Chapter 6.4, Problem 33WE

To determine

To Solve:

  $\begin{array}{l}If\\ f(x)=\frac{x}{x+1},findf(1-\sqrt{2})\end{array}$

Answer to Problem 33WE

  $f\left(1-\sqrt{2}\right)=-\frac{\sqrt{2}}{2}$

Explanation of Solution

Given:

The expression is :

  $\begin{array}{l}f(x)=\frac{x}{x+1}\\ and\\ x=1-\sqrt{2}\end{array}$

Formula Used:

• $(a+b)(a-b)=a^2-b^2$

Calculation:

Given:

  $\begin{array}{l}f(x)=\frac{x}{x+1}\\ and\\ x=1-\sqrt{2}\\ f\left(1-\sqrt{2}\right)\\ =\frac{1-\sqrt{2}}{1-\sqrt{2}+1}\\ =\frac{1-\sqrt{2}}{2-\sqrt{2}}\\ =\frac{1-\sqrt{2}}{2-\sqrt{2}}\cdot \frac{2+\sqrt{2}}{2+\sqrt{2}}\mathrm{......}\left[\begin{array}{l}\text{Rationalize the denominator by multiplying the numerator}\\ \text{and denominator by the conjugate of the denominator}\text{.}\end{array}\right]\\ =\frac{\left(1-\sqrt{2}\right)\left(2+\sqrt{2}\right)}{\left(2-\sqrt{2}\right)\left(2+\sqrt{2}\right)}\\ =\frac{\left(1-\sqrt{2}\right)\left(2+\sqrt{2}\right)}{{\left(2\right)}^2-{\left(\sqrt{2}\right)}^2}\mathrm{.............}[(a+b)(a-b)=a^2-b^2]\\ =\frac{1\cdot 2+1\cdot \sqrt{2}+\left(-\sqrt{2}\right)\cdot 2+\left(-\sqrt{2}\right)\cdot \sqrt{2}}{4-2}\mathrm{...}[\text{Use FOIL method]}\\ \text{=}\frac{2+\sqrt{2}-2\sqrt{2}-2}{2}\\ =\frac{-\sqrt{2}}{2}\mathrm{....................................................}[\text{Combine like terms]}\\ \Rightarrow f\left(1-\sqrt{2}\right)=-\frac{\sqrt{2}}{2}\end{array}$