UPSKILL MATH PLUS
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Learn moreIn 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:
- The denominator (bottom) of a fraction cannot be zero.
- 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 \([-3,3]\).
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