I am having trouble with this question. I don't have any calculus background.
I think $f^{-1}$ means the inverse of $f(x)$
Does that mean I have to find the inverse of $f(x)$ first? But again, I am trapped when operating the e because there is also $x^3$ and $x$. I cannot think of a way to isolate them.
On
A function needs to have domain and Range (co-domain) defined. But here these are not given so we can take it as $f:R\to R$, it being increasing as $f'(x)=e^x+3x^2+1>0, \forall x \in R$. it is one to one. As $f(-\infty)=-\infty, f(\infty)=+\infty$, it is onto. Hence, it is a bijection so the inverse exists, but it is not obtainable. Noting that $f(0)=-2$ we can claim that $f^{-1}(-2)=0$. $f^{-1}(e-1)=1$ but we cannot find $f^{-1}(2)$ by hand, though it exist and it is numerically obtainable as $f^{-1}(2)=1.04084...$ similarly one can find $f^{-1}(0)=0.68723$. See the plot of $f(x)$ below
We can just nuzzle around and find some $x$ wifh $f(x)=-2$, then $f^{-1}(-2)=x$. $x=0$ works – and, as it turns out, the function is increasing everywhere so this is the only solution.