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

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. \(x\neq0\). --- b) We have \[ (f-g)(x)=f(x)-g(x)=\left(1+\frac{1}{x}\right)-\frac{1}{x}=1. \] The domain remains \(x\neq0\). --- 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 \(x\neq0\). --- 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). --- e) Substitute \(x=3\) into \((f+g)(x)=1+\frac{2}{x}\): \[ (f+g)(3)=1+\frac{2}{3}=\frac{5}{3}. \] --- f) Substitute \(x=2\) into \((fg)(x)=\frac{1}{x}+\frac{1}{x^2}\): \[ (fg)(2)=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}. \] --- 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 \(x\neq 0\). --- 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 \(x\neq -1\).