Question

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

Explanation:

Step1: Replace \( f(x) \) with \( y \)

We start with the function \( y = \frac{1}{x + 3} \).

Step2: Swap \( x \) and \( y \)

After swapping, we get \( x = \frac{1}{y + 3} \).

Step3: Solve for \( y \)

First, multiply both sides by \( y + 3 \): \( x(y + 3) = 1 \).
Then, expand the left side: \( xy + 3x = 1 \).
Subtract \( 3x \) from both sides: \( xy = 1 - 3x \).
Finally, divide both sides by \( x \) (assuming \( x
eq 0 \)): \( y = \frac{1 - 3x}{x} \), which can also be written as \( y = \frac{1}{x} - 3 \).

Answer:

\( \frac{1}{x} - 3 \) (or \( \frac{1 - 3x}{x} \))