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.
$\bullet (1)$ takes $-1$ and $0$ to the same place, so it isn't injective. I doubt it's surjective, either, because we need a solution to $y=x^2+x$ for any $y$. Take $y=-1$, say. $x^2+x+1$ has no real roots, since the discriminant is negative. Complete the square and you get $y=(x+1/2)^2-1/4$. Thus you can graph and "see" that nothing less than $-1/4$ is hit. It's a parabola (opening upwards), after all.
$\bullet(3)$ isn't surjective: not every natural is a fourth power
$\bullet (4)$ is of course not injective, but surjective: two different sets can have the same $\inf$; there is a set with any natural as $\inf$
Best Answer
You allready received nice answers focused on injectivity and surjectivity. Here a slightly different route.
Can you find a function $g:\mathbb R\rightarrow \mathbb R$ such that $f(g(x))=x$ for each $x\in\mathbb R$?
If you have found such $g$ then check whether it is also true that $g(f(x))=x$ for each $x\in\mathbb R$.
If so then you are ready because you have shown that $f$ is "invertible" (i.e. has an inverse). A function is bijective if and only if it is invertible.
You could also take the opposite route: finding a function $g:\mathbb R\rightarrow \mathbb R$ such that $g(f(x))=x$ for each $x\in\mathbb R$ and checking whether it is also true that $f(g(x))=x$ for each $x\in\mathbb R$.