If $g(x)=3+x+e^x$, then how do I find $g^{-1}(4)$?

Question

If $g(x)=3+x+e^x$, then how do I find $g^{-1}(4)$?

I took $g(x)=y$ and tried to solve the problem, but i could not get the solution.So, please help me by providing me the solution to my question.

Answer 1

$g(0)=3+0+e^0=3+0+1=4\iff g^{-1}(4)=0$

And since $g'(x)=0+1+e^x>0\ ,\ \forall x\in\mathbb{R}\iff g(x)$ is strictly increasing on $\mathbb{R}$, hence $x=0$ is the only solution.

Answer 2

I don't really think there is an algorithmic way of doing it.

The guess $$ x = 0 $$

will show you that $$g^{-1}(4) = 0\\ \because \ g(0) = 4 $$

Answer 3

The inverse of your function is given using the product log function $W(z)$:

$$g^{-1} = -W(e^{x - 3}) + x - 3$$

So your best bet is to find a unique $x_0$ such that $g(x_0) = 4$, if one such $x_0$ exists. Then we have that $g^{-1}(4) = x_0$.

We can do that by simply noting that $$g(0) = 3 + 0 + e^0 = 4 $$ Furthermore, $0$ is the only $x_0 \in \mathbb R$ such that $g(x_0) = 4,\,$ since $g$ is a strictly increasing function. Hence, $$ \;g^{-1}(4) = 0$$

Answer 4

$g^{-1}(4)$ is the solution to $g(x)=x+3+e^x=4$

Since $g'(x)=1+e^x>0$,$g(x)=4$ has only real root.

Notice $g(0)=4\implies g^{-1}(4)=0$