How do I find the inverse function of $f(x)=\sin(x)+2x+1$

Question

$$f(x) = \sin(x) + 2x + 1$$

After setting

$$x = \sin(y) + 2y +1$$

I don't know how to solve for $y$.

For context, the problem is asking to find $g'(x)$, knowing that $g(x)$ is the inverse of $f(x)$.

I thought maybe I could differentiate and find the inverse of $f'(x)$. Would that be $g'(x)$?

Thanks!

Answer 1

The point of this exercise is the Inverse Function Theorem, which allows you to find the derivative of the inverse function at particular points under certain conditions (the link noted above in the comments explains this exactly so I will not state all that). It is particularly useful when you cannot find the inverse $g(x)$ explicitly using algebra as is the case here, and furthermore you should not generally expect to obtain $g'(x)$ as an explicit function.

A simpler example where we can find the inverse explicitly will help illustrate. Consider $f(x)=\frac 1{x-1}$. Then the inverse $g(x)$ is given by $g(x)=\frac 1x +1$. Obviously we have to exclude $x=1$ but this is fine: the IVT says we can operate in a small interval around any $x \ne 1$.

Since $g(f(x)) = x$, the chain rule gives that:

$g'(f(x))f'(x) =1$ or $g'(f(x)) = \frac 1{f'(x)}$.

You can already see that this generally will not allow you to find $g'(x)$ as an explicit function; however this formula does hold at individual points.

Back in the example, we have $f(2)=1$ and $g(1)=2$.

This tells us that $g'(1)$ should be $1/f'(2)$. Since $f'(x) = \frac {-1}{(x-1)^{2}}$, $f'(2)=-1$.

In this case we can check: $g'(x) = -\frac 1{x^2}$ so $g'(1)=-1$.

Note that that $g'(x) \ne \frac 1{f'(x)}$ as functions which is a common misconception.

You should be able to apply this in your example now to get a formula for $g'(f(x))$.

Answer 2

Inverse functions are found by doing all of the steps that were done to the original function in reverse order. For example, to solve 2x + 3 = 11, think about how the function was constructed using PEMDAS. X was multiplied by 2 (using a "getting dressed" analogy, "put on your socks") then 3 was added ("put on your shoes"). To solve it, subtract 3 ("take off your shoes") then divide by 2 ("take off your socks"). So if in the original function, the instructions are "take the sine, then add 2x, then add 1", to find the inverse, subtract 1, then subract 2x, then take the inverse sine/arcsine of this answer. So it should be $f^{-1}(x) = \arcsin(x-1-2x) = \arcsin(-x-1)$. I think.