Chapter 1.4, Problem 11E

To determine

To calculate: the domain of the expression $\frac{x^2+1}{x^2-x-2}$ .

Answer to Problem 11E

The domain of the given expression is $R-(\{-1\},\{2\})$

Where R is real number

Explanation of Solution

Given information:

The given expression is provided below

  $\frac{x^2+1}{x^2-x-2}$

Formula used:

The denominator part of the expression does not be $0$ .

And the quadratic equation formula

  $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Where a,b,c is the constant

Calculation:

Consider the expression as,

  $\frac{x^2+1}{x^2-x-2}$

Recall that the denominator should not be equal to 0

Therefore

  $x^2-x-2\ne 0$

Recall the quadratic equation formula then

a=1,b=-1,c=-2

  $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

  $\begin{array}{l}\frac{-(-1)\pm \sqrt{{(-1)}^2-\mathrm{4.1.}(-2)}}{2.1}\ne 0\\ \frac{1\pm \sqrt{1+8}}{2}\ne 0\\ \frac{1\pm \sqrt{9}}{2}\ne 0\\ \frac{1\pm 3}{2}\ne 0\\ \left(\frac{1+3}{2},\frac{1-3}{2}\right)\ne 0\\ \left(\frac{4}{2},\frac{-2}{2}\right)\ne 0\\ (2,-1)\ne 0\end{array}$

Therefore th value of x will not be $\left\{2\right\},\{-1\}$

So the domain of the expression will be $R-(\{-1\},\{2\})$

Where $R$ belongs to real number.