Find inverse of a function $f(x)=x^3+3x^2+4x-1$ If $f(x)=x^3+3x^2+4x-1$,  then $f(x)$ is one to one as it is increasing function for $x \in R$. I want to find $x$ such that

$\frac{-1}{4}x-\frac{7}{16}=2f^{-1}(\frac{-x}{2}-1)$

my attempt:
One option is that calculate $f^{-1}$ , inverse of $f$, and then changes $x$ to $\frac{-x}{2}-1$  in $f^{-1}$ and then solve the equation.
I don't know how to find inverse here
Could someone help me with this?
 A: Knowing that an inverse exists tells us that the equation $\frac{-x}{4} - \frac{7}{16} = 2 f^{-1}\bigl(\frac{-x}{2} - 1\bigr)$ is equivalent to
$$
f\Bigl(\frac{-x}{8} - \frac{7}{32}\Bigr) = \frac{-x}{2} - 1.
$$
Or $f(y) = 4 y - \frac{1}{8}$, where $y = \frac{-x}{8} - \frac{7}{32}$, which can be rewritten as
$$\begin{align}
y^3 + 3 y^2 - 7/8 &= 0 \\
8y^3 + 24y^2 - 7 &= 0.
\end{align}$$
We can avoid using the cubic formula by testing, according to the rational root theorem, the numbers $\pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{8}, \pm 7, \pm \frac{7}{2}, \pm \frac{7}{4}, \pm \frac{7}{8}$. This gives $y = 1/2$ as a rational root, and dividing by $y - 1/2$ and using the quadratic formula gives the remaining two roots, $-7/4 \pm \sqrt{21}/4$.
Using the formula $x = -8y - 7/4$ on these three numbers gives the three solutions to our original equation
$$
x = -\frac{23}{4}, \frac{49}{4} \pm 2\sqrt{21}.
$$
A: Hint: Since the range of $f$ is $\Bbb{R}$, it is equivalent to solve
$$\frac{1}{2}f(t) + \frac{1}{16} = 2t$$
Then the desired $x$ is given by $x=-2(f(t)+1)$
