Finding the inverse of a function/algebraic manipulation

Question

Sorry in advance if this is a little basic. It's just some simple algebra, but I'm repeatedly getting an answer that differs from the one in my textbook. I can see that the answer is incorrect, but I can't see where I'm going wrong in my working - even though it must be something pretty elementary. I have that:

$$f(x) = \frac{x-3}{x+2}$$

i.e.

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

So:

\begin{align*} y(x+2) & = x - 3\\ y(x+2) + 3 & = x \end{align*}

Multiplying those brackets out, I get:

$$x = yx + 2y + 3$$

And it should follow that:

$$f^{-1}(x) = x^2 + 2x + 3$$

But clearly $f^{-1}(f(x)) \neq x$.

The answer in my textbook (which I know to be correct) is $f^{-1}(x) = \frac{3+2x}{1-x}$. I can follow how they've arrived at this answer, but I can't see what I did incorrectly. Can someone point out my error?

Answer 1

Let's start from here:

You concluded that: $x = yx + 2y + 3$, you should continue simplification:

$x = yx + 2y + 3 \to x-xy=2y+3 \to x(1-y)=2y+3 \to x=\dfrac{2y+3}{1-y}$

$\to f^{-1}(x)=\dfrac{2x+3}{1-x}$

Answer 2

Note that

$$y=\frac{x-3}{x+2}\iff xy+2y-x+3=0\iff x(y-1)=-2y-3\iff x=\frac{-2y-3}{y-1}$$

and thus

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

Your mistake was in the following step

$$x = y\color{red}x + 2y + 3 \not \Rightarrow f^{-1}(x) = x^2 + 2x + 3$$

since you still had an $x$ term on the RHS.

Answer 3

HINT: At this point $x = yx + 2y + 3$, we have $$x-yx = 2y+3 \implies x(1-y) = 2y+3 \implies x = \frac{2y+3}{1-y}$$

Answer 4

How did you go from $$x = yx + 2y + 3$$ to

$$f^{−1}(x) = x^2 + 2x + 3?$$

There should not be an $x^2$.

Note that from $$x = yx + 2y + 3$$ we get $$x(1-y)=2y+3$$

Thus $$x= \frac {2y+3}{1-y}$$ which implies $$ f^{-1}(x)=\frac {2x+3}{1-x}$$

Answer 5

$$f(x) = \frac{x-3}{x+2} = \left(1 - \frac{5}{2 + x}\right) =y$$ So: $$f(x)=y = \left(1 - \frac{5}{2 + x}\right)\implies f^{-1}(y) = \frac{5}{1-y}-2 = x$$
For more detail: \begin{align} y = 1 - \frac{5}{2 + x}\\ y + 1 = - \frac{5}{2 + x}\\ 1- y = \frac{5}{2 + x}\\ (2 + x)(1- y) = 5\\ 2 + x = \frac{5}{(1- y)}\\ x = \frac{5}{(1- y)}-2\\ \end{align} \begin{align} \tag*{$\Box$} \end{align}