# Verify if the function f=2x/(x^2+1) has one or two extreme values.

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### 3 Answers

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

To find the extreme values, first we need to determine 1st derivative f'(x)"

Let f(x) = u/v such that:

u= 2x ==> u'= 2

v= x^2 + 1) ==> v'= 2x

==> f'(x) = (u'v-uv')/v^2

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

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

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

Now to find crititcal values, we need to find the derivative's zeros.

==> 2-2x^2 = 0

==> 2x^2 = 2

==> x^2 = 1

==> x1 = 1

==> f(x1) = f(1) = 2/2=1

==> x2= -1

==> f(x2) = f(-1) = -2/2= -1

Then the function has two extreme value at the points:

(1,1 ) and (-1,-1)

To establish the number of extreme values of the given function, we have to apply the first derivative test.

We notice that the function is a ratio, so we'll calculate the first derivative using the quotient rule:

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

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

In order to calculate the extreme values of the function, we have to determine the roots of the first derivative.

f'(x)=0

The denominator is a sum of squares, so, it will be positive (it won't be zero) for any value of x.

So, only the numerator could have roots, if the delta>0.

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 set each factor as zero.

1-x=0, x=1

1+x=0,x=-1

So, there are 2 extreme values of f(x) and the values are:

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

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

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

The extreme values are given by the solutions of f/(x) = 0.

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

={2(x^2+1) - 2x*2x}/(x^2+1)^2

= {2x^2+2 -4x^2}/(x^2+1)^2

f'(x) =(2-2x^2)/(x^2+1)^2= 2(1-x^2)/(x^2+1)^2. Setting this to zero, we get: 1-x^2 =0 or x^2 =1. Or x= +1 or -1.

Since there are 2 critical values for x . So there are 2 extreme values for x.

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

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