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

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

First we will determine the derivative f'(x)

Let f(x)= u/v

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

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

Reduce similar:

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

Now we sill find critical values:

f'(x)=0 ==> x=1 is a critical value

The extreme value is:

f(1)= 2/2 =1

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neela | Student

To determine the extrme values of f(x) = 2x/x^2+1)

Solution:

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

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

So f'(x) = 0 gives: 1-x^2 = 0 Or x^2= 1. Or x=1 or x=-1.

f''(x) =  (1-x^2)(-4x)/(1+x^2)^3  -2x/ (1+x^2)^2

f"(1) = -2/4 positive.

f'(-1) = 2/4 positive.

So f(x) has extreme values at x= 1. Maximum at f(1) = 2/2 =1.

and f(x) has another extreme value at  x=-1 which is minimum , f(-1) = -2/{1+(-1)^2} = -1

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giorgiana1976 | Student

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

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