Let
\[
f(x)=\frac{1}{3} x-5, \quad 4 \leq x \leq 9
\]
The domain of $f^{-1}$ is the interval $[A, B]$ where $A=$ and $B=$

Answer

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Answer

$\boxed{[A, B] = \left[-\frac{11}{3}, -2\right]}$

Steps

Step 1 ：Let $f(x)=\frac{1}{3} x-5, \quad 4 \leq x \leq 9$

Step 2 ：The domain of the inverse function $f^{-1}$ is the range of the original function $f$.

Step 3 ：To find the range of $f$, we substitute the endpoints of the domain of $f$ into the function $f$.

Step 4 ：When $x=4$, $f(x)=\frac{1}{3} \times 4 - 5 = -\frac{11}{3}$

Step 5 ：When $x=9$, $f(x)=\frac{1}{3} \times 9 - 5 = -2$

Step 6 ：So, the domain of $f^{-1}$ is the interval $[A, B]$ where $A=-\frac{11}{3}$ and $B=-2$

Step 7 ：$\boxed{[A, B] = \left[-\frac{11}{3}, -2\right]}$