b) $f(x)=\frac{1}{2}(3)^{-2 x-8}-1$
Answer
However, this equation is transcendental and cannot be solved algebraically. We can only solve it numerically or graphically.
Step 1 ：Substitute \(f^{-1}(x)\) into our expression for \(f\) to get \(f(f^{-1}(x))=\frac{1}{2}(3)^{-2f^{-1}(x)-8}-1\).
Step 2 ：Since \(f(f^{-1}(x))=x\) for all \(x\) in the domain of \(f^{-1}\), we have \(x=\frac{1}{2}(3)^{-2f^{-1}(x)-8}-1\).
Step 3 ：Rearrange the equation to find \(f^{-1}(x)\), we get \(f^{-1}(x)=\frac{-\log_3(2x+2)+8}{2}\).
Step 4 ：We want to solve the equation \(f(x) = f^{-1}(x)\), so \(\frac{1}{2}(3)^{-2x-8}-1=\frac{-\log_3(2x+2)+8}{2}\).
Step 5 ：However, this equation is transcendental and cannot be solved algebraically. We can only solve it numerically or graphically.
