Finding the Domain of a Composite Function Involving Radicals

Find the domain of

(f \circ g) (x) where f (x) = \sqrt{x + 2} and g (x) = \sqrt{3 - x}

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
Because we cannot take the square root of a negative number, the domain of $g$ is $(- \infty, 3] .$ Now we check the domain of the composite function

$(f \circ g) (x) = \sqrt{\sqrt{3 - x} + 2}$
For $(f \circ g) (x) = \sqrt{\sqrt{3 - x} + 2}, \sqrt{3 - x} + 2 \geq 0,$ since the radicand of a square root must be positive. Since square roots are positive, $\sqrt{3 - x} \geq 0,$ or, $3 - x \geq 0,$ which gives a domain of $(-\infty, 3]$ .

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Find the domain of

(f \circ g) (x) where f (x) = \frac{1}{x - 2} and g (x) = \sqrt{x + 4}

How did it go?