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

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Homework Help Overview

The discussion revolves around finding the range of the function 1/{(x-1)(x+2)} over the interval [0, 6]. Participants are exploring the implications of the function's domain and the behavior of its reciprocal.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the domain of the function and its implications on the range, noting invalid operations at certain points. There are attempts to find the range of the quadratic (x-1)(x+2) as a step towards understanding the reciprocal function. Questions arise about how the behavior of the quadratic affects the reciprocal, particularly near critical points.

Discussion Status

Some participants have provided insights into the relationship between the ranges of the quadratic and its reciprocal. Others are seeking clarification on specific concepts and the implications of negative values in the context of the reciprocal function. The discussion is ongoing, with various interpretations being explored.

Contextual Notes

Participants are navigating the complexities of the function's behavior near points of discontinuity and are questioning their assumptions about the ranges involved. There is an acknowledgment of confusion regarding the implications of negative values and the behavior of the function as it approaches certain limits.

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Homework Statement



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

Homework Equations


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.
 
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solve said:

Homework Statement



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

Homework Equations


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.
Can you find the range of (x-1)(x+2) for [0, 1) ∪ (1, 6] ?
 
SammyS said:
Can you find the range of (x-1)(x+2) for [0, 1) ∪ (1, 6] ?

Hopefully I am right: [-2,0) and (0,40]
 
solve said:
Hopefully I am right: [-2,0) and (0,40]
Yes, that's correct.

Do you see how that's related to the range of 1/((x-1)(x+2)) ?
 
SammyS said:
Yes, that's correct.

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


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

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


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

[-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:
solve said:
Mentallic said:
If the values of a variable k range between 2 to 5, then what are the possible values of 1/k?
[-0.5, infinity) and [0.025, infinity) ?
Well no :-p but I see we're just going to disregard my example question since you already have the hang of it.

solve said:
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.

Well for the quadratic (x-1)(x+2) with [itex]x\in [0,1)[/itex], 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 [itex]-\infty[/itex] rather than [itex]+\infty[/itex]
 
Mentallic said:
Well for the quadratic (x-1)(x+2) with [itex]x\in [0,1)[/itex], 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 [itex]-\infty[/itex] rather than [itex]+\infty[/itex]

Whoa, this is WAY above my head. [itex]x\in [0,1)[/itex] for one. Have no idea what that's supposed to mean. Can you, please, tell me what that is? Thank You.
 
  • #10
solve said:
Whoa, this is WAY above my head. [itex]x\in [0,1)[/itex] for one. Have no idea what that's supposed to mean. Can you, please, tell me what that is? Thank You.

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
Mentallic said:
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.

Thank You very much, Mentallic. Will look into it a bit closer.
 
  • #12


solve said:
[-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.
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:
  • #13


SammyS said:
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 ?

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:

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