Question

find (h^{-1}(x)), given the function (h(x) = \frac{2x-4}{x+3}).

\\(h^{-1}(x) = \frac{x+3}{2x-4}\\)

\\(h^{-1}(x) = \frac{x-3}{2x-4}\\)

\\(h^{-1}(x) = \frac{3x+4}{x-2}\\)

\\(h^{-1}(x) = \frac{-3x-4}{x-2}\\)

Explanation:

⚡ Using what you learned: inverse functions: definition and notation

Step 1: Swap variables

Set \( y = h(x) \) and swap \( x \) and \( y \):

$$ y = \frac{2x-4}{x+3} \implies x = \frac{2y-4}{y+3} $$

Step 2: Solve for y

Multiply both sides by \( y+3 \):

$$ x(y+3) = 2y-4 $$

$$ xy + 3x = 2y-4 $$

Group all terms containing \( y \) on one side:

$$ xy - 2y = -3x - 4 $$

Factor out \( y \):

$$ y(x-2) = -3x - 4 $$

Divide by \( x-2 \):

$$ y = \frac{-3x-4}{x-2} $$

Thus, the inverse function is:

$$ h^{-1}(x) = \frac{-3x-4}{x-2} $$

Answer:

\( h^{-1}(x) = \frac{-3x-4}{x-2} \)