Find the points of inflection of the curve #y=(1+x)/(1+x^2)#?

Question

Charles Anctil · Accepted Answer

#y={1+x}/{1+x^2}#

By Quotient Rule,

#y'={1cdot(1+x^2)-(1+x)cdot2x}/{(1+x^2)^2} ={1-2x-x^2}/{(1+x^2)^2}#

By Quotient Rule,

#y''={(-2-2x)cdot(1+x^2)^2-(1-2x-x^2)cdot2(1+x^2)(2x)}/{(1+x^2)^4}#

#={2(1+x^2)(-1-x^2-x-x^3-2x+4x^2+2x^3)}/{(1+x^2)^4}#

#={2(x^3+3x^2-3x-1)}/{(1+x^2)^3}#

#={2(x-1)(x^2+4x+1)}/{(1+x^2)^3}#

By setting #y''=0#,

#{(x-1=0 Rightarrow x=1),(x^2+4x+1=0 Rightarrow x=-2pm sqrt{3}):}#

Using the x-values found above to split #(-infty,infty)# into intervals

#(-infty,-2-sqrt{3}),(-2-sqrt{3},-2+sqrt{3}),(-2+sqrt{3},1), " and " (1,infty)#.

Using sample points: #x=-4,-2,0#, and #2# for the intevals, respectively,

#y''(-4)<0#, #y''(-2)>0#, #y''(0)<0#, and #y''(2)>0#,

which indicate concavity changes at each point; therefore, there are inflection points at #x=1, -2pmsqrt{3}#.

The inflection points are

#(1,y(1))=(1,1)#,

#(-2-sqrt{3},y(-2-sqrt{3}))=(-2-sqrt{3},{1-sqrt{3 }}/4)#,

and

#(-2+sqrt{3},y(-2+sqrt{3}))=(-2+sqrt{3},{1+sqrt{3}}/4)#.

I hope that this was helpful.

Isabella Ackerman · Answer

undefined The points of inflection of the curve $ y = \frac{1+x}{1+x^2} $ occur at $ x = -1 $ and $ x = 1 $.