Question

find the inverse of the function $f(x)=\frac{7x - 4}{x + 3}$.
(1 point)
\bigcirc this function is not one - to - one and therefore has no inverse.
\bigcirc $f^{-1}(x)=\frac{-3x - 4}{x - 7}$
\bigcirc $f^{-1}(x)=\frac{x + 3}{7x - 4}$
\bigcirc $f^{-1}(x)=\frac{3x + 4}{x + 7}$

Explanation:

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

$y = \frac{7x - 4}{x + 3}$

Step2: Swap $x$ and $y$

$x = \frac{7y - 4}{y + 3}$

Step3: Multiply both sides by $y+3$

$x(y + 3) = 7y - 4$

Step4: Expand left-hand side

$xy + 3x = 7y - 4$

Step5: Group $y$-terms on one side

$xy - 7y = -3x - 4$

Step6: Factor out $y$

$y(x - 7) = -3x - 4$

Step7: Solve for $y$

$y = \frac{-3x - 4}{x - 7}$

Step8: Replace $y$ with $f^{-1}(x)$

$f^{-1}(x) = \frac{-3x - 4}{x - 7}$

Answer:

$\boldsymbol{f^{-1}(x)=\frac{-3x - 4}{x - 7}}$ (matches the second option)