Domain of a function

When substituting $$x$$ by a real number in the analytical expression of a function, the result is not always another real number.

Let's consider for example the function $$f(x)=\displaystyle \sqrt{x-3}$$. To be able to compute images we need $$x$$ to be greater or equal to zero, since the root of a negative number is not a real number.

Therefore, only the real numbers have images for $$x$$ greater or equal than $$3$$.

The real domain of a function $$f$$ is the set of real numbers that have image for $$f$$. $$Dom$$ is denote $$d(f)$$ or $$D(f)$$.

Find the real domain of the following functions:

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

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

  4. We should note that the image of any real number $$x$$ is another real number. Therefore $$Dom (f) = \mathbb{R}$$
  5. As in the previous case, the image of any real number $$x$$ is another real number. Therefore $$Dom (f) = \mathbb{R}$$
  6. In this case, the image of any real number is another real number except for zero, where the function is not defined. We then have, $$Dom (f) =\mathbb{R} - \{0\}$$

Domains calculation

To calculate the domain of a function we first have to think that any number of the real line $$(\mathbb{R})$$ is possible, and then start restricting the set depending on the function. To do these restrictions we must identify the "weak" points of our functions or rather, the non defined points. Next we list the sets where some main functions are not defined:

Function Set of not definition
$$f(x)=log(g(x))$$ $$\{ x | g(x) \leq 0\} =$$ the values of $$x$$ such that $$g(x)$$ is zero or negative
$$f(x)=\sqrt{g(x)}$$ $$\{x | g(x) < 0 \} =$$ the values of $$x$$ such that $$g (x)$$ is negative
$$f(x)=\dfrac{g(x)}{h(x)}$$ $$\{ x | h(x) = 0 \} =$$ the values of $$x$$ such that $$h (x)$$ is zero
$$f(x)=\sqrt[2n]{g(x)}$$ $$\{ x | g(x) < 0 \} =$$ the values of $$x$$ such that $$g (x)$$ is negative

Let's see an example:

If we take the function $$f(x)=\Big( \dfrac{2x+1}{x-4}-ln(x+8)\Big) \cdot \sqrt{x^2+1}$$ and we want to find its domain, we first must think that it is the whole real line and then restrict it as we identify points or intervals where the function is not defined.

In this case, we observe that we have $$3$$ possible intervals:

  1. when $$x-4$$ is zero $$\Rightarrow x-4=0 \Rightarrow x=4$$ the function is not defined.
  2. when $$x+8$$ is negative or zero $$\Rightarrow x+8\leq 0 \Rightarrow x\leq -8$$ the function is not defined.
  3. when $$x^2+1$$ is negative $$\Rightarrow x^2+1 < 0 \Rightarrow x^2<-1$$ and this cannot happen since $$x$$ square is always positive, therefore the function does not have any problems in this part.

Then, we can conclude that the domain of our function will be $$Dom(f)=(-8,4)\cup(4,\infty)$$