In Problems 63–72, for the given functions f and g, find the following. For parts (a)–(d), also find the domain.

$$\begin{gathered} \left(a\right)(f+g)\left(x\right)\left(b\right)(f-g)\left(x\right)\left(c\right)(f·g)\left(x\right)\left(d\right)\left(\frac{f}{g}\right)\left(x\right) \\ \left(e\right)(f+g)\left(3\right)\left(f\right)(f-g)\left(4\right)\left(g\right)(f·g)\left(2\right)\left(h\right)\left(\frac{f}{g}\right)\left(1\right) \end{gathered}$$

$f\left(x\right)=1+\frac{1}{x};g\left(x\right)=\frac{1}{x}$

In the given problems we have to solve the given functions f and g, find the following. For parts (a)–(d), we have to also find the domain.

$$\begin{gathered} \left(a\right)(f+g)\left(x\right)\left(b\right)(f-g)\left(x\right)\left(c\right)(f·g)\left(x\right)\left(d\right)\left(\frac{f}{g}\right)\left(x\right) \\ \left(e\right)(f+g)\left(3\right)\left(f\right)(f-g)\left(4\right)\left(g\right)(f·g)\left(2\right)\left(h\right)\left(\frac{f}{g}\right)\left(1\right) \end{gathered}$$

The given function is$f\left(x\right)=1+\frac{1}{x};g\left(x\right)=\frac{1}{x}$

This requires that:

$x\ne 0$

So the domain of both $f\mathrm{and}g$are$\left\{x\right|x\ne 0\}$

$$\begin{gathered} (f+g)\left(x\right)=1+\frac{1}{x}+\frac{1}{x} \\ (f+g)\left(x\right)=1·\frac{x}{x}+\frac{1}{x}+\frac{1}{x} \\ (f+g)\left(x\right)=\frac{x+1+1}{x} \\ (f+g)\left(x\right)=\frac{x+2}{x} \end{gathered}$$

The domain of $f+g$ consists of those numbers x that are in the domains of both fandg. Therefore, the domain of $f+g$is $\left\{x\right|x\ne 0\}$.

Putting the value of $f\mathrm{and}g$

$$\begin{gathered} (f-g)\left(x\right)=1+\frac{1}{x}-\frac{1}{x} \\ (f-g)\left(x\right)=1 \end{gathered}$$

The domain of $f-g$ consists of those numbers x that are in the domains of both . Therefore, the domain of $f-g$is $\left\{x\right|x\ne 0\}$.

Putting the value of $f\left(x\right)\mathrm{and}g\left(x\right)$

$$\begin{gathered} (f·g)\left(x\right)=\left(1+\frac{1}{x}\right)·\frac{1}{x} \\ (f·g)\left(x\right)=1·\frac{1}{x}+\frac{1}{x}·\frac{1}{x} \\ (f·g)\left(x\right)=\frac{1}{x}·\frac{x}{x}+\frac{1}{x^2} \\ (f·g)\left(x\right)=\frac{x}{x^2}+\frac{1}{x^2} \\ (f·g)\left(x\right)=\frac{x+1}{x^2} \end{gathered}$$

Using the commutative property

The domain of $f·g$consists of those numbers x that are in the domains of both $f\mathrm{and}g$. Therefore, the domain of $f·g$is $\left\{x\right|x\ne 0\}$.

Firstly solving $f\left(x\right)=1+\frac{1}{x}$

$$\begin{gathered} f\left(x\right)=1·\frac{x}{x}+\frac{1}{x} \\ f\left(x\right)=\frac{x+1}{x} \end{gathered}$$

Putting the value of $f\left(x\right)\mathrm{and}g\left(x\right)$

role="math" localid="1646139464564" $$\begin{gathered} \left(\frac{f}{g}\right)\left(x\right)=\frac{\frac{x+1}{x}}{\frac{1}{x}} \\ \left(\frac{f}{g}\right)\left(x\right)=\frac{x+1}{x}·\frac{x}{1} \\ \left(\frac{f}{g}\right)\left(x\right)=x+1 \end{gathered}$$

Since $g\left(x\right)\ne 0$when

The domain of $\frac{f}{g}$is$\left\{x\right|x\ne 0\}$.

Putting $x=3$in the value of $(f+g)\left(x\right)$

$$\begin{gathered} (f+g)\left(3\right)=\frac{3+2}{3} \\ (f+g)\left(3\right)=\frac{5}{3} \end{gathered}$$

Putting $x=4$in the value of $(f-g)\left(x\right)$

$(f-g)\left(4\right)=1$

Putting $x=2$in the value of $(f·g)\left(x\right)$

$$\begin{gathered} (f·g)\left(2\right)=\frac{2+1}{{\left(2\right)}^2} \\ (f·g)\left(2\right)=\frac{3}{4} \end{gathered}$$

Putting $x=1$in the value of $\left(\frac{f}{g}\right)\left(x\right)$

localid="1646139923257" $$\begin{gathered} \left(\frac{f}{g}\right)\left(1\right)=1+1 \\ \left(\frac{f}{g}\right)\left(1\right)=2 \end{gathered}$$