Answer
a) $D=[0,\infty)$; $R=[0,\infty)$
b) $D=(-\infty,2)\cup(2,\infty)$; $R=(-\infty,0)\cup(0,\infty)$;
c) $D=(-\infty, -1]\cup[3,\infty)$; $R=[0,\infty)$
Work Step by Step
a) $f(x)=x^5+\sqrt x$
The radical is defined for $x\geq 0$.
The domain of $f$ is:
$D=[0,\infty)$
As $x$ is positive, the range of the function is:
$R=[0,\infty)$
b) $g(y)=\dfrac{1}{y-2}$
The rational expression is defined for the values of $y$ for which the denominator is not zero:
$y-2\not=0\Rightarrow y\not=2$
$D=(-\infty,2)\cup(2,\infty)$
Determine the range:
$\dfrac{1}{y-2}=x$
$x(y-2)=1$
$xy-2x=1$
$xy=2x+1$
$y=\dfrac{2x+1}{x}$
$y=2+\dfrac{1}{x}$
$R=(-\infty,0)\cup(0,\infty)$
c) $h(z)=\sqrt{z^2-2z-3}$
The radical is defined for the values of $z$ for which we have:
$z^2-2z-3\geq 0$
$(z^2-2z+1)-4\geq 0$
$(z-1)^2-4\geq 0$
$(z-1-2)(z-1+2)\geq 0$
$(z-3)(z+1)\geq 0$
$D=(-\infty, -1]\cup[3,\infty)$
As the radical is positive, the range of $h$ is:
$R=[0,\infty)$
