Answer:
The relative maximum value is [tex]\frac{2}{3}[/tex]
Step-by-step explanation:
The given function is
[tex]g(x)=\frac{2}{x^2+3}[/tex]
We differentiate to obtain;
[tex]g'(x)=-\frac{4x}{(x^2+3)^2}[/tex]
At turning points [tex]g'(x)=-\frac{4x}{(x^2+3)^2}=0[/tex]
[tex]\implies x=0[/tex]
[tex]g''(x)=\frac{16x^2}{(x^2+3)^3}- \frac{4}{(x^2+3)^2}[/tex]
We apply the second derivative test to obtain:
[tex]g''(0)=\frac{16(0)^2}{((0)^2+3)^3}- \frac{4}{((0)^2+3)^2}=-\frac{4}{9}[/tex]
Since the second derivative is negative, there is a relative maximum at x=0.
We substitute x=0 into the original function to obtain the relative maximum value.
[tex]g(0)=\frac{2}{(0)^2+3}=\frac{2}{3}[/tex]
