Suppose that $f(x) = x^3 - 1$. (a) Sketch the graph of $f$. (b) Use the graph of $f$ to sketch the graph of $f^{-1}$. (c) Find $f^{-1}$.
a.) The graph of $f(x) = x^3 - 1$ is obtained by shifting the graph of $y = x^3$ one unit downward.
b.) To sketch the graph of $f^{-1}$, we interchange the values of $x$ and $y$.
c.) To find $f^{-1}$, we set $y = f(x)$
$
\begin{equation}
\begin{aligned}
y =& x^3 - 1
&& \text{Solve for $x$, add 1}
\\
\\
x^3 =& y + 1
&& \text{Take the cube root}
\\
\\
x =& \sqrt[3]{y + 1}
&& \text{Interchange $x$ and $y$}
\\
\\
y =& \sqrt[3]{x + 1}
&&
\end{aligned}
\end{equation}
$
Thus, the inverse of $f(x) = x^3 - 1$ is $f^{-1} (x) = \sqrt[3]{x + 1}$
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