Question

Given g(x)= (x^2)/(x^2-4)

a) Find all critical values for 𝑔(𝑥).

b) Find 𝑔 ′′(𝑥). Then, use your answers from part (a) and the 2nd derivative test to determine if each critical value represents a relative maximum, minimum, or neither

c) On what interval(s), if any, is 𝑔(𝑥) concave down. Justify.

Answer 1

Consider the function $g(x)= \frac{x^2}{x^2-4}$

a) Let us find the derivative of $g(x):$

$g'(x)=\frac{2x(x^2-4)-x^22x}{(x^2-4)^2}=-8\frac{x}{(x^2-4)^2}$

In the points $-2$ and $2$ the function $g(x)$ doesn’t exist. If $g'(x)=0$, then $x=0.$ Therefore, $x=0$ is a critical value for $𝑔(𝑥)$.

b) Let us find $𝑔''(𝑥)$:

$g''(x)=(-8\frac{x}{(x^2-4)^2})'=-8\frac{(x^2-4)^2-x\cdot 2(x^2-4)2x}{(x^2-4)^4}= -8\frac{x^2-4-4x^2}{(x^2-4)^3}=8\frac{3x^2+4}{(x^2-4)^3}$

Since $g''(0)=-0.5<0$, we conclude that the critical value $x=0$ represents a relative maximum.

c) Since $g''(x)<0$ for $x\in(-2,2)$, we conclude that on the interval $(-2,2)$ the function $𝑔(𝑥)$ is concave down.