Find the relative extrema of the functions.

f(x)=4x/(x^2+1)

f(x)=1/(x-2)

take the derivative, set to zero.

the first as an example:

f(x)= 4x (x^2 +1)^-1

f'(x)= 4 (x^2 +1)^-1 -4x(2x)(x^2 +1)^-2
set to zero. Multiply through by (x^2+1)^2

0=4(x^2+1) - 8x^2
4x^2=1
x= +- 1/2

what about the second question
f(x)1/(x-2)

To find the relative extrema of the function f(x) = 1/(x-2), we need to take the derivative and set it equal to zero.

First, let's find the derivative:

f(x) = 1/(x-2)

Using the quotient rule, we can differentiate f(x):

f'(x) = [1*(x-2) - 1*(1)] / (x-2)^2
= -1 / (x-2)^2

Now, we can set f'(x) equal to zero:

-1 / (x-2)^2 = 0

To solve this equation, we can multiply both sides by (x-2)^2:

-1 = 0

However, there is no solution to this equation. It means that f'(x) has no critical points or relative extrema.

Therefore, the function f(x) = 1/(x-2) does not have any relative extrema.