Thread: Inverse of an absolute value

I've whittled it down to:
y = (e^x+1)/2
and y = (-e^x-1)/2
For positive and negative takes on the absolute value, but which do I choose?
What's more, as it says "state the rule", should I present my answer as f^-1-infinity,a), etc? Presenting the entire rule like they do in the question?

Oh, and also, in the previous question we worked out a = 1/2.

Originally Posted by Naur

I've whittled it down to:
y = (e^x+1)/2
and y = (-e^x+1)/2
For positive and negative takes on the absolute value, but which do I choose?
What's more, as it says "state the rule", should I present my answer as f^-1-infinity,a), etc? Presenting the entire rule like they do in the question?

Oh, and also, in the previous question we worked out a = 1/2.

i suppose the base is "e", i couldn't see it that well.

anyway, take both. you can define a piece-wise function

Oh yes, the base is e. My mistake.
And you're right, that's a mistake, should be a positive.

Here's the answer as it's given. They have a way of defining which one to use. I guess it kind of makes sense, when y is negative it's on the negative side of the y-axis, which fits the domain, so it must be that one.
Is that it though?
It can be hard to accurately judge the rule they've used when all they show is the maths involved.

Originally Posted by Naur

Oh yes, the base is e. My mistake.
And you're right, that's a mistake, should be a positive.

Here's the answer as it's given. They have a way of defining which one to use. I guess it kind of makes sense, when y is negative it's on the negative side of the y-axis, which fits the domain, so it must be that one.
Is that it though?
It can be hard to accurately judge the rule they've used when all they show is the maths involved.

ok, they gave you a range. yes, then that answer is correct. you have to make sure the domains match up. since x < 1/2 for the function, you need y < 1/2 for the inverse function

Ah okay. I'll try to remember that, thanks very much