How to find inverse of the function $f(x)=\sin(x)\ln(x)$

Question

My friend asked me to solve it, but I can't.

If $f(x)=\sin(x)\ln(x)$, what is $f^{-1}(x)$?

I have no idea how to find the solution. I try to find

$$\frac{dx}{dy}=\frac{1}{\frac{\sin(x)}{x}+\ln(x)\cos(x)}$$

and try to solve it for $x$ by some replacing and other things, but I failed.

Can anyone help? Thanks to all.

Answer 1

The function fails the horizontal line test for one, very badly in fact. One to one states that for any $x$ and $y$ in the domain of the function, that $f(x) = f(y) \Rightarrow x = y$. that is to each point in the domain there exists a unique point in the range. Many functions can be made to be one to one $(1-1)$ by restricting the interval over which values are taken, for example the inverse trig functions and the square root function (any even root). We typically only take the positive square root because otherwise the function would have two answers and each x wouldn't have a unique y value (The so called vertical line test, which is why generally if functions fail the horizontal line test they don't typically have an inverse). If you graph this function it looks very much like a growing sinusoidal shape this cannot be restricted uniquely in a manner that makes an inverse definable.

Answer 2

This function isn't one-to-one, so you can't inverse it.

Answer 3

In fact it has an infinity of solutions for any real value of $y=f(x)$ as may be seen on this picture by Alpha (nothing special happens near $0$ because of the equivalence to $x\ln(x)$) :

You may find solutions numerically by iterations.

For example using Newton-Raphson applied to $\log(x)\sin(x)=y$ we get the first solution illustrated (at the left) in the case $y=1$ : $$x_{n+1}=x_n-\frac{\sin(x_n)\log(x_n)-y}{\cos(x_n)\log(x_n)+\sin(x_n)/x_n},\quad x_0=7$$ $$s_1\doteq 6.83056530451751861265\cdots$$

Answer 4

I think you should also note that generally $f(x)=g(x)\sin(x)$ cannot have an inverse if $g$ itself is not a periodic function. Let's look at your example of $\ln(x)\sin(x)$. (I'm using the sign $<$ to represent greater than or equal to) Since for $h(x)=\sin(x)$, $-1<\sin(x)<1$, then for $f(x)=\ln(x)\sin(x)$ we have $-\ln(x)<\sin(x)<\ln(x)$, and since $\ln(x)$ is increasing monotonically, the sinusoid will keep growing and it will not repeat.