Find $\displaystyle f''\left( x \right)=0$ when $\displaystyle f\left( x \right)=\frac{\left( x-1 \right)^{2}}{1+x^{2}}$
Attempt: https://dl.dropbox.com/u/586465/2012...2016.48.57.jpg
the thing is: I get +3 instead of -3.
Find $\displaystyle f''\left( x \right)=0$ when $\displaystyle f\left( x \right)=\frac{\left( x-1 \right)^{2}}{1+x^{2}}$
Attempt: https://dl.dropbox.com/u/586465/2012...2016.48.57.jpg
the thing is: I get +3 instead of -3.
first derivative:
$\displaystyle f'(x) = \frac{(1+x^2)(2(x-1)) - (x-1)^2(2x)}{(1+x^2)^2}$
$\displaystyle = \frac{2x^3 - 2x^2 + 2x - 2 - (2x^3 - 4x^2 + 2x)}{(1+x^2)^2}$
$\displaystyle = \frac{2x^2 - 2}{(1+x^2)^2}$.
second derivative:
$\displaystyle f''(x) = \frac{(1+x^2)^2(4x) - (2x^2 - 2)(2(1+x^2)(2x))}{(1+x^2)^4}$
$\displaystyle = \frac{(1+x^2)(4x) - (2x^2 - 2)(4x)}{(1+x^2)^3}$
$\displaystyle = \frac{4x^3 + 4x - 8x^3 + 8x}{(1+x^2)^3}$
$\displaystyle = \frac{(4x)(3 - x^2)}{(1+x^2)^3}$.
now if f''(x) = 0, the numerator must be 0, so:
$\displaystyle (4x)(3-x^2) = 0$.
thus either 4x = 0, and x = 0, or: 3-x^{2} = 0, in which case:
$\displaystyle x = \pm \sqrt{3}$
(your error was in taking -2x^{2} + x^{2} to be x^{2}, instead of -x^{2}).
perhaps so, but future readers of this thread may not click on, or view the image link you posted.
also, i'm no computational prodigy, but a bit more like a turtle: slow and sure wins the race. so to keep things straight in my mind, i have to verify. every. single. step (i discovered several computational errors before i finished my post). derivatives involving quotients are particularly nasty, because of the many opportunities for sign errors involved.