Function $f$ has an inverse. The graph of $f^{-1}$ passes through $(3,-6)$. Define $g(x) = 2f(3x) - 5$. Function $g$ also has an inverse. What point must the graph of $g^{-1}$ pass through?
Since $f$ has an inverse, we know that it is one-to-one and onto. Thus, we can write:
y = f(x) <==> x = f^{-1}(y)
Let $(3, -6)$ be a point on the graph of $f^{-1}$. Then we have:
-6 = f^{-1}(3) ⟺3=f(−6)
Now, let's look at the function $g(x) = 2f(3x) - 5$. To find its inverse, we can use the following steps:
\begin{align*} y &= 2f(3x) - 5 \ \frac{y+5}{2} &= f(3x) \ x &= f^{-1}\left(\frac{y+5}{2}\right) \ \frac{x}{3} &= f^{-1}\left(\frac{y+5}{6}\right) \end{align*}
Therefore, the inverse of $g(x)$ is given by:
g^{-1}(x) = 1/3*f^{-1}((x + 5)2/2)
To find the point that the graph of $g^{-1}$ passes through, we need to find a point of the form $(a,b)$ such that $g^{-1}(a) = b$. Let's substitute $a = 3$ and use the fact that $f^{-1}(3) = -6$:
\begin{align*} g^{-1}(3) &= \frac{1}{3}f^{-1}\left(\frac{3+5}{2}\right) \ &= \frac{1}{3}f^{-1}(4) \ &= \frac{1}{3}(f^{-1}(2)+f^{-1}(2)) \ &= \frac{1}{3}(-3-9) \ &= -4 \end{align*}
Therefore, the graph of $g^{-1}$ passes through the point $(3,-4)$.
