Injective, exhaustive and bijective functions
In the graph of a function we can observe certain characteristics of the functions that give us information about its behaviour.
Observe the graphs of the functions $f(x)=x^2$ and $g(x)=2x$
For the function $f$, we observe that we can trace at least one horizontal straight line ($y$ = constant) that the cuts the graph in more than one point.
For example, if we consider the horizontal straight line $y = 4$, we see that there are two points in the domain of $f$, $x = 2$, and $x = -2$, with have the same image $f (x) = 4$.
On the other hand, any horizontal straight line traced on the graph of the function $g$ will cut its graph only once. Therefore, there aren't any two elements in the domain of $g$ that have the same image.
A function $f$ is injective if every two different elements in its domain have two different images, that is, if it is satisfied that: $$x_1\neq x_2 \Longrightarrow f(x_1)\neq f(x_2)$$
Therefore we observe that the function $g$ is injective while $f$ is not.
If we consider again the functions $f$ and $g$, we observe that: The graph of the function $f$ lies on the positive side of the $y$-axis, thus $Im(f)=[0,+\infty)$.
On the other hand, the graph of the function $g$ is on all the real numbers, thus $Im(g)=\mathbb{R}$.
A function $f$ is exhaustive if its graph coincides with the set of the real numbers, that is, if we have that:$$Im (f)=\mathbb{R}$$
We have therefore that the function $f$ is not exhaustive and that the function $g$ is exhaustive.
Finally
A function is bijective if it is injective and exhaustive simultaneously.
This way the function $g$ of the example is bijective while the function $f$ is not.
Determine if the function $f$ represented in the following figure is injective, exhaustive or bijective:
First, we should realize that the function is not injective since we can trace the straight line $y = 1$, cutting the graph in more than one point. This means that different values of the independent variable $x$ have the same image.
On the other hand, the function $f$ is exhaustive since all the real numbers are in its image, that is $Im (f)=\mathbb{R}$.
Obviously the function will not be bijective since it is not injective.