Find the domain and range of the function \( f(x) = \sqrt{4-x} \)
Answer
For the function \( f(x) = \sqrt{4-x} \), since the smallest value inside the square root is 0 (when \( x = 4 \)) and the square root of a number is always non-negative, the smallest possible output value is 0. And the largest possible output value occurs when \( x \) is the smallest, i.e., when \( x \) goes to negative infinity, but since \( x \) is less than or equal to 4, the largest output value is \( f(4) = \sqrt{4-4} = 0 \). Therefore, the range of \( f \) is \( y \geq 0 \).
Step 1 ：The domain of a function \( f \) is the set of all possible input values (often the \( x \) variable), which produce a valid output from a particular function.
Step 2 ：The square root function \( \sqrt{x} \) is only defined for values of \( x \) greater than or equal to 0. Therefore, in the function \( f(x) = \sqrt{4-x} \), the value inside the square root (\( 4 - x \)) must be greater than or equal to 0.
Step 3 ：To find this, we solve the inequality \( 4 - x \geq 0 \), which gives us \( x \leq 4 \). Therefore, the domain of \( f \) is \( x \leq 4 \).
Step 4 ：The range of a function \( f \) is the set of all possible output values (often the \( y \) variable), which are the result of the function from within a particular domain.
Step 5 ：For the function \( f(x) = \sqrt{4-x} \), since the smallest value inside the square root is 0 (when \( x = 4 \)) and the square root of a number is always non-negative, the smallest possible output value is 0. And the largest possible output value occurs when \( x \) is the smallest, i.e., when \( x \) goes to negative infinity, but since \( x \) is less than or equal to 4, the largest output value is \( f(4) = \sqrt{4-4} = 0 \). Therefore, the range of \( f \) is \( y \geq 0 \).
