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.
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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.
The derivative is correct. $\displaystyle -2x^2+x^2+3 = 3 - x^2$.
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-x2 = 0, in which case:
$\displaystyle x = \pm \sqrt{3}$
(your error was in taking -2x2 + x2 to be x2, instead of -x2).
the last line, "your error was in taking -2x2 + x2 to be x2, instead of -x2", would've sufficed, but gee… Thank you so much!
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.
