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 $B=$

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