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Calculus/please help me in this question

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Expert: - 7/25/2009


Question
given 2 separate functions of,
k(x)={x^4-1  x<1  and, h(x)={x+1              x<-1
    {x^3-8  x>2           ={(x+1)(x-1)(x-2)  x>-1

a)Find (h/k)(x)
b)Domain for (h/k)(x)

Answer
By what's given we have to define h(x)/k(x) by the values of x<-1 and x>-1.
For x<-1, the value of h(x)/k(x) is (x+1)/(x^4 - 1);
for x>-1, the value of h(x)/k(x) is (x+1)(x-1)(x-2)/(x^3 - 8).

We have the critical points -1, 1, and 2.

For x<-1
We have (x+1)/(x^4 - 1)
Note that x^4 - 1 = (x^2 + 1)(x^2 - 1).  x^2 - 1 = (x+1)(x-1).
That means that x^4 - 1 = (x^2+1)(x+1)(x-1).

Therefore (x+1)/(x^4 - 1) = (x+1)/(x^2+1)(x+1)(x-1) = 1/[(x^2 + 1)(x-1).
Note that the x value of -1 is not in the interval, so we don't have to worry about it.

For -1<x<1
We have (x^4 - 1)/[(x+1)(x-1)(x-2)].
That is the same as (x^2 + 1)(x-1)(x+1)/[(x+1)(x-1)(x-2)].
Note that (x+1)and (x-1) cancel.  

This leaves (x^2 + 1)/(x-2).
Note that 1 and -1 are at the endpoints, so the function is not defined there anyway.

For 1<x
Here, the function would be (x^3 - 8)/[(x+1)(x-1)(x-2)].
Since x^3 - 8 = (x-2)(x^2 + 2x + 4), so the x-2 cancels.

This leaves (x^2 + 2x + 4)/[(x+1)(x-1)].
Note that for this one, we have to throw out the value 2 since we had the value x-2,
and that is in the domain.

b) The domain would be all x except the values [1,2], since that is where k is undefined.
By [1,2], that means from 1 to 2.  The [ and ] mean including the endpoints.

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