Exercise 1.3.35 (35-40)

Question

Find the functions (a) $f\circ g$ (b) $g\circ f$, (c) $f\circ f$, and (d) $g\circ g$ and their domains.
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35. $f(x)=x^3+5, \quad g(x)=\sqrt[3]{x}$
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36. $f(x)=\frac{1}{x}, \quad g(x)=2 x+1$
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37. $f(x)=\frac{1}{\sqrt{x}},\quad g(x)=x+1$
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38. $f(x)=\frac{x}{x+1},\quad g(x)=2 x-1$
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39. $f(x)=\frac{2}{x} \quad g(x)=\sin x$
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40. $f(x)=\sqrt{5-x},\quad g(x)=\sqrt{x-1}$.

Khagan Gulusoy · Accepted Answer

\begin{enumerate}
    \item [35.] \begin{enumerate}
        \item $$f\circ g(x)=(\sqrt[3]{x})^3+5=x+5.$$
        The domain of the function $f\circ g$ is $(-\infty, +\infty).$
        \item $$g\circ f(x)=\sqrt[3]{x^3+5}.$$
        The domain of the following function is $(-\infty, +\infty).$
        \item $$f\circ f(x)=(x^3+5)^3+5=x^9+15x^6+75x^3+125+5=x^9+15x^6+75x^3+130.$$
        The domain of the following function is $(-\infty, +\infty).$
        \item $$g\circ g(x)=\sqrt[3]{\sqrt[3]{x}}=\sqrt[9]{x}.$$
        The domain of the following function is $(-\infty, +\infty).$
        \end{enumerate}
    \item [36.] \begin{enumerate}[label=(\alph*)]
        \item $$f\circ g(x)=\frac{1}{2x+1}.$$
        Since the de(nominator of a fraction cannot be equal to zero, the domain of the function $f\circ g$ is $\{x\mid 2x+1
eq 0\}=\{x\mid x
eq -0.5\}=(-\infty, -0.5)\cup (-0.5, +\infty).$
        \item $$g\circ f(x)=2\cdot \frac{1}{x}+1=\frac{2}{x}+1.$$
        Since the denominator of a fraction cannot be equal to zero, the domain of the function $g\circ f$ is $\{x\mid x
eq 0\}=(-\infty,0)\cup (0,+\infty).$
        \item $$f\circ f(x)=\frac{1}{1/x}=x.$$
        Since the denominator of a fraction cannot be equal to zero, the domain of the function $g\circ f$ is $\{x\mid x
eq 0\}=(-\infty,0)\cup (0,+\infty).$
        \item $$g\circ g(x)=2(2x+1)+1=4x+2+1=4x+3.$$
        The domain of the function is $(-\infty,+\infty).$
    \end{enumerate}
    \item [37.] \begin{enumerate}[label=(\alph*)]
        \item $$f\circ g(x)=\frac{1}{\sqrt{x+1}}.$$
        Since the denominator of a fraction cannot be equal to zero and the term under the quadratic root must be bigger than or equal to zero, the domain of the function $f\circ g$ is $\{x\mid \sqrt{x+1}
eq 0	ext{ and } x+1\geq 0\}=\{x\mid x+1
eq 0	ext{ and } x\geq -1\}=\{x\mid x> -1\}=(-1,+\infty).$
        \item $$g\circ f(x)=\frac{1}{\sqrt{x}}+1.$$
        Since the denominator of a fraction cannot be equal to zero and the term under the quadratic root must be bigger than or equal to zero, the domain of the function $g\circ f$ is $\{x\mid \sqrt{x}
eq 0	ext{ and }x\geq 0\}=\{x\mid x
eq 0	ext{ and }x\geq 0\}=(0, +\infty).$
        \item $$f\circ f(x)=\frac{1}{\sqrt{\frac{1}{\sqrt{x}}}}=\frac{1}{\frac{1}{\sqrt[4]{x}}}=\sqrt[4]{x}.$$
        Since the term under the quadratic root must be bigger than or equal to zero, the domain of the function $f\circ f$ is:
        $$\{x\mid x\geqslant 0\}=[0,+\infty).$$
        \item $$g\circ g(x)=(x+1)+1=x+2.$$
        The domain of this function is $(-\infty, +\infty).$
    \end{enumerate}
    \item [38.] \begin{enumerate}[label=(\alph*)]
        \item $$f \circ g(x)=\frac{2 x-1}{2 x-1+1}=\frac{2 x-1}{2 x}=\frac{2 x}{2 x}-\frac{1}{2 x}=1-\frac{1}{2 x}. $$
        Recall that the denominator of a fraction cannot be equal to zero. Thus, the domain of the function $f\circ g$ is $\{x\mid 2x
eq 0\}=\{x\mid x
eq 0\}=(-\infty,0)\cup (0,+\infty).$
        \item $$g \circ f(x)=2 \cdot \frac{x}{x+1}-1=\frac{2 x}{x+1}-1 .$$
        Since the denominator of a fraction cannot be equal to zero, the domain of the function $g\circ f$ is $\{x\mid x+1
eq 0\}=\{x\mid x
eq -1\}=(-\infty,-1)\cup (-1,+\infty).$
        \item $$f \circ f(x)=f(f(x))=f(\frac{x}{x+1}) = \frac{\frac{x}{x+1}}{\frac{x}{x+1}+1}.$$
        Since the denominator of a fraction cannot be equal to zero, the domain of the function $f\circ f$ is 
        
