Given f(x) = 2 - x2, g(x) = sqrt(x + 2) f(x) = sqrt(x), g(x) = sqrt(1 - x) (a) write formulas for f@g and g@f and find the (b) domain and (c) range of each.

(a) $f\circ g(x)$
${\displaystyle \Rightarrow f\left(\sqrt{x+2}\right)}$
${\displaystyle \Rightarrow 2-{\left(\sqrt{x+2}\right)}^2}$
${\displaystyle \Rightarrow 2-(x+2)}$
${\displaystyle \Rightarrow -x}$
${\displaystyle g\circ f\left(X\right)}$
${\displaystyle \Rightarrow g(2-x^2)}$
${\displaystyle \Rightarrow \sqrt{2-x^2+2}}$
${\displaystyle \Rightarrow \sqrt{4-x^2}}$
(b) Now, find the domain of the both composite functions.
The domain of the function f(x) consists of all real and the domain of the function g(x) is ${\displaystyle [-2,\mathrm{\infty }).}$
Now check the domain of the composite function ${\displaystyle f\circ g}$ (x) = -x. The domain of this function consists of all real values.
To find the domain of ${\displaystyle f\circ g}$ (x), there are some restrictions. But ${\displaystyle [-2,\mathrm{\infty })}$ is a proper subset of the domain of ${\displaystyle f\circ g}$ (x).
Therefore, the domain of ${\displaystyle f\circ g}$ (x) will be ${\displaystyle [-2,\mathrm{\infty })}$.
Similarly, check the domain of the composite function ${\displaystyle g\circ f}$ (x). The domain of this function is [-2, 2].
To find the domain of ${\displaystyle g\circ f}$ (x), there are some restrictions. But [-2, 2] is a proper subset of the domain of ${\displaystyle g\circ f}$ (x).
Therefore, the domain of ${\displaystyle g\circ f}$ (x) will be [-2, 2].
(c) Now, find the range of the both composite functions.
The range of the composite function ${\displaystyle f\circ g}$ (x) will be all real values.
i.e. Range = ${\displaystyle (-\mathrm{\infty },\mathrm{\infty })}$
The range of the composite function ${\displaystyle g\circ f}$ (x) will be 0 to 2.
i.e. Range = [0, 2].