If the function f(x) = (2x)/(x^2 + 1) determine the extreme values of the function .

For calculating the extreme values of a function we have to calculate the first derivative of a function. In this case, we have to calculate the firts derivative of a ratio:

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

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

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

f'(x)= (-2x^2+2)/(x^2+1)^2

We'll factorize and we'll divide by 2:

f'(x)= (1-x^2)/(x^2+1)^2

Now, we have to calculatethe roots of the first derivative.

f'(x)=0

Because the denominator is positive, always, we'll calculate the roots of the numerator, only.

1-x^2 it's a difference between squares:

a^2-b^2=(a-b)(a+b)

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

(1-x)(1+x)=0

We'll put each factor as zero.

1-x=0, x=1

1+x=0,x=-1

So, the extreme values of the function are:

f(1)=2*1/(1^2+1)=2/2=1

f(-1)=2*(-1)/(-1^2+1)=-2/2=-1