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calc problem

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Your teacher is a b----. Let’s leave it at that. Hehehehe

Alright let’s get down to business.

Finding C
Each of the three pieces of information gives us clues as to what the different aspects of f(x) are. #1 looks like a bummer so we’ll wait on that. #2 says that as x approaches 2 from the right side, the values of f(x) go up to +∞, which is an automatic clue that there is a vertical asymptote at x=2. Because of this, we can find the denominator of the function, (x^2-c). Only one value of c satisfies the difference of squares (x+2)(x-2), and that is 4. So c=4. Try putting any other value for c and you’ll never get (x-2) in the denominator, which we’re trying to get. Now, our function is (ax+b)/(x^2-4).

Finding A
I thought about this for a while… hmm… if the function is symmetrical with respect to the y-axis, then all values of f(x) should equal all values of f(-x). Draw a graph of f(x)=|x| and you’ll see that x=1 and x=-1 have the same value and so on. Thus, because f(x)=f(-x), (ax+b)/(x^2-c) = (a(-x)+b)/((-x)^2-c), or (ax+b)/(x^2-c) = (-ax+b)/(x^2-c). Then (ax+b) = (-ax+b) because the denominators cancel. Simplifying, we get 2ax = 0. So for any value of x, 2ax = 0, which means that a=0.

Finding B.
Yessirreee we’re gonna differentiate this mofo.
f(x) = (ax+b)/(x^2-c)

by property of differentiating quotients, we get:
f`(x) = [(x^2-c)(ax+b)’ – (ax+b)(x^2-c)’ ]/(x^2-c)^2
f`(x) = [(x^2-c)(a) – (ax+b)(2x)]/(x^2-c)^2
f`(x) = (ax^2 – ac – 2ax^2 – 2xb)/(x^2-c)^2
f`(x) = (-ax^2 – ac – 2xb)/(x^2-c)^2

here’s property #3 where f`(1) = -2
(-a(1)^2- ac – 2(1)b)/(x^2-c)^2 = -2
(-a – ac – 2b)/(1-c)^2 = -2

and because we established that c=4
(-a – a(4) – 2b)/(1-(4))^2 = -2
(-5a – 2b)/9 = -2
so b= (-5a+18)/2 and because a=0
b=9

SO...
A) a=0, b=9, c=4
B) Look at the denominator of the function to determine the vertical asymptotes. (x^2-4) = 0 where there is an asymptote because… well there are asymptotes that approach positive or negative infinity where the function isn’t defined. So, we have two values of x, -2 and 2. Vertical asympotes: x=-2, x=2

For horizontal asympote… we have the final function of f(x) = 9/(x^2-4). Notice how f(x) approach zero as x approaches positive or negative ∞. So when you go all the way to the left or all the way to the right, x^2 in the denominator becomes huge and f(x) approaches 0. Horizontal asymptote: y=0.

Let’s see what handy Ti-89 shows… what a sexy graph, f(x) is symmetrical about the y-axis, the limit as x approaches 2 from the right is +∞, and the slope at f(1) certainly looks like -2. And the vertical and horizontal asymptotes are clearly visible.


Hopefully I didn’t make it too excrutiatingly comprehensive :/

heyy.....i checked it over....n it all works out....
thanx muchh again

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woah bckp nice writing there man, took a lot of time out to write that. very cool. glad you got your problem fixed az0nd2r3

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