        \begin{align*}
            \{x\mid x+1
eq 0	ext{ and }\frac{x}{x+1}+1
eq 0\}&=\{x\mid x
eq -1	ext{ and }x
eq -x-1\}\
            &=\{x\mid x
eq -1	ext{ and } x
eq -0.5\}\
            &=(-\infty,-1)\cup (-1,-0.5)\cup (-0.5,+\infty).
        \end{align*}
        \item $$g\circ g(x)=2(2 x-1)-1=4 x-2-1=4 x-3 . $$
        The domain of the function $g\circ g$ is $\mathbb{R}.$
    \end{enumerate}
    \item [39.] \begin{enumerate}[label=(\alph*)]
    \item $$f \circ g(x)=\frac{2}{\sin x} .$$
    Since the denominator of a fraction cannot be equal to zero, the domain of the function $f\circ g$ is $\{x\mid \sin x
eq 0\}=\{x\mid x
eq n\pi 	ext{ for some $n\in \mathbb{Z}$}\}.$
    \item $$g \circ f(x)=\sin \frac{2}{x} .$$
    Since the denominator cannot be equal to zero, the domain of the function $g\circ f$ is $(-\infty,0)\cup (0,+\infty).$
    \item $$f \circ f(x)=\frac{x}{2 / x}.$$
    The domain of the function $f\circ f$ is $\mathbb{R}/\{0\}$ since the denominator of a fraction cannot be equal to zero.
    \item $$g \circ g(x)=\sin (\sin x) .$$
    The domain of the function $g\circ g$ is $\mathbb{R}.$
    \end{enumerate}
    \item [40.] \begin{enumerate}[label=(\alph*)]
\item
$$
f \circ g(x)=\sqrt{5-\sqrt{x-1}}
$$
Since the domain of a square root is the set of non-negative real numbers, the domain of the function $f \circ g$ is:
$$
\begin{aligned}
& \{x \mid x-1 \geqslant 0 	ext { and } 5-\sqrt{x-1} \geqslant 0\}=\{x \mid x \geqslant 1 	ext { and } \sqrt{x-1} \leqslant 5\}= \
& =\{x \mid x \geqslant 1 	ext { and } x-1 \leq 25\}=[1,26] 	ext {. } \
&
\end{aligned}
$$
\item
$$
g \circ f(x)=\sqrt{\sqrt{5-x}-1}
$$
Since the domain of a square root is the set of non-negative real numbers, the domain of the function $g \circ f$ is:
$$	ext{$\{x \mid 5-x \geq 0$ and $\sqrt{5-x}-1 \geqslant 0\}=\{x \mid x \leqslant 5$ and $\sqrt{5-x} \geqslant 1\}=$}$$
$$
=\{x \mid x \leqslant 5 	ext { and } 5-x \geqslant 1\}=\{x \mid x \leqslant 5 	ext { and } x \leqslant 4\}=(-\infty, 4]
$$
\item $$ f \circ f(x)=\sqrt{5-\sqrt{5-x}}$$
Since the domain of the square root is the set of non-negative real numbers, we have:
\begin{align*}
\{x \mid 5-x \geqslant 0 	ext{ and } 5-\sqrt{5-x} \geqslant 0\}&=\{x \mid x \leq 5 	ext{ and } \sqrt{5-x} \leqslant 5\}\
&=\{x \mid x \leqslant 5	ext{ and } 5-x \leqslant 25\}\
&=\{x \mid x \leqslant 5	ext{ and } x \geqslant-20\}\
&=[-20,5].
\end{align*}
\item $$g \circ g(x)=\sqrt{\sqrt{x-1}-1}$$
Since the domain of the square root is the set of non-negative real numbers, the domain of the function $g\circ g$ is:

\begin{align*}
    \{x \mid x-1 \geqslant 0	ext{ and } \sqrt{x-1}-1 \geqslant 0\}&=\{x \mid x \geqslant 1	ext{ and } \sqrt{x-1} \geqslant 1\}\
    &=\{x \mid x \geqslant 1	ext{ and } x-1 \geqslant 1\}\
    &=\{x \mid x \geqslant 1	ext{ and }x \geqslant 2\}\
    &=[2,+\infty).
\end{align*}
\end{enumerate}

\end{enumerate}

Exercise 1.3.35 (35-40)

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Exercise 1.3.35 (35-40)

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