# Thread: [SOLVED] Domain and range confusion

1. ## [SOLVED] Domain and range confusion

Okay so I have the function f(x) = square root of 9-x^2 and g(x) = square root x+3

I must find f of g

This is square root 9 - (square root x+3)^2

Now I must find the domain and range. Here is my attempt:

Simplified I have square root 6-x. Since x cannot be 6 or greater the domain must be (-infinity, 6). The range is (6, +infinity).

2. Hello, thekrown!

Given: .$\displaystyle \begin{array}{cccc} (1) & f(x) &=& \sqrt{9-x^2} \\ (2) & g(x) &=& \sqrt{x+3} \end{array}$

Find $\displaystyle f\circ g$

$\displaystyle f\circ g \;=\;\sqrt{9-(\sqrt{x+3})^2} \;=\;\sqrt{9-x-3} \;=\;\sqrt{6-x}$

The domain of $\displaystyle f(x)$ is: .$\displaystyle -3 \,\leq x \,\leq\,3$

The domain of $\displaystyle g(x)$ is: .$\displaystyle x \,\geq\,-3$

. . Hence, the domain of $\displaystyle f \circ g$ is: .$\displaystyle [-3 ,\;3]$

Hence, $\displaystyle f\circ g$ ranges between: .$\displaystyle \sqrt{6-(\text{-}3)}\,\text{ and }\,\sqrt{6-3}$

. . Therefore, the range of $\displaystyle f\circ g$ is: . $\displaystyle \left[ \sqrt{3},\:3\right]$

3. Got it!

Okay, now the second half of this question is to find g of f

Would the domain of this g of f be the same as f of g?