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

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

bobboxx said:
I have a vertical asymptote at X=1 and a horizontal at y=-1
Both correct.

bobboxx said:
My derivative of the function is (x-5)/(x-1)^3 but I am not sure f this is right either.
The derivative is correct as well.

bobboxx said:
I don't know which points to use to tell if it is increasing or decreasing-do I use five and 1?
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).
 
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.
 
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?
 
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.
 
There is indeed a minimum at x = 5, that's right :smile:

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.
 
Thanks appreciate it
 
  • #10
You're welcome, if there's anything else - do ask :smile:
 
  • #11
TD said:
You're welcome, if there's anything else - do ask :smile:

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".
 
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