Is Sechx Invertible on [0,inf)?

[Jimbo57]

Homework Statement

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

Sechx on [0,inf)

Homework Equations

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

 

[Mark44]

Homework Statement

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

Sechx on [0,inf)

Homework Equations

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 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.

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

Any help would be greatly appreciated!
Jim

 

[STEMucator]

Homework Statement

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

Sechx on [0,inf)

Homework Equations

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

Mark beat me to it.

 

[I like Serena]

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

 

[STEMucator]

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

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.

 

[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.

 

[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?

 

[Dick]

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?

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.

 

[Jimbo57]

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