfrac{\sqrt{3}}{3}, f''(x)>0, f(x)$ is concave upward.

If $x>\dfrac{\sqrt{3}}{3}, f''(x)<0, f(x)$ is concave downward.

$f(x)$ is concave upward on $(-\dfrac{\sqrt{3}}{3}, \dfrac{\sqrt{3}}{3}).$

$f(x)$ is concave downward on $(-\infin,-\dfrac{\sqrt{3}}{3})\cup ( \dfrac{\sqrt{3}}{3}, \infin).$