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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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
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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