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) * (x-1/sqrt(3)) + 3/4
Edited on February 24, 2016, 3:10 pm
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Posted by Charlie
on 2016-02-24 15:09:22 |
solution