Question

let
$f(x)=\frac{x + 1}{x + 6}$
$f^{-1}(-8)=square$
question help: video

Explanation:

Step1: Recall the property of inverse - function

If \(y = f(x)\) and \(x = f^{-1}(y)\), then \(f(f^{-1}(y))=y\). So, if \(y=-8\), we need to solve the equation \(f(x)= - 8\) for \(x\), since \(f^{-1}(-8)=x\) when \(f(x)=-8\).
Set \(\frac{x + 1}{x + 6}=-8\).

Step2: Cross - multiply

Multiply both sides of the equation \(\frac{x + 1}{x + 6}=-8\) by \(x + 6\) (assuming \(x
eq - 6\)) to get \(x + 1=-8(x + 6)\).
Expand the right - hand side: \(x + 1=-8x-48\).

Step3: Solve for \(x\)

Add \(8x\) to both sides: \(x+8x + 1=-48\), which simplifies to \(9x+1=-48\).
Subtract 1 from both sides: \(9x=-48 - 1=-49\).
Divide both sides by 9: \(x=-\frac{49}{9}\).

Answer:

\(-\frac{49}{9}\)