# Is it Invertible?

1. May 20, 2013

### Jimbo57

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

Indicate whether the function is invertible on the given interval, explain.

Sechx on [0,inf)

2. Relevant equations

3. The attempt at a solution

So, I know a function is invertible if it's one-to-one. I can figure that out by the horizontal line test or by the first derivative test (increasing or decreasing on the interval). I know that by the first derivative test that this function has an absolute max at f(0) and is decreasing. I thought I read somewhere that a one-to-one function has to be increasing or decreasing on the interval, that f'(x) < or > 0 for all x on the interval, and that it can't have any extrema. Is this true?

The interval is the only part throwing me off here, when x=0.

Any help would be greatly appreciated!
Jim

2. May 20, 2013

### Staff: Mentor

The last part isn't true. If the domain is some proper subset of the real line, as it is in your problem, the maximum occurs at x = 0, and the maximum value is 1.

Possibly what you're thinking of is the situation where a maximum occurs at an interior point of the domain, such as, for example, y = 1 - x2 on the restricted domain [-1, 2]. The maximum occurs at (0, 1). Since y' > 0 for x $\in$ [-1, 0) and y' < 0 for x $\in$ (0, 2], this function doesn't have an inverse.

3. May 20, 2013

### Zondrina

Mark beat me to it.

4. May 20, 2013

### I like Serena

Erm... all true... except that sech(x) does have an inverse on [0,∞) ...

5. May 20, 2013

### Zondrina

You've shown that $sech(x)$ is monotone decreasing on [0,∞) by applying the first derivative test.

This tells you that $sech(x)$ is indeed one to one.

6. May 20, 2013

### I like Serena

Actually, it has to be strictly decreasing with the possible exception of a countable number of points, which it is.
That is, it is not allowed to be constant in some interval.

7. May 21, 2013

### Jimbo57

Thanks a lot for the input but I seem to be getting two very different and very confident answers guys. Unless I'm misunderstanding something, I still feel like I'm on the fence. So, since the maxima is at the end of the interval, is it still invertible, or does that make it not invertible?

8. May 21, 2013

### Dick

If a function is strictly decreasing over an interval, then it's invertible. Having a derivative being zero at a single point doesn't make it not strictly decreasing.

Last edited: May 21, 2013
9. May 21, 2013

### Jimbo57

Your second sentence cleared up any confusion I had with regards to extrema Dick. Thanks all!

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