Find the domain and range of the function \( f(x) = \sqrt{x^2 - 9} \)
Answer
The range of the function is all possible output values. Since we are taking the square root of a number, the output values will always be greater than or equal to zero.
Step 1 ：The function \( f(x) = \sqrt{x^2 - 9} \) is a square root function, thus the expression under the square root must be greater than or equal to zero.
Step 2 ：We set the expression under the square root equal to zero and solve for x: \( x^2 - 9 = 0 \) which gives us \( x = -3 \) or \( x = 3 \). This indicates that the function is defined for \( x \leq -3 \) or \( x \geq 3 \).
Step 3 ：The range of the function is all possible output values. Since we are taking the square root of a number, the output values will always be greater than or equal to zero.
