$$ \begin{array}{ll} ext { Exercise } 5 & ext { Let } f(x)=1+\frac{1}{x} ext { and } g(x)=\frac{1}{x} ext {. Find the following: } \ ext { a) }(f+g)(x) ext { and its domain. } \ ext { b) }(f-g)(x) ext { and its domain. } \ ext { c) }(f g)(x) ext { and its domain. } & f(x)=1+1 / x \ ext { d) }\left(\frac{f}{g} ight)(x) ext { and its domain. } \ ext { e) }(f+g)(3) \ ext { f) }(f g)(2) \ ext { g) }(f \circ g)(x) ext { and its domain. } & g(x)=\frac{1}{x} \ ext { h) }(g \circ f)(x) ext { and its domain. } & \end{array} $$

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$$ \begin{array}{ll}	ext { Exercise } 5 & 	ext { Let } f(x)=1+\frac{1}{x} 	ext { and } g(x)=\frac{1}{x} 	ext {. Find the following: } \ 	ext { a) }(f+g)(x) 	ext { and its domain. } \ 	ext { b) }(f-g)(x) 	ext { and its domain. } \ 	ext { c) }(f g)(x) 	ext { and its domain. } & f(x)=1+1 / x \ 	ext { d) }\left(\frac{f}{g}ight)(x) 	ext { and its domain. } \ 	ext { e) }(f+g)(3) \ 	ext { f) }(f g)(2) \ 	ext { g) }(f \circ g)(x) 	ext { and its domain. } & g(x)=\frac{1}{x} \ 	ext { h) }(g \circ f)(x) 	ext { and its domain. } & \end{array} $$

UpStudy AI · Accepted Answer

a) We have
$$
(f+g)(x)=f(x)+g(x)=\left(1+\frac{1}{x}\right)+\frac{1}{x}=1+\frac{2}{x}.
$$
The domain is all $$x$$ such that the expressions are defined, i.e. $$x\neq0$$.

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b) We have
$$
(f-g)(x)=f(x)-g(x)=\left(1+\frac{1}{x}\right)-\frac{1}{x}=1.
$$
The domain remains $$x\neq0$$.

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c) We have
$$
(fg)(x)=f(x)\cdot g(x)=\left(1+\frac{1}{x}\right)\cdot \frac{1}{x}=\frac{1}{x}+\frac{1}{x^2}.
$$
The domain is $$x\neq0$$.

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d) We have
$$
\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{1+\frac{1}{x}}{\frac{1}{x}}.
$$
Multiplying numerator and denominator by $$x$$ (where $$x\neq0$$) gives
$$
\frac{(1+\frac{1}{x})x}{1} = x+1.
$$
The domain is $$x\neq0$$ (since $$g(x)=\frac{1}{x}$$ must be nonzero).

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e) Substitute $$x=3$$ into $$(f+g)(x)=1+\frac{2}{x}$$:
$$
(f+g)(3)=1+\frac{2}{3}=\frac{5}{3}.
$$

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f) Substitute $$x=2$$ into $$(fg)(x)=\frac{1}{x}+\frac{1}{x^2}$$:
$$
(fg)(2)=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}.
$$

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g) We have
$$
(f\circ g)(x)=f(g(x))=f\left(\frac{1}{x}\right)=1+\frac{1}{\frac{1}{x}}=1+x.
$$
Here $$g(x)=\frac{1}{x}$$ is defined when $$x\neq 0$$. Thus the domain is $$x\neq 0$$.

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h) We have
$$
(g\circ f)(x)=g(f(x))=g\left(1+\frac{1}{x}\right)=\frac{1}{1+\frac{1}{x}}.
$$
Combine the terms in the denominator:
$$
\frac{1}{1+\frac{1}{x}}=\frac{1}{\frac{x+1}{x}}=\frac{x}{x+1}.
$$
The function $$f(x)=1+\frac{1}{x}$$ is defined for $$x\neq0$$, and we must also have 
$$
1+\frac{1}{x}\neq0.
$$
Setting this equal to zero:
$$
1+\frac{1}{x}=0 \quad\Rightarrow\quad \frac{1}{x}=-1 \quad\Rightarrow\quad x=-1.
$$
Thus, the domain is $$x\neq0$$ and $$x\neq -1$$.
a) $$ (f+g)(x) = 1 + \frac{2}{x} $$ with domain $$ x \neq 0 $$. b) $$ (f-g)(x) = 1 $$ with domain $$ x \neq 0 $$. c) $$ (fg)(x) = \frac{1}{x} + \frac{1}{x^2} $$ with domain $$ x \neq 0 $$. d) $$ \left(\frac{f}{g}\right)(x) = x + 1 $$ with domain $$ x \neq 0 $$. e) $$ (f+g)(3) = \frac{5}{3} $$. f) $$ (fg)(2) = \frac{3}{4} $$. g) $$ (f \circ g)(x) = 1 + x $$ with domain $$ x \neq 0 $$. h) $$ (g \circ f)(x) = \frac{x}{x+1} $$ with domain $$ x \neq 0 $$ and $$ x \neq -1 $$.

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