Chapter 1.2, Problem 10QR

(a)

To determine

To find: The formula for $f(-x)$

Answer to Problem 10QR

The formula : $f(-x)=\left(x+\frac{1}{x}\right)\sin (x)$

Explanation of Solution

Given information: The function is: $f(x)=\left(x+\frac{1}{x}\right)\sin (x)$

Calculation:

The given function is:

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

Replace $x$ by $-x$ in the given function.

  $f(-x)=\left(-x+\frac{1}{-x}\right)\sin (-x)$

  $\sin (-x)=-\sin (x)$ since the function $\sin x$ is odd. To make things simpler, spread the negative inside the parentheses.

  $\begin{array}{c}f(-x)=-\left(-x-\frac{1}{x}\right)\sin (x)\\ =\left(x+\frac{1}{x}\right)\sin (x)\end{array}$

Therefore the required formula is $f(-x)=\left(x+\frac{1}{x}\right)\sin (x)$

(b)

To determine

To find: The formula for $f(1/x)$

Answer to Problem 10QR

The formula: $f\left(\frac{1}{x}\right)=\left(\frac{1}{x}+x\right)\sin \left(\frac{1}{x}\right)$

Explanation of Solution

Given information: The function is: $f(x)=\left(x+\frac{1}{x}\right)\sin (x)$

Calculation:

The given function is:

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

In the given function, swap out $x$ for $\frac{1}{x}$

  $\begin{array}{c}f\left(\frac{1}{x}\right)=\left(\frac{1}{x}+\frac{1}{\frac{1}{x}}\right)\sin \left(\frac{1}{x}\right)\\ =\left(\frac{1}{x}+x\right)\sin \left(\frac{1}{x}\right)\end{array}$

Therefore the required formula is $f\left(\frac{1}{x}\right)=\left(\frac{1}{x}+x\right)\sin \left(\frac{1}{x}\right)$