# 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.

Given
$f\left(x\right)=2-x2,g\left(x\right)=\sqrt{x+2}$
$f\left(x\right)=\sqrt{x},g\left(x\right)=\sqrt{1-x}$
(a) write formulas for $f\circ g$ and $g\circ f$ and find the
(b) domain and
(c) range of each.
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brawnyN

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