A function is bijective if it is injective (one-to-one) and surjective (onto).
You can show $f$ is injective by showing that $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$.
You can show $f$ is surjective by showing that for each $y \in \mathbb{R} - \{2\}$, there exists $x \in \mathbb{R} - \{-1\}$ such that $f(x) = y$.
If $f(x_1) = f(x_2)$, then
\begin{align*}
\frac{4x_1 + 3}{2x_1 + 2} & = \frac{4x_2 + 3}{2x_2 + 3}\\
(4x_1 + 3)(2x_2 + 2) & = (2x_1 + 2)(4x_2 + 3)\\
8x_1x_2 + 8x_1 + 6x_2 + 6 & = 8x_1x_2 + 6x_1 + 8x_2 + 6\\
8x_1 + 6x_2 & = 6x_1 + 8x_2\\
2x_1 & = 2x_2\\
x_1 & = x_2
\end{align*}
Thus, $f$ is injective.
Let $y \in \mathbb{R} - \{2\}$. We must show that there exists $x \in \mathbb{R} - \{-1\}$ such that $y = f(x)$. Suppose
$$y = \frac{4x + 3}{2x + 2}$$
Solving for $x$ yields
\begin{align*}
(2x + 2)y & = 4x + 3\\
2xy + 2y & = 4x + 3\\
2xy - 4x & = 3 - 2y\\
(2y - 4)x & = 3 - 2y\\
x & = \frac{3 - 2y}{2y - 4}
\end{align*}
which is defined for each $y \in \mathbb{R} - \{2\}$. Moreover, $x \in \mathbb{R} - \{-1\}$. To see this, suppose that
$$-1 = \frac{3 - 2y}{2y - 4}$$
Then
\begin{align*}
-2y + 4 & = 3 - 2y\\
4 & = 3
\end{align*}
which is a contradiction.
The inverse function is found by interchanging the roles of $x$ and $y$. Hence, the inverse is
$$y = \frac{3 - 2x}{2x - 4}$$
To verify the function
$$g(x) = \frac{3 - 2x}{2x - 4}$$
is the inverse, you must demonstrate that
\begin{align*}
(g \circ f)(x) & = x && \text{for each $x \in \mathbb{R} - \{-1\}$}\\
(f \circ g)(x) & = x && \text{for each $x \in \mathbb{R} - \{2\}$}
\end{align*}
\begin{align*}
(g \circ f)(x) & = g\left(\frac{4x + 3}{2x + 2}\right)\\
& = \frac{3 - 2\left(\dfrac{4x + 3}{2x + 2}\right)}{2\left(\dfrac{4x + 3}{2x + 2}\right) - 4}\\
& = \frac{3(2x + 2) - 2(4x + 3)}{2(4x + 3) - 4(2x + 2)}\\
& = \frac{6x + 6 - 8x - 6}{8x + 6 - 8x - 8}\\
& = \frac{-2x}{-2}\\
& = x\\
(f \circ g)(x) & = f\left(\frac{3 - 2x}{2x - 4}\right)\\
& = \frac{4\left(\dfrac{3 - 2x}{2x - 4}\right) + 3}{2\left(\dfrac{3 - 2x}{2x - 4}\right) + 2}\\
& = \frac{4(3 - 2x) + 3(2x - 4)}{2(3 - 2x) + 2(2x - 4)}\\
& = \frac{12 - 8x + 6x - 12}{6 - 4x + 4x - 8}\\
& = \frac{-2x}{-2}\\
& = x
\end{align*}
Hence, $g = f^{-1}$, as claimed.
Here's another idea,
$$(-\infty,0]\cap[0,\infty)=\{0\}$$
What can you say about the cardinality of $(-\infty,0]$ and $[0,\infty)$?
Best Answer
There is an overloading here, of the symbol $f^{-1}$.
If $R$ is a binary relation, then $R^{-1}=\{(a,b)\mid (b,a)\in R\}$, and when $f$ is an injective function we can show that $f^{-1}$ is an injective function as well (here a function is just a set of ordered pairs with a particular property). In the case we call $f^{-1}$ the inverse function.
But if $f$ is not an injective function then $f^{-1}$ is not a function, it is a binary relation. In that case we use $f^{-1}$ to denote the preimage function, which is a function mapping subsets of the range (or codomain) to subsets of the domain. Namely, $f^{-1}(A)=\{x\in X\mid f(x)\in A\}$.
Note, that if $f$ is indeed injective, then for every singleton, $\{b\}$ its preimage is at most a $\{a\}$ for some $a$ in the domain.
You are asked to find the preimage, which is the set of all values being mapped into the given set. So writing $g^{-1}(\{-1,0,1\})=\{-1,0,1\}$ is blatantly wrong, since there are many numbers being mapped to those three values, not just the three.