Question

Identify the relative maximum value of g(x) for the function shown below.

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

Answer 1

Answer:

The maximum value of g(x) = 2/3 at x = 0

Step-by-step explanation:

* Lets find the maximum value of a function using derivative of it

- The function g(x) = 2/(x² + 3)

- 1st step use the negative power to cancel the denominator

∴ g(x) = 2(x² + 3)^-1

- 2nd use derivative of g(x) to find the value of x when g'(x) = 0

* How to make the derivative of a function

# If f(x) = a(h(x))^n, then f'(x) = an[h(x)^(n-1)](h'(x))

∵ $g(x)=2(x^{2}+3)^{-1}$

∴ $g'(x) = 2(-1)(x^{2}+3)^{-2}(2x)=-4x(x^{2}+3)^{-2}$

# Put g'(x) = 0

∴ $-4x(x^{2}+3)^{-2}=0====\frac{-4x}{(x^{2}+3)^{2}}=0$

∴ $-4x=(0)(x^{2}+3)^{2}====-4x = 0$

∴ x = 0

* The maximum value of g(x) at x = 0

- Substitute the value of x in g(x)

∴ g(0) = 2/(0 + 3) = 2/3

* The maximum value of g(x) = 2/3 at x = 0