Most tangent lines to the curve y=1/(x^2+1) reintersect the curve at another point. Three of these tangent lines do not. One of them is the trivial line y=1. Find the other two.
The other two are the inflection points of the curve, where the second derivative is zero.
y = (x^2 + 1)^1
y' = ((x^2+1)^2) * 2x
y" = 2(3x^2  1) / (x^2+1)^3
Solving 3x^2  1 = 0,
x = +/1/sqrt(3)
Therefore the tangents go through points (+/1/sqrt(3), 3/4).
Equations of the tangents:
y = (3*sqrt(3)/8) * (x+1/sqrt(3)) + 3/4
y = (3*sqrt(3)/8) * (x1/sqrt(3)) + 3/4
Edited on February 24, 2016, 3:10 pm

Posted by Charlie
on 20160224 15:09:22 