2.12 Algebraic Combination of Functions

(a) The domain of $f$ is $\{x|\ x\neq3\}$ or $\mathbb{R}-\{3\}$ because division by zero is not defined. The square root of negative numbers is not defined [square root of a negative number becomes imaginary], thus the domain of $g$ is $\{x|\ x\geq2\}$. The intersection of the domains of $f$ and $g$ is
\[
\{x|\ x\geq2\text{ and }x\neq3\}=[2,3)\cup(3,\infty).
\] Thus we have
\begin{align*}
(f+g)(x) & =f(x)+g(x)=\frac{1}{x-3}+\sqrt{x-2}\\
(f-g)(x) & =f(x)-g(x)=\frac{1}{x-3}-\sqrt{x-2}\\
(fg)(x) & =f(x)\cdot g(x)=\frac{\sqrt{x-2}}{x-3},
\end{align*}
and the domains of all of these functions are
\[
\{x|\ x\geq2\text{ and }x\neq3\}=[2,3)\cup(3,\infty).\qquad\text{(try them!)}
\] To determine the domain of $(f/g)(x)$, from this set of numbers we need to exclude 2 because $g(2)=0.$ That is, the domain of $(f/g)(x)$ is
\[
\{x|\ x>2\text{ and }x\neq3\}=(2,3)\cup(3,\infty),
\] and
\[
\left(\frac{f}{g}\right)(x)=\frac{1}{(x-3)\sqrt{x-2}}.
\] Because $f$ does not vanish ($f(x)\neq0$) in its domain,* the domain of $(g/f)(x)$ is the same as $Dom(f)\cap Dom(g)=[2,3)\cup(3,\infty)$,and
\[
\left(\frac{g}{f}\right)(x)=\frac{\sqrt{x-2}}{\frac{1}{x-3}}=(x-3)\sqrt{x-2}.
\]

(b) Because $x=5$ is in the domain of each function, each of these values exists:
\begin{align*}
(f+g)(5) & =f(5)+g(5)=\frac{1}{5-3}+\sqrt{5-2}=\frac{1}{2}+\sqrt{3}\\
(f-g)(5) & =f(5)-g(5)=\frac{1}{5-3}-\sqrt{5-2}=\frac{1}{2}-\sqrt{3}\\
(fg)(5) & =f(5)\cdot g(5)=\frac{1}{5-3}\cdot\sqrt{5-2}=\frac{\sqrt{3}}{2}\\
\left(\frac{f}{g}\right)(5) & =\frac{f(5)}{g(5)}=\frac{\frac{1}{5-3}}{\sqrt{5-2}}=\frac{1}{2\sqrt{3}}\\
\left(\frac{g}{f}\right)(5) & =\frac{g(5)}{f(5)}=\frac{\sqrt{5-2}}{\frac{1}{5-3}}=2\sqrt{3}
\end{align*}

*  You may sketch the graph of $y=1/(x-3)$ to see that. Note that the graph of $y=1/(x-3)$ is the graph of $y=1/x$ shifted horizontally 3 units to the right. See Section 2.11 for more details.

We have already determined the domain of this function in Example 1 in the Section on Natural Domain and Range of a Function. But now, let’s use what we learned in this section. Let $f(x)=\sqrt{x-1}$, $g(x)=\sqrt{2-x}$, and $H=f+g$. The function $f$ is defined for all $x$ such that $x-1\geq0$. That is, $Dom(f)=\{x|\ x\geq1\}=[1,\infty)$. The domain of $g$ consists of all $x$ for which $2-x\geq0$, i.e. $Dom(g)=\{x|\ x\leq2\}=(-\infty,2]$. The intersection of the domains of $f$ and $g$ is
\begin{align*}
Dom(H) & =\{x|\ x\leq2\text{ and }x\geq1\}\\
& =\{x|\ 1\leq x\leq2\}\\
& =[1,2],
\end{align*}
or
\[
Dom(H)=Dom(f)\cap Dom(g)=[1,\infty)\cap(-\infty,2]=[1,2],
\] which is exactly the same as we saw in Example 1 in Section 2.3.

The functions $h_{1}(x)$ and $h_{2}(x)$, and $u_{1}(x)$ and $u_{2}(x)$ can be equal functions if they have the same (natural) domains. [Note that if $r$ and $s$ are both nonnegative, then $rs\geq0$. Therefore if $A=\sqrt{r}\sqrt{s}$, then we can also write $A$ as
\[A=\sqrt{rs}.\] However, if $B=\sqrt{pq}$, without knowing that $p$ and $q$ are both nonnegative, we cannot write $B$ as $\sqrt{p}\sqrt{q}.$ For example, if $p=-2$, and $q=-3$, then we can calculate $\sqrt{(-2)\cdot(-3)}$
but $\sqrt{-2}$ or $\sqrt{-3}$ are not defined.] To determine the domains of these functions, we note that:
\[
\sqrt{\square}\text{ is defined if the sign of }\square\text{ is not negative}
\] and
\[
\text{sign }(\square\times\lozenge)=\text{sign }(\square)\times\text{sign }(\lozenge)
\] (a) First, we should determine the signs of all expressions under $\sqrt{.}$ 
From what we learned in this section:
\begin{align*}
Dom(h_{1}) & =Dom(\sqrt{x-1})\cap Dom(\sqrt{2-x})\\
& =[1,\infty)\cap(-\infty,2]=[1,2],
\end{align*}
and from the above table:
\[
Dom(h_{2})=Dom(\sqrt{(x-1)(2-x)}=[1,2].
\] Therefore $h_{1}(x)=h_{2}(x)$ everywhere they are defined. In other words, they are equal functions.

(b) Similarly, using the following table: 

\begin{align*}
Dom(u_{1}) & =Dom(\sqrt{x-1})\cap Dom(\sqrt{x-2})\\
& =[1,\infty)\cap[2,\infty)=[2,\infty)
\end{align*}
but
\[
Dom(u_{2})=(-\infty,1]\cup[2,\infty).
\] Therefore $u_{1}(x)$ and $u_{2}(x)$ are equal only when $x\in[2,\infty)$. Because the domains of $u_{1}(x)$ and $u_{2}(x)$ are not equal, they are not equal functions.

Recall that $\log x$ is defined when $x>0$ and
\[
\log x=0\Rightarrow x=1
\] (a)
\begin{align*}
Dom(f) & =Dom(\sqrt{1-x})\cap Dom(\log x)\cap\{x|\log x\neq0\}\\
& =(-\infty,1]\cap(0,\infty)\cap\{x|x\neq1\}\\
& =(0,1).
\end{align*}
(b)
\begin{align*}
Dom(g) & =Dom(e^{u})\cap Dom(\sqrt{|1+u|})\cap\{u|\ \sqrt{|1+u|}\neq0\}\\
& =\mathbb{R}\cap\mathbb{R}\cap\{u|\ |1+u|\neq0\}\\
& =\mathbb{R}\cap\mathbb{R}\cap\{u|\ u\neq\pm1\}\\
& =\{u|\ u\neq\pm1\}\\
& =\mathbb{R}-\{-1,1\},
\end{align*}
which can also be written as
\[
Dom(g)=(-\infty,-1)\cup(-1,1)\cup(1,\infty).
\]