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In this topic, we are going to consider the function as a 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 value 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.
 
The 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.
 
The 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 equation, 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 a 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].