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What is the domain and range of f(x) = 1/ln(x^2)? watch

1. any help is appreciated
2. (Original post by zezno)
any help is appreciated
Well typically domain should be given...but I assume it just wants you to write which values the function can be defined on. Well, remember that we can't divide by 0, and that lnx is defined for x > 0. As for the range, for what values of k can you solve the equation 1/ln(x^2) = k?
3. (Original post by 13 1 20 8 42)
Well typically domain should be given...but I assume it just wants you to write which values the function can be defined on. Well, remember that we can't divide by 0, and that lnx is defined for x > 0. As for the range, for what values of k can you solve the equation 1/ln(x^2) = k?
the Question comes with the domain x is equal to or greater than 0.

My thought process:

The denominator can't be 0 as you said.

so x^2 > 0

so would the domain be (-infinity, 0), (0, infinity)?

Would the range be (0, infinity)? no idea lol
4. I believe the domain would be x≠1, as that would give ln(1), which equals 0 and would leave f(x) undefined.
I think the range would be f(x) > 0.
5. (Original post by zezno)
the Question comes with the domain x is equal to or greater than 0.

My thought process:

The denominator can't be 0 as you said.

so x^2 > 0

so would the domain be (-infinity, 0), (0, infinity)?

Would the range be (0, infinity)? no idea lol
Well, for one, if the question gives that x >/= 0, then you can't include anything from (-infinity, 0). But otherwise, we have ln(1) = 0, not ln(0) = 0 (though of course ln0 is undefined anyhow so 0 can't be included for that reason..)

Well, the range should contain that interval, but remember ln can be negative as well.
6. (Original post by 13 1 20 8 42)
Well, for one, if the question gives that x >/= 0, then you can't include anything from (-infinity, 0). But otherwise, we have ln(1) = 0, not ln(0) = 0 (though of course ln0 is undefined anyhow so 0 can't be included for that reason..)

Well, the range should contain that interval, but remember ln can be negative as well.
so what would be the domain and range? My head hurts lol
7. (Original post by zezno)
so what would be the domain and range? My head hurts lol
Okay, whatever. So if the question indeed gives x >/=0, you can't have x = 0, or x = 1. Those are your only restrictions.
As for 1/ln(x^2); well, suppose 1/ln(x^2) = k where k is a real number. Then we can solve; we get ln(x^2) = 1/k, so x^2 = e^(1/k), and then x = e^(1/2k). Of course, this doesn't work when k = 0. But otherwise it works, right, be k negative or positive. So that should tell you the range.
8. (Original post by 13 1 20 8 42)
Okay, whatever. So if the question indeed gives x >/=0, you can't have x = 0, or x = 1. Those are your only restrictions.
As for 1/ln(x^2); well, suppose 1/ln(x^2) = k where k is a real number. Then we can solve; we get ln(x^2) = 1/k, so x^2 = e^(1/k), and then x = e^(1/2k). Of course, this doesn't work when k = 0. But otherwise it works, right, be k negative or positive. So that should tell you the range.
ln(x^2) cannot be equal to 0

therefore x^2 =e^0
therefore x^2 = 1
so x = +- root 1
so x cannot be equal to +-1

So I think the domain will be (0, infinity), but x cannot be equal to +-1 ?

please could you tell me if this is correct and what the range is, it's my first time learning this so I'm getting confused a lot
9. (Original post by zezno)
ln(x^2) cannot be equal to 0

therefore x^2 =e^0
therefore x^2 = 1
so x = +- root 1
so x cannot be equal to +-1

So I think the domain will be (0, infinity), but x cannot be equal to +-1 ?

please could you tell me if this is correct and what the range is, it's my first time learning this so I'm getting confused a lot
Yes, if x has to be greater than or equal to zero, then indeed the domain is (0, infinity) but missing out 1 (note that you don't need to say that x =/= -1, because -1 is not in this interval)

As for the range, I mean, you can always solve the equation 1/ln(x^2) = k for k =/= 0 right? So the range is just all real numbers apart from zero.
10. (Original post by 13 1 20 8 42)
Yes, if x has to be greater than or equal to zero, then indeed the domain is (0, infinity) but missing out 1 (note that you don't need to say that x =/= -1, because -1 is not in this interval)

As for the range, I mean, you can always solve the equation 1/ln(x^2) = k for k =/= 0 right? So the range is just all real numbers apart from zero.

I can just write the range as (-infinity, infinity) except for 0 right? as this is writing all real numbers
11. (Original post by zezno)

I can just write the range as (-infinity, infinity) except for 0 right? as this is writing all real numbers
Yeah. The classy way is R \ {0} or (-infinity, infinity) \ {0}
12. (Original post by 13 1 20 8 42)
Yeah. The classy way is R \ {0} or (-infinity, infinity) \ {0}
Now for this question:

x^-1>0
(x-1)(x+1)>0

So would the domain be:

(-infin, -1), (1, infin)

should I also include x=0 in the domain because of the numerator?

also, what would the range be?
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13. (Original post by 13 1 20 8 42)
Yeah. The classy way is R \ {0} or (-infinity, infinity) \ {0}
.

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