Question

Identify the relative maximum value of g(x) for the function shown below g(x) = 2/ x^2 +3

Answer 1

Answer:

The relative maximum value is $\frac{2}{3}$

Step-by-step explanation:

The given function is

$g(x)=\frac{2}{x^2+3}$

We differentiate to obtain;

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

At turning points $g'(x)=-\frac{4x}{(x^2+3)^2}=0$

$\implies x=0$

$g''(x)=\frac{16x^2}{(x^2+3)^3}- \frac{4}{(x^2+3)^2}$

We apply the second derivative test to obtain:

$g''(0)=\frac{16(0)^2}{((0)^2+3)^3}- \frac{4}{((0)^2+3)^2}=-\frac{4}{9}$

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.

$g(0)=\frac{2}{(0)^2+3}=\frac{2}{3}$

Answer 2

Answer:

they're right, it is 2/3

Step-by-step explanation: