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:

  1. $$f (x) = 2x - 1$$
  2. $$f(x)=3x^2$$

  3. $$f(x)=\displaystyle \frac{1}{x}$$

  4. The function can take any real number as an image. Therefore, $$Im (f) =\displaystyle \mathbb{R}$$.

  5. In this case the function only has positive images or $$0$$, since the square of a number cannot be a negative. Therefore $$Im (f) = [0, +\infty)$$

  6. 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, $$Im (f) =\mathbb{R}-{0}$$.