Recall the definition of the inverse function $f^{-1} $. $f^{-1}(y)$ is a number; it is the input value $x$ of $f$ such that $f(x)=y$. In other words, $f^{-1}(y)$ is the input value for the function $f$ that produces the output value $y$.
In functional notation: $f(x)=y,\quad \text{ if and only if }\quad f^{-1}(y)=x.$
Let's look at your function, which I assume to be $f(x)=x^3+1$. If you set $y=f(x)$, then you have $\tag{1} y=x^3+1 $
In the above, $x$ is the input value for $f$ that gives the output value $y$. Equation $(1)$ gives a formula for determining the output value $y$ of $f$ when you know what the input value $x$ is. Equation $(1)$ is the rule for the function $f$. For instance if the input were $x=1$, the output would be $y=1^3+1=2$. In functional notation, you would write $f(1)=2$.
The inverse function sort of "reverses things". You specify an output value of $f$, such as $y=2$, and the inverse function tells you what the input to $f$ was. For example, if the output of $f$ was 2, then the input had to have been $1$. In functional notation $f^{-1}(2)=1$.
It's worth repeating: in general if $f(x)=y$, then $x=f^{-1}(y)$.
What you need to do in your problem is to find the rule for $f^{-1}$. That is, you need to find a general formula that tells you how to compute the input $x$ if you know what the output $y$ is. Well, we almost have one in hand: we just need to solve equation $(1)$ for $x$ in terms of $y$.
That is, we need to isolate $x$ on one side of equation $(1)$ and insure that there are "no $x$'s" on the other side. Then on one side $x$ will be by itself, and the other side of the equation will be the rule for $f^{-1}(y)$.
Let's do this: Start with equation ${1}$. $ y=x^3+1. $ We want to get $x$ by itself on one side of the equation and no $x$'s on the other. We may (and do) subtract 1 from both sides of the above equation. This gives: $ y-1=x^3.$ We don't want $x^3$ on the right hand side, we just want $x$. Take the cube root of both sides to obtain: $\root 3\of{ y-1}=x.$ And just to make things pretty, write the above equation in the opposite order: $\tag{2} x = \root 3\of{ y-1} $ This is what we want. If we know $y$, then we can use equation $(2)$ to figure out what $x$ was.
For instance, if $y=28$, what was the input value $x$ for $f$? The formula above tells you: $x=f^{-1}(28)=\root 3\of{28-1}=3$.
More generally formula $(2)$ tells you that $f^{-1}(y) = \root 3\of{ y-1}.$ In more conventional notation (that is, using $x$ to denote the independent variable): $f^{-1}(x) = \root 3\of{ x-1}.$ and we're done.
More generally:
Keep in mind that not all functions have inverses. Only one-to-one functions do.
If $f$ is one-to-one it has an inverse. To find the rule for $f^{-1}(x)$, you can do the following:
- Write down the equation $y=f(x)$ (where $f(x)$ is replaced by the rule you were given ($x^3-1$ in your example).
- Solve the equation for $x$. You want an equation of the form x=\text{stuff with only } y{\text{'s}}.
Write down f^{-1}(y) = \text{stuff with only } y{\text{'s}}.
Replace in the formula all $y$'s with $x$'s.