Question

let
$f(x)=\frac{x + 3}{x + 8}$
$f^{-1}(-8)=$

Explanation:

Step1: Recall inverse - function property

If \(y = f(x)\) and \(x = f^{-1}(y)\), then to find \(f^{-1}(-8)\), we set \(f(x)=-8\). So, \(\frac{x + 3}{x + 8}=-8\).

Step2: Cross - multiply

Multiply both sides by \(x + 8\) (assuming \(x
eq - 8\)) to get \(x + 3=-8(x + 8)\).

Step3: Expand the right - hand side

Expand \(-8(x + 8)\) to \(-8x-64\). So the equation becomes \(x + 3=-8x-64\).

Step4: Solve for \(x\)

Add \(8x\) to both sides: \(x+8x + 3=-64\), which simplifies to \(9x+3=-64\). Then subtract 3 from both sides: \(9x=-64 - 3=-67\). Divide both sides by 9: \(x=-\frac{67}{9}\).

Answer:

\(-\frac{67}{9}\)