How do you find the inflection point of #(x+1)/(x^(2)+1)#?

Question

Charles Arocha · Accepted Answer

We get the Points
#P_1(1;1),P_2(-2-sqrt(3),1/4*(1-sqrt(3))),P_3(-2+sqrt(3),1/4(1+sqrt(3)))# By the Quotient rule we get #f'(x)=(-x^2-2x+1)/(x^2+1)^2# #f''(x)=(2x^3+6x^2-6x+2)/(x^2+1)^3# so we have to solve #2(x^3+3x^2-3x-1)=0# This is #(x-1)(x^2+4x+1)=0# Solutions are #x_1=1# #x_2=-2-sqrt(3)# #x_3=-2+sqrt(3)#

Scarlett Allison · Answer

#x_1=1#
#x_2=-2-sqrt(3)#
#x_3=-2+sqrt(3)# For a point of inflection #g#: #f''(g)=0# #f'''(g)ne0# #f(x)=(x+1)/(x^2+1)# Using the Quotient rule: #f'(x)=(1*(x^2+1)-(x+1)(2x))/(x^2+1)^2# #=(x^2+1-2x^2-2x)/(x^2+1)^2# #=(-x^2-2x+1)/(x^2+1)^2# #=-((x+1)^2-2)/(x^2+1)^2# Again, using the Quotient rule: #f''(x)=-(2*(x+1)*(x^2+1)^2-((x+1)^2-2)(2*2x(x^2+1)))/(x^2+1)^4# #=-((2x+2)* (x^2+1)^2-4x(x+1)^2(x^2+1)-4x(x^2+1))/(x^2+1)^4# #=-((2x+2)*(x^2+1)-4x(x+1)^2-4x)/(x^2+1)^3# #f''(x)=0# #0=-((2x+2) * (x^2+1)-4x(x+1)^2-4x)/(x^2+1)^3|* (x^2+1)^3# #0=-(2x+2) * (x^2+1)-4x(x+1)^2-4x# #0=-2x^3+2x-2x^2-2+4x^3+8x^2-4x-4x# #0=2x^3+6x^2-6x-2# #x_1=1# #x_2=-2-sqrt(3)# #x_3=sqrt(3)-2# Again, using the Quotient rule: #f'''(x)=-(6(x^4+4x^3+4x^2-4x-1))/(x^2+1)^4# #f'''(1)=-3/2# #f'''(-2-sqrt(3))=3/32*(4*sqrt(3)-7)~~-0.0067# #f'''(-2+sqrt(3))=-3/32*(4*sqrt(3)+7)~~-1.31#

Charles Arocha · Answer

undefined To find the inflection point of $ \frac{{x + 1}}{{x^2 + 1}} $:

1. Find the second derivative of the function.
2. Set the second derivative equal to zero and solve for $ x $.
3. Determine the $ y $-coordinate of the inflection point by plugging the found $ x $-value into the original function.