A step by step tutorial, with detailed solutions, on how to find the domain of square root functions is presented. Matched problems to the exercises with solutions at the bottom of the page are also presented.
For a function \( f \) defined by an expression with variable \( x \), the implied domain of \( f \) is the set of all real numbers variable \( x \) can take such that the expression defining the function is real. The domain can also be given explicitly.
For a square root function given by
\( f(x) = \sqrt{x} \)
to have real values, the radicand \( x \) must be positive or equal to zero.
also
Step by Step Calculator to Find Domain of a Function
For \( f(x) \) to have real values, the radicand \((x - 2)(x + 3)\) must be positive. Hence
\( (x - 2)(x + 3) \geq 0 \)
Solve the above inequality to obtain the solution set, which is also the domain, in interval form as follows:
\( (-\infty , -3] \cup [2 , + \infty) \)
For \( \sqrt{ x^2 + 2 x - 1 } \) to be real, the radicand must be positive or equal to 0. Hence the inequality
\( x^2 + 2x - 1 \geq 0 \)
The solution set of the above quadratic inequality, which is also the domain, is given in interval form as follows:
\( (-\infty , -1 - \sqrt{2}] \cup [-1 + \sqrt{2} , + \infty) \)
The domain of the given function is given by the interval \((- \infty , -1 - \sqrt{2}] \cup [-1 + \sqrt{2} , + \infty)\).
The domain of this function is the set of
all values of x such that \( \dfrac{2x - 1}{x + 3} \geq 0 \) which is an inequality to solve. The solution set of the above inequality which is also the domain is given by
\( (-\infty , -3) \cup [ \dfrac{1}{2} , +\infty) \)
The domain of this function is the set of
all values of x such that \( |-2 x - 6| \geq 0 \). We need to solve the inequality
\( |-2 x - 6| \geq 0 \)
Because of the absolute value, the expression \( |-2 x - 6| \) is greater than or equal to 0 for all real numbers. Hence the domain of the above function is given by:
\( (-\infty , +\infty) \)