Chapter 11.3, Problem 1ACYP

To determine

To State: The excluded values for rational expression.

Answer to Problem 1ACYP

  $x\ne \pm 9$

Explanation of Solution

Given: $\frac{5x}{x^2-81}$

Given: $\frac{5x}{x^2-81}$

Exclude the values for which $x^2-81=0$

The denominator cannot be zero.

  $x^2-81=0$

Factoring

  $\begin{array}{l}(x-9)(x+9)=0\\ x=\pm 9\end{array}$

Thus

The excluded values are $\pm 9$

Hence

  $x\ne \pm 9$