algebra minus

**jacob93** · September 14th 2012, 06:52 AM

Find $f''\left( x \right)=0$ when $f\left( x \right)=\frac{\left( x-1 \right)^{2}}{1+x^{2}}$

Attempt: https://dl.dropbox.com/u/586465/2012...2016.48.57.jpg

the thing is: I get +3 instead of -3.

**emakarov** · September 14th 2012, 07:12 AM

The derivative is correct. $-2x^2+x^2+3 = 3 - x^2$ .

**Deveno** · September 14th 2012, 08:48 AM

first derivative:

$f'(x) = \frac{(1+x^2)(2(x-1)) - (x-1)^2(2x)}{(1+x^2)^2}$

$= \frac{2x^3 - 2x^2 + 2x - 2 - (2x^3 - 4x^2 + 2x)}{(1+x^2)^2}$

$= \frac{2x^2 - 2}{(1+x^2)^2}$ .

second derivative:

$f''(x) = \frac{(1+x^2)^2(4x) - (2x^2 - 2)(2(1+x^2)(2x))}{(1+x^2)^4}$

$= \frac{(1+x^2)(4x) - (2x^2 - 2)(4x)}{(1+x^2)^3}$

$= \frac{4x^3 + 4x - 8x^3 + 8x}{(1+x^2)^3}$

$= \frac{(4x)(3 - x^2)}{(1+x^2)^3}$ .

now if f''(x) = 0, the numerator must be 0, so:

$(4x)(3-x^2) = 0$ .

thus either 4x = 0, and x = 0, or: 3-x² = 0, in which case:

$x = \pm \sqrt{3}$

(your error was in taking -2x² + x² to be x², instead of -x²).

**jacob93** · September 14th 2012, 09:03 AM

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!

**Deveno** · September 14th 2012, 09:14 AM

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.

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