Solution

(i) $g(x)=5-x$

• Domain: all real numbers.
• Range: all real numbers.

(ii) $g(x)=\sqrt{x+2}$

For real outputs, require $x+2\ge 0\Rightarrow x\ge -2$.

(iii) $g(x)=\begin{cases}6x+7,&x\le -2\\4-3x,&x>-2\end{cases}$

• Domain: both formulas are defined on their intervals, so $D_g=\mathbb{R}$.

Range of first piece ($x\le -2$):

• At $x=-2$, $g(-2)=6(-2)+7=-5$.
• As $x\to -\infty$, $6x+7\to -\infty$.

So range of first piece is $(-\infty,-5]$.

Range of second piece ($x>-2$):

• As $x\to -2^+$, $g(x)\to 4-3(-2)=10$.
• As $x\to \infty$, $4-3x\to -\infty$.

So range of second piece is $(-\infty,10)$.

Combined range is $(-\infty,\infty)=\mathbb{R}$.

(iv) $g(x)=|x-5|$

• Domain: all real numbers.
• Range: $|x-5|\ge 0$.

(v) $g(x)=\dfrac{x+2}{3-x}$

Domain: denominator $3-x\ne 0\Rightarrow x\ne 3$.

For range, let $y=\dfrac{x+2}{3-x}$ and solve for $x$:

So $1+y\ne 0\Rightarrow y\ne -1$.