Exercise 1.3.33 (33-34)

Question

Find (a) $f+g,$ (b) $f-g$, (c) $f g$, and (d) $f / g$ and state their domains.
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33. $f(x)=\sqrt{25-x^2}, g(x)=\sqrt{x+1}$
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34. $f(x)=\frac{1}{x-1}, g(x)=\frac{1}{x}-2$

Khagan Gulusoy · Accepted Answer

\begin{enumerate} 
\item[33.] 
\begin{enumerate}[label=$\alph*$]
    \item 
    
    $$(f+g)(x)=f(x)+g(x)=\sqrt{25-x^2}+\sqrt{x+1}.$$
    The domain of $f+g$ is:
    $$\{x\mid 25-x^2\geq 0\ 	ext{and} \ x+1\geq 0\}=\{x\mid -5\leq x\leq 5\ 	ext{and}\ x\geq -1\}=[-1,5].$$
    \item $$(f-g)(x)=f(x)-g(x)=\sqrt{25-x^2}-\sqrt{x+1}$$
    The domain of $f-g$ is:
    $$\{x\mid 25-x^2\geq 0\ 	ext{and} \ x+1\geq 0\}=\{x\mid -5\leq x\leq 5\ 	ext{and}\ x\geq -1\}=[-1,5].$$
    \item $$f g(x)=f(x) \cdot g(x)=\sqrt{25-x^2} \cdot \sqrt{x+1}=\sqrt{(5-x)(5+x)(x+1)} .$$Since the domain of the square root is the set of non-negative real numbers, the factors under the root must be positive (since there are three factors). Thus, the domain of the function $fg$ is:
$$	ext{$\{x \mid 5-x \geqslant 0$ and $5+x \geqslant 0$ and $x+1 \geqslant 0\}=\{x \mid x \leqslant 5 	ext { and } x \geqslant-5 	ext { and } x \geqslant-1\}=[-1,5]$}$$
\item $$\frac{f}{g}(x)=\frac{\sqrt{25-x^2}}{\sqrt{x+1}}=\sqrt{\frac{25-x^2}{x+1}}$$Since the domain of the square soot is the set of non-negative real numbers and the denominator of the fraction must be positive, the domain of the function $\frac{f}{g}$ is:
$$	ext{$\{x \mid 5-x \geqslant 0$ and $5+x \geqslant 0$ and $x+1 \geqslant 0$ and $x+1 
eq 0\}=$}$$ 
$$	ext{$=\{x \mid x \leqslant 5$ and $x \geqslant-5$ and $x \geqslant-1$ and $x 
eq-1\}=$}$$
$$=(-1,5).$$
\end{enumerate}

\item[34.] Since the denominator of a fraction cannot be equal to zero, the domain of $f$ is $(-\infty,1)\cup (1,+\infty)$ and the domain of $g$ is $(-\infty,0)\cup(0,+\infty).$

\begin{enumerate}[label=(\alph*)]
    \item The domain of $f+g$ is the intersection of the domains of $f$ and $g.$ Thus, the domain of $f+g$ is $(-\infty,0)\cup (0,1)\cup (1,+\infty).$
    $$\forall x\in(-\infty,0)\cup (1,+\infty):\quad (f+g)(x)=f(x)+g(x)=\frac{1}{x-1}+\frac{1}{x}-2 .$$
    \item The domain of $f-g$ is also the intersection of the domains of $f$ and $g.$, i.e. $(-\infty,0)\cup (0,1)\cup (1,+\infty).$
    $$\forall x\in(-\infty,0)\cup (1,+\infty):\quad (f-g)(x)=\frac{1}{x-1}-\frac{1}{x}+2=\frac{x-x+1}{x^2-x}+2=\frac{1}{x^2-x}+2 .$$
    \item The domain of $fg$ is also the intersection of the domains of $f$ and $g.$, i.e. $(-\infty,0)\cup (0,1)\cup (1,+\infty).$
    $$\forall x\in(-\infty,0)\cup (1,+\infty):\quad f g(x)=f(x) \cdot g(x)=\frac{1}{x-1} \cdot\left(\frac{1}{x}-2ight)=\frac{1}{x^2-x}-\frac{2}{x-1} .$$
    \item The domain of $f/g$ is the intersection of the domains of $f$ and $g$ and the area where $g$ is non-zero:
    \begin{align*}
        \{x\mid x 
eq 0 	ext{ and } x 
eq 1 	ext{ and }  \frac{1}{x}-2
eq 0\}
        &=(-\infty,0)\cup (0,0.5)\cup (0.5,1)\cup (1,+\infty).
    \end{align*}
    $$\frac{f}{g}(x)=\left(\frac{1}{x-1}ight):\left(\frac{1-2 x}{x}ight)=\frac{x}{(x-1)(1-2 x) .}$$
    
\end{enumerate}

\end{enumerate}

Exercise 1.3.33 (33-34)

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Exercise 1.3.33 (33-34)

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