The way to verify something like that is to check the definitions one by one and see if $g(x)$ satisfies the needed properties.
Recall that $F\colon A\to B$ is a bijection if and only if $F$ is:
- injective: $F(x)=F(y)\implies x=y$, and
- surjective: for all $b\in B$ there is some $a\in A$ such that $F(a)=b$.
Assuming that $R$ stands for the real numbers, we check.
Is $g$ injective?
Take $x,y\in R$ and assume that $g(x)=g(y)$. Therefore $2f(x)+3=2f(y)+3$. We can cancel out the $3$ and divide by $2$, then we get $f(x)=f(y)$. Since $f$ is a bijection, then it is injective, and we have that $x=y$.
Is $g$ surjective?
Take some $y\in R$, we want to show that $y=g(x)$ that is, $y=2f(x)+3$. Subtract $3$ and divide by $2$, again we have $\frac{y-3}2=f(x)$. As before, if $f$ was surjective then we are about done, simply denote $w=\frac{y-3}2$, since $f$ is surjective there is some $x$ such that $f(x)=w$. Show now that $g(x)=y$ as wanted.
Alternatively, you can use theorems. What sort of theorems? The composition of bijections is a bijection. If $f$ is a bijection, show that $h_1(x)=2x$ is a bijection, and show that $h_2(x)=x+2$ is also a bijection. Now we have that $g=h_2\circ h_1\circ f$ and is therefore a bijection.
Of course this is again under the assumption that $f$ is a bijection.
If for a given $x\in X$ there is no $y\in Y$ s.t. $(x,y) \in f$ then $x$ isn't in the domain of $f$. At that point saying $f:X\rightarrow Y$ is somewhat misleading. The domain of $f(x)=\frac{1}{x}$ is $\mathbb R\backslash \{0\}$.
Best Answer
If you have $f: X \to Y$, $g : Y \to X$ with $g(f(x)) = x$ for all $x \in X$, it is still possible for $f$ to not be bijective. However, $f$ will be a bijection onto its image; i.e., $f$ is a bijection from $X$ to $f(X)$. In other words, $f$ is injective.
If you additionally require that $f(g(y)) = y$ for all $y \in Y$ (i.e. $g$ is a "two-sided inverse"), then $f$ is a bijection from $X$ to $Y$.