Chapter 11.2, Problem 63PPE

To determine

Find the values of x for which the expression $\frac{2x^2-5x-12}{6x}÷\frac{-3x-12}{x^2-16}$ is undefined. Explain you reasoning.

Answer to Problem 63PPE

  $\frac{2x^2-5x-12}{6x}÷\frac{-3x-12}{x^2-16}=-\frac{(2x+3){(x-4)}^2}{18x}$and undefined at $x\ne 0$

Explanation of Solution

Given:

Concept Used:

Rational expression is in the form $\frac{\text{P}}{\text{Q}}$ , where $\text{Q}\ne 0$ , denominator cannot be zero in any rational expression.

Excluded values are also called points of discontinuity or the expression is undefined. These are the values that make the denominator equal to zero and are not part of the domain.

In any rational function, the denominator cannot be zero.

To find the excluded value of the function, make denominator equal to zero and simplify

Calculation:

  $\begin{array}{l}\frac{2x^2-5x-12}{6x}÷\frac{-3x-12}{x^2-16}\\ \\ \frac{2x^2-5x-12}{6x}·\frac{x^2-16}{-3x-12}\\ \\ \frac{(2x+3)(x-4)}{6x}·\frac{(x+4)(x-4)}{-3(x+4)}\\ \\ \frac{(x+4)}{(x+4)}·\frac{(2x+3)(x-4)(x-4)}{6x·-3}=-\frac{(2x+3){(x-4)}^2}{18x}\\ \\ \frac{2x^2-5x-12}{6x}÷\frac{-3x-12}{x^2-16}=-\frac{(2x+3){(x-4)}^2}{18x}\end{array}$ Undefined at $x\ne 0$

Thus, $\frac{2x^2-5x-12}{6x}÷\frac{-3x-12}{x^2-16}=-\frac{(2x+3){(x-4)}^2}{18x}$ and undefined at $x\ne 0$