Derivative of f(x) = (-x-2)/(x-2)³ using quotient rule

[PhysicsinCalifornia]

Please Help Me !

Hello,

I REALLY need help deriving a function

if $$f(x) = \frac{-x-2}{(x-2)^3}$$
what would its derivative be?

Here's what I did so far::
Using the quotient rule,
$$f'(x) = \frac{(x-2)^3(-1) - (-x-2)(3(x-2)^2(1))}{((x-2)^3)^2)}$$
$$= \frac{-(x-2)^3 + (x+2)(3(x^2-4x+4))}{(x-2)^6}$$
$$= \frac{-(x^2 -4x+4)(x-2) + (x+2)(3x^2 -12x +12)}{(x-2)^6}$$
$$= \frac{-(x^3 -4x^2 + 4x -2x^2 +8x -8) + (3x^3 -12x^2 + 12x +6x^2 -24x +24)}{(x-2)^6}$$
$$= \frac{-x^3 + 4x^2 -2x + 2x^2 -8x + 8 + 3x^3 - 12x^2 + 12x + 6x^2 -24x +24}{(x-2)^6}$$
$$f'(x) = \frac{2x^3 - 24x + 32}{(x-2)^6}$$

Is this the derivative, and if so, how can I find the critical numbers? (When the derivative equals 0 or DNE(does not exist))

 

[whozum]

You took the derivative correctly, and nice job with the latex, but Ican't check through allthat simplification in my head. To find critical points, set the expression equal to zero and solve. It equals zero whenever the numerator equals zero (and the denominator isnt).

 

[PhysicsinCalifornia]

You took the derivative correctly, and nice job with the latex, but Ican't check through allthat simplification in my head. To find critical points, set the expression equal to zero and solve. It equals zero whenever the numerator equals zero (and the denominator isnt).

I first tried to solve when $$2x^3 -24x +32 = 0$$
I thought it was easier to factor the numerator first

Getting: $$2(x-2)^2 (x+4) = 0$$
There, much easier.

Now it's clear that x=2 and x=-4

Thanks for your help Whozum

 

[dextercioby]

You needn't have done all those additions & multiplications,u should have factored $x^{2}-4x+4$ in the 3-rd line and it would have been simpler.

Daniel.

 

[HallsofIvy]

I first tried to solve when
I thought it was easier to factor the numerator first

Getting:
There, much easier.

Now it's clear that x=2 and x=-4

Warning! Those are not the only critical points. A "critical point" is, as you said, a point where the derivative is 0 (i.e. the numerator is 0) or where the derivative does not exist (i.e. where the denominator is 0).

 

[dextercioby]

The function is not defined in the point $x=2$,so you could simplify through the monom $(x-2)$ in the final expression of the derivative.

Daniel.

 

[uart]

I got the following

6(x-2)^-4 + 3x(x-2)^-4 - (x-2)^-3

= (3x + 6 - x + 2) (x-2)^-4

= (2x+8)(x-2)^-4

I think my way is easier.

 

[dextercioby]

It looks okay.So the only critical point is $x=-4$.A second derivative computation will reveal its nature.

Daniel.