Image of a function
Not all the elements of the domain need to belong to the real numbers.
Consider the function $f(x)=\displaystyle \sqrt{x-3}$ where we take the positive solution of the square root, only has positive images.
We will only consider those $x$ in the domain that belong to the real number, in which case all the real numbers that are greater than or equal to $0$.
We will call the image of a function $f$ the set of real numbers that are an image of $f$ of the elements in its real domain. It will be denoted by $Im (f)$.
Therefore the image of the function $f(x)=\sqrt{x-3}$ is $Im (f) = [0, +\infty)$
Find the image of the following functions:
$f (x) = 2x - 1$
$f(x)=3x^2$
$f(x)=\displaystyle \frac{1}{x}$
The function can take any real number as an image. Therefore, $Im (f) =\displaystyle \mathbb{R}$.
In this case the function only has positive images or $0$, since the square of a number cannot be a negative. Therefore $Im (f) = [0, +\infty)$
Finally, the function can take any real value except $0$, since it $f(x)=\displaystyle \frac{1}{x}$ only when $x=0$ we have that the image is not defined in $\mathbb{R}$.
Therefore, $Im (f) =\mathbb{R}-{0}$.