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))
∵ [tex]g(x)=2(x^{2}+3)^{-1}[/tex]
∴ [tex]g'(x) = 2(-1)(x^{2}+3)^{-2}(2x)=-4x(x^{2}+3)^{-2}[/tex]
# Put g'(x) = 0
∴ [tex]-4x(x^{2}+3)^{-2}=0====\frac{-4x}{(x^{2}+3)^{2}}=0[/tex]
∴ [tex]-4x=(0)(x^{2}+3)^{2}====-4x = 0[/tex]
∴ 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
