Find the inflection point of (x+1)/(x^(2)+1)?

Question

sirm3d · Answer

we know that the first derivative is $$\huge \frac{ x^2+2x-1 }{ (x^2+1)^2 }$$
just take the second derivative and find the critical numbers of the numerator.

anonymous · Answer

i found the second derivative. its really long and tedious. i don't know how to solve for x for it

anonymous · Answer

[(2x-4x+2)(x^2+1)^2-(x^2+1-2x^2+2x)2(x^2+1)(2x)]/[(x^2+1)^2]^2

anonymous · Answer

this is the second derivative

sirm3d · Answer

the first factor (2x-4x+2) isn't right. i'm still checking the other factors in the numerator.

anonymous · Answer

for the first factor, i took the derivative of (x^2+1-2x^2+2x). i found it's derivative to be (2x+0-4x+2) = (2x-4x+2)

sirm3d · Answer

my mistake. let's simplify first the numerator of the first derivative and get $$\huge \frac{-x^2-2x+1}{(x^2+1)^2}$$

anonymous · Answer

ok

anonymous · Answer

hey you mind if i change the question a little bit to (3x-2)/(3x+1)?

anonymous · Answer

i'll find the derivatives

sirm3d · Answer

the numerator is -x^2 - 2x +1, NOT -x^2 PLUS 2x +1

sirm3d · Answer

(3x-2)/(3x+1) is fine with me. what do you need to find?

anonymous · Answer

the same thing. inflection point

sirm3d · Answer

already have an answer or just starting?

anonymous · Answer

i'm finding the derivative. give me a second

anonymous · Answer

no inflection point

sirm3d · Answer

i'll change my drawing. |dw:1352195987399:dw|