Why this function has an inverse function

[Tevion]

Ok, i need to explain why this function has an inverse function. fr^-1

[math]fr(x) = 1.5/sinx[/math]

and state its image set & domain.

Is that (0,1/2Pi]

Thanks :)

 

[Mark44]

You realize that 1.5/sin(x) = 1.5csc(x), right?

On what interval(s) is 1.5 csc(x) one-to-one? There are a bunch, so pick one. If you haven't already graphed y = 1.5 csc(x), you should do that now, since it will give you a good start on determining domain and range for the inverse.

 

[Tevion]

here we are plotted:

so how do we get the image set and domain

 

[Tevion]

how does this explain that is has an inverse function

 

[Mark44]

You should increase the size of the interval to, say, -pi to pi. You're getting only a small part of the graph of y = 1.5 cscx.

When you find an interval for which the graph is one-to-one, then the domain for your inverse will be the range of y = 1.5 csc x, and the range for the inverse will be the domain for y = 1.5 csc x.

 

[Mark44]

how does this explain that is has an inverse function

If you have a function that is strictly increasing or strictly decreasing, it is one-to-one. Every one-to-one function has an inverse.

What you're trying to do is to find an interval for which y = 1.5 csc(x) is one-to-one, and on which there is an inverse for this function.

 

[Tevion]

ok Thanks - so how do i go about do that? basically from my attachment, i need to explain why fr has fr^-1, stating the domain and image set a hand would be great with it please

 

[Mark44]

I'll say it again:
If you have a function that is strictly increasing or strictly decreasing, it is one-to-one. Every one-to-one function has an inverse.

What you're trying to do is to find an interval for which y = 1.5 csc(x) is one-to-one, and on which there is an inverse for this function.

 

[Tevion]

i hear what you're saying bud, but any advice how to take that further? I don't have a clue :(

 

[Mark44]

You have a graph of y = 1.5 csc x. See post #5 and what I said about using a larger interval. Can you look at a graph and tell whether a function is one-to-one, either overall or over some part of its domain?

 

[HallsofIvy]

You were told in your problem that the domain of fr is $(0, \pi/2]$. Do you see why 0 is not included? For x arbitrarily close to 0, fr(x) is arbitrarily large. What is $fr(\pi/2)$? What is the range of fr?

Taking the inverse of a function "swaps" domain and range.

 

[Tevion]

the range is the function is 0 < x < 1/2pi

Thanks for helping

 

[Mark44]

The principal domain of fr(x) = 1.5*csc(x) is [-pi/2, 0) U (0, pi/2]. The range of this function is (-infinity, -1.5) U (1.5, infinity). In the original post of this thread, you guessed that the domain of fr was (0, pi/2], but the domain is actually quite a bit larger.

 

[Tevion]

so do we make it?

for the inverse the image set (range)

(0,1/2pi]

for the domain

(1.5, infinity)

 

[Mark44]

Yes, but those domain/range sets are only half of the story. The inverse function is defined at, for example, -2.

Did you ever get a better graph of y = 1.5csc(x)? The one you had in post #3 was too limited.