NEED HELP with double derivative(long)

[PhysicsinCalifornia]

This is another HW prob I have to finish by Thursday(5/12/05)


For the function $$f(x) = \frac{x}{( x - 2 )^2}$$
I'm supposed to find:
a) domain
b) the coordinates of the x-intercepts, if any
c) the coordinates of the y-intercepts, if any
d) the horizontal asymptotes, if any
e) the vertical asymptotes, if any
f) the intervals on which f(x) is increasing
g) the intervals on which f(x) is decreasing
h) the coordinates of the maximum and minimum points, if any, and state whether they are absolute or relative
i) the intervals oh which f(x) is concave upward
j) the intervals on which f(x) is concave downward
k) the coordinates of any inflection points

(whew)

Now, I found from a-h
a) domain is $$x \neq 2$$
b) x-int:0 (0,0)
c) y-int:0 (0,0)
d) For this, I set $$\lim_{x\rightarrow \pm\infty} f(x)$$
both approach 0, so H.A. when y=0

e)$$\lim_{x\rightarrow \pm 2} f(x)$$
since the denominator would be 0, both approach infinity, so there is a V.A. when x=2


f) I found $$f '(x) = \frac{-(x + 2)}{( x - 2)^3}$$ (this IS right, right?)

and set the top and bottom equal to 0 (critical points)
and got x=-2 and x=2
So f(x) is increasing from (-2,2) by testing points
g) Using the information I got from f),
f(x) is decreasing from $$(-\infty, -2) \cup (2, \infty)$$
h) There is an absolute and relative minimum when $$f(-2) = -\frac{1}{8}$$
Proven by the fact that f '(x) crosses 0 from negative to positive at (-2, -1/8)
i) This is where I'm stuck. But I do know I have to find the derivative of f '(x), or aka f "(x)

i got $$f ''(x) = \frac{2x^3 - 24x + 32}{( x - 2)^6}$$

I'm not sure if this is even the double derivative

Thanks for your help in advance, and sorry for being so long

 

[monet A]

For the first derivative I get (x-2-2x)/(x-2)^3 using the quotient rule for the whole function and the general power rlue to differentiate the bottom, are you familiar with these methods or are you using the definition of the derivative for your working?

The second derivative will go in a similar fashion use the quotient rule overall and get the derivative of the denominator using the general power rule.

The quotient rule is - set numerator = f(x) and denominator = g(x) then the derivative of this is (f '(x)*g(x) - g'(x) * f(x))/ (g(x))^2)

The general power rule is if f(x) = (g(x))^c then f '(x) = (c(g(x))^(c-1))*g'(x)

Well at least I think I am right, if you don't mind me being so bold, I am not a list mentor or anything..

 

[whozum]

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

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

Simplifies to

$$f'(x) = \frac{1}{(x-2)^2} - \frac{2x}{(x-2)^3}$$

Quotient rule:

$$\left(\frac{f(x)}{g(x)}\right)' = \frac{f'(x)g(x) - g'(x)f(x)}{g(x)^2}$$

Just apply it twice to f'(x) (once for each fraction) and that will give you f''(x).

 

[monet A]

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

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

Simplifies to

$$f'(x) = \frac{1}{(x-2)^2} - \frac{2x}{(x-2)^3}$$

Hey Whozum, may I ask how you got this simplification, I factored out one degree of (x-2) from the top and bottom to get my simplification and that's all I have done, yours looks like a useful form to use sometimes, could you demonstrate how you got it for me?

 

[inha]

well there's nothing special to the simplification. (a+b)/c=a/c+b/c

 

[monet A]

well there's nothing special to the simplification. (a+b)/c=a/c+b/c

Well great, I'll probably forget that again a few times before it sticks. I guess I should say thankyou, right?

 

[PhysicsinCalifornia]

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

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

Simplifies to

$$f'(x) = \frac{1}{(x-2)^2} - \frac{2x}{(x-2)^3}$$

Quotient rule:

$$\left(\frac{f(x)}{g(x)}\right)' = \frac{f'(x)g(x) - g'(x)f(x)}{g(x)^2}$$

Just apply it twice to f'(x) (once for each fraction) and that will give you f''(x).

Whozum, Isn't this another way of writing what I got, or am I supposed to use what you got for the second derivative?

Also, how do I write the way you did above??
Thanks

 

[whozum]

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

Your expression for the first derivative is incorrect, when deriving a quotient, you must use the quotient rule which I gave.

For the latex, click on my text, it should open a window showing the code, for more information click the link in the window.

 

[PhysicsinCalifornia]

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

Your expression for the first derivative is incorrect, when deriving a quotient, you must use the quotient rule which I gave.

For the latex, click on my text, it should open a window showing the code, for more information click the link in the window.

I did NOT use the above for my first derivative.

I DID use the quotient rule, and got what you got. But using what you got
$$\frac{-(x + 2)}{( x - 2)^2} = \frac{1}{(x-2)^2} - \frac{2x}{(x-3)^3}$$

I found the LCD to be (x-2)^3, so i distributed the numerator, and i added/subtracted like terms, resulting in -(x+2)

 

[whozum]

What I'm gathering here is that my expression simplifies to yours? Let's see


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

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

*I'm sorry, your right, they are equal. x-2-2x = -x-2 = -(x+2). Just make sure that your denominator is cubed not squared, like you said.

 

[PhysicsinCalifornia]

its ok Whozum. I just didn't want to confuse myself.

so, after i use the quotient rule on the $$f'(x)$$,

the second derivative would be:
$$f''(x) = \frac { 2x^3 -24x +32} {(x - 2)^6}$$

The $$x^2$$ will be gone.

Is this right??

 

[PhysicsinCalifornia]

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

Your expression for the first derivative is incorrect, when deriving a quotient, you must use the quotient rule which I gave.

For the latex, click on my text, it should open a window showing the code, for more information click the link in the window.

Sorry whozum, you were right

I had to the power of 2. I already edited it in the original

Thanks