Question

Let y=f(x)=(3x+7)/(x-2) find a rule for f^-1

Answer 1

Answer:

$f^{-1}(x)=\frac{2x+7}{x-3}$.

Step-by-step explanation:

The given function is $f(x)=\frac{3x+7}{x-2},\:x\ne2$.

We want to find a rule for $f^{-1}(x)$.

Let $y=\frac{3x+7}{x-2}$.

Interchange x and y.

$x=\frac{3y+7}{y-2}$.

Solve for y.

$x(y-2)=3y+7$.

Expand

$xy-2x=3y+7$.

Group y-terms

$xy-3y=7+2x$.

Factor

$(x-3)y=7+2x$.

$y=\frac{2x+7}{x-3}$.

$\therefore f^{-1}(x)=\frac{2x+7}{x-3},\:x\ne3$.