If $f^{-1}(g(x))=x^3-1$ and $g$ has an inverse, find $g^{-1}(f(7))$.
We know that $f^{-1}(u)=v$ is the same as $u=f(v)$.  Therefore $f^{-1}(g(x))=x^3-1$ is the same as  \[g(x)=f(x^3-1).\]We can also use that $g(s)=t$ is equivalent to $s=g^{-1}(t)$ to say \[x=g^{-1}(f(x^3-1)).\]This gives an expression containing $g^{-1}\circ f$.

Therefore, $g^{-1}(f(7))$ is the value of $x$ such that $x^3 - 1 = 7$.  Solving for $x$, we find $x = \boxed{2}$.
2
Level 3
Intermediate Algebra
Problem: If $f^{-1}(g(x))=x^3-1$ and $g$ has an inverse, find $g^{-1}(f(7))$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find \( g^{-1}(f(7)) \), we start with the equation \( f^{-1}(g(x)) = x^3 - 1 \). This implies that \( g(x) = f(x^3 - 1) \) when we apply \( f \) to both sides and use the property of inverse functions.

Next, we need to evaluate \( g^{-1}(f(7)) \). To do this, we first set \( y = g^{-1}(f(7)) \). By the definition of the inverse function, this means \( g(y) = f(7) \).

From our earlier derivation, we know that \( g(y) = f(y^3 - 1) \). Therefore, we can set up the equation:

\[
f(y^3 - 1) = f(7)
\]

Since \( f \) is a function, we can apply the property of injectivity (assuming \( f \) is one-to-one) to conclude that:

\[
y^3 - 1 = 7
\]

Solving for \( y \), we add 1 to both sides:

\[
y^3 = 8
\]

Taking the cube root of both sides gives:

\[
y = 2
\]

Thus, we find that \( g^{-1}(f(7)) = 2 \). The final answer is \( 2 \).<|END_OF_LATENT|> The final answer is $2$.