finding intervals on which f is a continuous inverse

Question

I'm having trouble wrapping my head around this problem. I'm given a function f(x) - x + sinx and told to find all the intervals on which f has a continuous inverse. I honestly really have no idea where to start this problem, and it probably stems from my lack of understanding of inverses other than the fact that an inverse function's domain is the range of the original function. Even just a good stating point would be very helpful. Thanks!

Answer 1

Where to Start

I always start with a pictorial representation of the problem. In this case a graph of the existing function is straightforward; not what your looking for, but when you are stuck just start with what you know even if it seems obvious.

Correction

I assume that your function is

f(x) = x + sin (x)

and not

f(x) - x + sin (x)

... right?

Graphing the Given

If you don't understand why the graph of f(x) = x + sin(x) is what these images show, then you might try looking at the components that comprise your given function. Simply, [f(x) = x][2] and [f(x) = sin(x). Those are probably familiar; hopefully the graphics are intuitive hy they combine the way that they do.

hxxp://www.wolframalpha.com/input/?i=x hxxp://www.wolframalpha.com/input/?i=sin+x

Can I Give You the Answer

I'm not sure what the rules are here regarding me just flat out answering the whole problem. I'm going to walk all the way through it. Hopefully, the steps I describe (although they seem dumb) are truly the key to get yourself unstuck.... just start with what you know and make pictures or some alternate representation of the problem that makes sense to you (a table of values, a plot/graph, the answer from your neighbors worksheet... NO not that last one, just seeing if you are paying attention).

What's an Inverse

I prefer to think of the inverse of a function as simply that function raised to the exponent -1.

The inverse of f(x) is f(x)^(-1)

Is that notation familiar to you? Do you remember what negative exponents do? Do you recall that negative exponents can be written as 1 over the function raised to the positive value of that same exponent. For example.

g(x)^(-3) is the same as 1 /( g(x)^3 )

and

f(x)^(-1) is the same as 1 / ( f(x)^1 ) or simply 1/f(x)

Make sense?

Take a step back

Have you ever looked at the graph of [the inverse of sin(x)]? You can see from the plots there that there are asymptotes at the points where the inverse of sin(x) becomes undefined, so 1/sin(x) continuous in between those undefined values.

Is that making sense? I hope so. We are not quite there yet, but hopefully you are seeing how to walk yourself out of "being stuck" but incrementally adding to what you know.

hxxp://www.wolframalpha.com/input/?i=sin%28x%29%5E-1

Before we finish this problem off; there are two important concepts that I want to highlight. The first is identifying boundry conditions (edge cases) for functions and writing them out in tabular form. The second concept you should know is the unit circle.

I can't draw the unit circle for you here, but I will make a quick table of values going around in steps of PI / 2 rad. (90 deg).

| rad | deg | x | y | cos (x) | sin (x) | | 0 | 0/360 | 1 | 0 | 1 | 0 | | PI/2 | 90 | 0 | 1 | 0 | 1 | | PI | 180 | 1 | 0 | 1 | 0 | | 3PI/4 | 270 | 0 | 1 | 0 | 1 |

Now consider our example

f(x) = 1 / sin(x)

Using the table above we can quickly make a table that identifies the edge cases (places where our function will become undefined) and write them in a table of values.

For example, when x = 0 then sin(x)=0 and 1 / sin(x) = 1 / 0 which is undefined! Got it? If you want you can find more values and add to the rows in the table below. The table of values should agree with the graph/plot you have made for the inverse of sin(x) (i.e. 1 / sin(x)).

| x | y= 1 / sin(x) | | 0 | undefined | | 1 | 1 |

Putting it all togeter

We are asked to find: * the inverse of f(x) = x + sin(x) * and define a range over which it is continuous.

We already done the first part actually, but this time we will substitute the right side of the equation into our inverse. So...

The inverse of f(x) is 1 / f(x)

That's simple. Let me add some additional notation that will help as you deal with more and more complex functions in the future. I'm just going to add some parentheses. They aren't required yet, but will make the following step a little more clear / definate.

The inverse 1 / f(x) can also be written 1 / ( f(x) )

and now the substitution

1 / ( f(x) ); substituting for f(x) equals 1 / (x + sin (x))

There; the parentheses clearly show that the quatity (x+sin(x)) is all in the denominator

THIS WOULD BE INCORRECT: 1 / x + sin(x)

The parenthesis ARE required once the substitution is written in.

Draw more Pitures

Plot and make a table of values (trying to identify edge cases) in the table of the our inverse function. To be clear;

Plot and make a table of edge cases for f(x)^(-1) = 1 / (x + sin(x))

Much like the the inverse of sin(x) your plot should show points where f(x) becomes undefined. Now you should be able to answer the final part; identify a region where f(x) is continuous.

Bonus

I'm going to predict that you teacher is going to give you almost this exact same problem on your test; there will just be one minor modification. The function will be:

g(x) = x - sin(x)

instead of

f(x) = x + sin(x)

Do you see it? It is not the "f(x) vs g(x)". The important change is "x MINUS sin(x)" instead of plus. The reason this is significant it that it creates new edge cases that didn't exist before; namely,

the points where

x = sin(x) 

and thus

x - sin = 0 

creating additional points in the domain where the function becomes undefined (can't divide by zero). I hope you caught that. I promise you that you will see some variation of this on your next test and/or final.

Answer 2

(I am assuming that $f(x)=x+\sin x$.)

Since $f^{\prime}(x)=1+\cos x\ge0$ for all x,

and $f^{\prime}(x)=0$ when $\cos x=-1$, so $x=(2n+1)\pi$ for some $n\in\mathbb{N}$,

$f$ is increasing and therefore 1-1 on all of $\mathbb{R}$.

Therefore if I is any interval in $\mathbb{R}$, $f$ has a continuous inverse on I.