Answer
a) $x^2+6+\sqrt{1-x}$
b) $x^2+6-\sqrt{1-x}$
c) $(x^2+6)\sqrt{1-x}$
d) $\dfrac{x^2+6}{\sqrt{1-x}}$
Domain: $\left(-\infty,1\right)$
Work Step by Step
We are given the functions:
$f(x)=x^2+6$
$g(x)=\sqrt{1-x}$
a) Determine $(f+g)(x)$:
$(f+g)(x)=f(x)+g)(x)=x^2+6+\sqrt{1-x}$
b) Determine $(f-g)(x)$:
$(f-g)(x)=f(x)-g)(x)=x^2+6-\sqrt{1-x}$
c) Determine $(fg)(x)$:
$(fg)(x)=f(x)g)(x)=(x^2+6)\sqrt{1-x}$
d) Determine $\left(\dfrac{f}{g}\right)(x)$:
$\left(\dfrac{f}{g}\right)(x)=\dfrac{x^2+6}{\sqrt{1-x}}$
The domain of $\dfrac{f}{g}$ is the set of all real numbers except the zeros of $g$ and the values of $x$ for which the radical is undefined:
$1-x\gt 0\Rightarrow x\lt 1$
The domain is:
$\left(-\infty,1\right)$
