# How do you find the range of 1/{(x-1)(x+2)} for [0, 6]?

1. Feb 12, 2012

### solve

1. The problem statement, all variables and given/known data

Need to find the range of 1/{(x-1)(x+2)} for [0, 6]

2. Relevant equations
3. The attempt at a solution

The domain is [0, 1) AND (1, 6] because 1/0 is an invalid operation. So I tried and this is what I got: [-0.5, ∞) and [0.025, ∞). Wrong, of course :)

How do I go about arriving to the right solution?

Thanks.

2. Feb 12, 2012

### SammyS

Staff Emeritus
Can you find the range of (x-1)(x+2) for [0, 1) ∪ (1, 6] ?

3. Feb 12, 2012

### solve

Hopefully I am right: [-2,0) and (0,40]

4. Feb 12, 2012

### SammyS

Staff Emeritus
Yes, that's correct.

Do you see how that's related to the range of 1/((x-1)(x+2)) ?

5. Feb 12, 2012

### solve

Honestly not. Can I get a hint or something :)

6. Feb 12, 2012

### Mentallic

If the values of a variable k range between 2 to 5, then what are the possible values of 1/k?

7. Feb 12, 2012

### solve

Re: How do you http://www.physicsforums.cfind [Broken] the range of 1/{(x-1)(x+2)} for [0, 6]?

[-0.5, infinity) and [0.025, infinity) ?

But it's supposed to be (-infinity, -0.5] instead of [-0.5, infinity) so I was wondering how 1/0 gives negative infinity. This is what originally confused me.

Last edited by a moderator: May 5, 2017
8. Feb 12, 2012

### Mentallic

Well no :tongue: but I see we're just going to disregard my example question since you already have the hang of it.

Well for the quadratic (x-1)(x+2) with $x\in [0,1)$, the range is negative. So for the reciprocal function 1/((x-1)(x+2)) it's still going to be negative, thus this is how you should quickly realize that as x approaches 1 from the left side (0.99... etc.), it should be approaching $-\infty$ rather than $+\infty$

9. Feb 12, 2012

### solve

Whoa, this is WAY above my head. $x\in [0,1)$ for one. Have no idea what that's supposed to mean. Can you, please, tell me what that is? Thank You.

10. Feb 12, 2012

### Mentallic

It's nothing special, it just means the values of x in that set (0 to 1, but not including 1). It means the same thing as what Sammy posted in post #2.

11. Feb 12, 2012

### solve

Thank You very much, Mentallic. Will look into it a bit closer.

12. Feb 12, 2012

### SammyS

Staff Emeritus
Re: How do you http://www.physicsforums.cfind [Broken] the range of 1/{(x-1)(x+2)} for [0, 6]?

For number, z, where z is between -2 and 0, what is 1/z? For one thing, isn't 1/z negative if z is negative.

... And if z is a negative number very close to zero, such as z=-0.001, isn't 1/z a very "large" negative number, such as -1000 ?

Last edited by a moderator: May 5, 2017
13. Feb 13, 2012

### solve

Re: How do you http://www.physicsforums.cfind [Broken] the range of 1/{(x-1)(x+2)} for [0, 6]?

Oh, now I understand the flaw in my thinking- I kept imagining a positive 1 between -2 and 0, for some reason. Thank You, SammyS.

Last edited by a moderator: May 5, 2017