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 $(-\infin,-\dfrac{\sqrt{3}}{3})\cup ( \dfrac{\sqrt{3}}{3}, \infin).$