Solve for x, f'(x)=0 if f(x)=x/(x^2+1).

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The given function is a ratio and we'll determine it's derivative using the quotient rule:

(u/v)'= (u'*v-u*v')/v^2

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

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

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

We've noticed that, at numerator, we have a difference of squares:

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

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

So, the solution of the equation f'(x)=0 are

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

We'll set each factor as zero:

1-x=0, x=1

1+x=0, x=-1

f(x) = x/(x^2+1), To solve for x if f'(x) = 0.

Solution:

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

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

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

= (x^2+1 - 2x^2)/(x^2+1) = (1-x^2)/(x^2+1). Setting f'(x) to zero we get:

(1-x^2)/(x^2+1)^2 = 0, Multiply by (x^2+1)^2.

1-x^2 = 0

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

1-x = 0 or 1+x = 0

x =1 or x =-1 are the solution to f'(x) = 0.

For h(x)=f(x)/g(x): h'(x)= (f'(x)*g(x)-f(x)*g'(x))/g(x)^2

Here f(x)= x and g(x)=(x^2+1)

h'(x)= [1*(x^2+1)-x*2x]/(x^2+1)^2= [x^2+1-2x^2]/(x^2+1)^2= (1-x^2)/(1+x^2)=0

Therefore 1-x^2=0 or (1-x)(1+x)=0

If 1-x=0, x=1

and if 1+x=0, x=-1

**So x is 1 or -1**

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

To calculate the dericative, first, let us rewrite f(x):

f(x) = u/v such that:

u= x ==> u'= 1

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

Now by dfinition:

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

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

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

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

No to find f'(x)= 0.

f'(x)= (1-x^2)/(x^2 +1)^2= 0

==> (1-x^2) = 0

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

==> x1= 1

==> x2= -1

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