Chapter 1.5, Problem 50E

(a)

To determine

To calculate: To determine domain of $f$

Answer to Problem 50E

Domain of f is set of all real numbers such that

  $x\ne 0$

Explanation of Solution

Given:

  $f(x)=\frac{1}{x}$

Formula Used:

The domain of a function is the set of its possible inputs, i.e., the set of input values where for which the function is defined.

Calculation:

Domain of $f$ is calculated as:

Given $f(x)=\frac{1}{x}$

f(x) is defined for all values of x except x=0

Therefore, Domain of f is set of all real numbers such that

  $x\ne 0$

Conclusion:

Hence Domain of f is set of all real numbers such that

  $x\ne 0$

(b)

To determine

To calculate: To determine domain of $g$

Answer to Problem 50E

Domain of g is set of all real numbers such that

  $x\ne 0$

Explanation of Solution

Given:

  $g(x)=\frac{1}{2x}$

Formula Used:

The domain of a function is the set of its possible inputs, i.e., the set of input values where for which the function is defined.

Calculation:

Domain of $g$ is calculated as:

Given $g(x)=\frac{1}{2x}$

g(x) is defined for all values of x except x=0

Therefore, Domain of g is set of all real numbers such that

  $x\ne 0$

Conclusion:

Hence domain of g is set of all real numbers such that

  $x\ne 0$

(c)

To determine

To calculate: To determine domain of $f\circ g$

Answer to Problem 50E

Domain of $f\circ g$ is set of all real numbers such that

  $x\ne 0$

Explanation of Solution

Given:

  $f(x)=\frac{1}{x}$

  $g(x)=\frac{1}{2x}$

Formula Used:

1. Composition of functions is when one function is inside of another function.

The notation used for the composition of functions looks like this, $\left(f\circ g\right)\left(x\right)$ .

The composition of the function f with g is defined as follows:

  $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$ , notice that in the case the function g is inside of the function

2. The domain of a function is the set of its possible inputs, i.e., the set of input values where for which the function is defined.

Calculation:

  $f\circ g$ is calculated as:

  $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$

Substitute: $g(x)=\frac{1}{2x}$

  $\left(f\circ g\right)\left(x\right)$ = $f\left(\frac{1}{2x}\right)$

Also, given $f(x)=\frac{1}{x}$

Thus, $\left(f\circ g\right)\left(x\right)$ is calculated as

  $\begin{array}{l}\left(f\circ g\right)\left(x\right)=\frac{1}{\left(\frac{1}{2x}\right)}\\ \left(f\circ g\right)\left(x\right)=2x\end{array}$

Above function is defined for all values of x

From part b: Domain of g is set of all real numbers such that

  $x\ne 0$

Therefore, domain of $f\circ g$ is set of all real numbers such that

  $x\ne 0$

Conclusion:

Hence, domain of $f\circ g$ is set of all real numbers such that

  $x\ne 0$