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

For the function [tex]f(x) = \frac{x}{( x - 2 )^2}[/tex]

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 [tex]x \neq 2[/tex]

b) x-int:0 (0,0)

c) y-int:0 (0,0)

d) For this, I set [tex]\lim_{x\rightarrow \pm\infty} f(x)[/tex]

both approach 0, so H.A. when y=0

e)[tex]\lim_{x\rightarrow \pm 2} f(x)[/tex]

since the denominator would be 0, both approach infinity, so there is a V.A. when x=2

f) I found [tex]f '(x) = \frac{-(x + 2)}{( x - 2)^3}[/tex] (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 [tex](-\infty, -2) \cup (2, \infty)[/tex]

h) There is an absolute and relative minimum when [tex]f(-2) = -\frac{1}{8}[/tex]

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 [tex]f ''(x) = \frac{2x^3 - 24x + 32}{( x - 2)^6}[/tex]

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

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

For the function [tex]f(x) = \frac{x}{( x - 2 )^2}[/tex]

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 [tex]x \neq 2[/tex]

b) x-int:0 (0,0)

c) y-int:0 (0,0)

d) For this, I set [tex]\lim_{x\rightarrow \pm\infty} f(x)[/tex]

both approach 0, so H.A. when y=0

e)[tex]\lim_{x\rightarrow \pm 2} f(x)[/tex]

since the denominator would be 0, both approach infinity, so there is a V.A. when x=2

f) I found [tex]f '(x) = \frac{-(x + 2)}{( x - 2)^3}[/tex] (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 [tex](-\infty, -2) \cup (2, \infty)[/tex]

h) There is an absolute and relative minimum when [tex]f(-2) = -\frac{1}{8}[/tex]

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 [tex]f ''(x) = \frac{2x^3 - 24x + 32}{( x - 2)^6}[/tex]

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

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

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