### Theory:

In this topic, we are going to consider the function as real valued function.
The domain of a function is the set of all possible input values that produce some output value range.
While finding the domain, do remember the following things:
1. The denominator (bottom) of a fraction cannot be zero.
2. The number under a square root sign must be non-negative.
Example:
Consider the following real values function and find its domain.

1. $$f(x) =$$ $$1-x$$

This function takes all possible real values of $$x$$. Thus, the domain of the function $$f$$ is the set of all real numbers $$\mathbb{R}$$.

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

This function is in fraction form.

So denominator of a fraction cannot be zero. That is, $$x$$ cannot be $$0$$.

The function takes all possible real values of $$x$$ except $$0$$. Thus, the domain of the function $$f$$ is the set of all real numbers $$\mathbb{R} - \{0\}$$.

3. $$f(x) =$$ $$\frac{1}{x+10}$$.

This function is in fraction form.

So denominator of a fraction cannot be zero. That is, $$x+10$$ cannot be $$0$$.

Suppose $$x+10 = 0$$.

Then $$x = -10$$.

Therefore, $$x$$ cannot be $$-10$$.

The function takes all possible real values of $$x$$ except $$-10$$. Thus, the domain of the function $$f$$ is the set of all real numbers $$\mathbb{R} - \{-10\}$$.

4. $$f(x) =$$ $$\frac{3x+2}{x^2+5x+6}$$.

This function is in fraction form.

So denominator of a fraction cannot be zero. That is, $$x^2+5x+6$$ cannot be $$0$$.

Suppose $$x^2+5x+6 = 0$$.

Factorising the equaiton, we have $$x = -2, x = -3$$.

Therefore, $$x$$ cannot be $$-2, -3$$.

The function takes all possible real values of $$x$$ except $$-2$$ and $$-3$$. Thus, the domain of the function $$f$$ is the set of all real numbers $$\mathbb{R} - \{-2,-3\}$$.

5. $$f(x) =$$ $$\sqrt{x^2-9}$$.

This function is in square root form.

So the number under square root should not be negative. That is, $$x^2-9$$ cannot be negative.

This results in negative only if the value of $$x$$ is greater than $$3$$ or less than $$-3$$.

The function takes all possible real values from $$-3$$ to $$3$$, Thus, the domain of the function $$f$$ is the set of all real numbers in the interval $$[-3,3]$$.