Determine the number of extreme values for the function 2x/(x^2+1).

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william1941 | College Teacher | (Level 3) Valedictorian

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The extreme values of an expression are the values of f(x) where x is determined by equating the derivative of f(x) to 0.

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

Now f'(x) for g(x)/ h(x) = [g'(x)h(x)- g(x)h'(x)] / [h(x)]^2

Therefore:

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

derivative of 2x is 2 and derivative for (1+x^2) is 2x

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

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

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

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

Now equating this to 0

=> 2*(1 - x^2) / (x^2+1)^2 =0

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

=> x = 1 and x=-1

For x= 1, f(1) = 2 / (1 +1) = 1

For x = -1 , f(-1) = -2 / (1 +1) = -1

Therefore there are two extremes, at x= 1 and x= -1

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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In order to determine the number of local extremes of the function, we'll have to find out the extreme values of a function, first.

We'll have to do the first derivative test of a function.

We'll apply 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

We'll simplify the ratio by 2:

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

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

f'(x)=0

It's obvious that the denominator is positive, for any value of x.

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

The numerator 1-x^2 is a difference of 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 0.

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

The number of local extremes of f(x) is 2.

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

To find the extreme values of 2x/(x^2+1).

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

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

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

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

Equating f'(x)  to zero we get:

numerator 2-4x^2 = 0

4x^2 = 2

x^2 = 1

x= +or- sqt1 .

x1 = 1 or

x2 = -1 are the extrme values.

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

f(-1) =  -1/({(-1)^2+1} = -1/2.

Therefore f(1) = 1/2  when x= 1. And f(-1) = -1/2 when are the extreme values when x= 1 and x = -1 respectively.

 

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