Find domain of (2+x-x^2)/((x-1)^2)

1. Oct 11, 2005

bobboxx

Could someone find the domain, intercepts, asymptotes, critical numbers, local min and max, absolute min and max, concavity, and inflection points of the function: (2+x-x^2)/((x-1)^2)

Thank you it would be much appreciated

2. Oct 11, 2005

Zurtex

We help out here, not do things for you. What do you know about these or what have you done so far?

3. Oct 11, 2005

bobboxx

I have found a critical number at x=5 but I don't know if that is correct.

I have a vertical asymptote at X=1 and a horizontal at y=-1

My derivative of the function is (x-5)/(x-1)^3 but I am not sure f this is right either.

I don't know which points to use to tell if it is increasing or decreasing-do I use five and 1?

4. Oct 11, 2005

TD

Since x = 5 is a zero of the derivative, so f'(5) = 0, it is indeed a stationary point. You still need to check whether this is an extremum (and then max or min) or not.

Both correct.

The derivative is correct as well.

A function is rising where its derivitive is positive, a function is decreasing where the derivative is negative. So check the sign of f'(x).

5. Oct 11, 2005

bobboxx

f'(5) is neither a minimum or a maximum

Why is this a critical number? The graph is increasing to the left of -1 and decreasing to the right of -1. The graph is not changing at 5.

6. Oct 11, 2005

TD

Since you derivative f'(x) = (x-5)/(x-1)^3, x = 5 is a zero of f'(x) so f'(5) = 0, therefore x = 5 is a stationary point which means the tangent line is parallel to the x-axis there. This may be a minimum, maximum or a point of inflection.

Do you know what it is?

7. Oct 11, 2005

bobboxx

I think I figured it out.
f(5) would be a local minimum

I think I messed up because I choose f(0) as a point to the left of f(5) instead of a point in between 1 and 5.

Do you have to include your vertical asymptotes when you are trying to find out if the function is increasing or decreasing?

If yes, this is where I messed up. Thanks for your help.

8. Oct 11, 2005

TD

There is indeed a minimum at x = 5, that's right

Well, I don't know what you have to do but when we had to find vertical asymptotes, we also had to check whether the function approached infinity or -infinity to the left and right of the asymptote.

9. Oct 11, 2005

bobboxx

Thanks appreciate it

10. Oct 11, 2005

TD

You're welcome, if there's anything else - do ask

11. Oct 11, 2005

Tom Mattson

Staff Emeritus
Yes, but ask in the Homework section of this site, which is located at the top.

The Math section is not for homework questions, as is clearly explained in the notice posted there entitled "Homework".