Please tell me know if my answer is right and whether the steps are correct? Thanks.
$f(x) = \frac{4x − 1}{2x + 3}$
Step 1: Write $y=f(x)$
$y=\frac{4x-1}{2x+3}$
Step 2: Solve this equation for $x$ in terms of $y$ (if possible)
2(a) Multiply both sides by $2x+3$
$(2x+3)\cdot(y)\ =\frac{4x-1}{2x+3}\cdot(2x+3)$
2(b) Distribute y term
$2xy+3y = 4x-1$
2(c) Isolate $x$ and $y$ terms
$2xy + 3y = 4x – 1 $
$-2xy+1 = -2xy$
$3y+1 = 4x-2xy$
$3y+1 = x (4-2y)$
2(d) Divide both sides by $(4-2y)$
$x=\frac{3y+1}{4-2y}$
$f^{-1}=\frac{3x+1}{4-2x} $
Best Answer
$y=\dfrac{4x-1}{2x+3}$
swap $ x,y $
$x=\dfrac{4y-1}{2y+3}$
Solve back $y$ in terms of $x$
$y=\dfrac{3x+1}{-2x+4}, $ done.
EDIT1:
It is an interesting bi-linear or fractional linear function. Notice that coefficients in the left diagonal got swapped and signs of right diagonal elements changed,
$$ \dfrac{a x + b }{c x + d} \rightarrow \dfrac{d x - b }{-c x + a} $$
leaving $ (a d - b c) $ unaltered.