Chapter 1.7, Problem 99E

To determine

To find:the values of the variable for which the given expression is defined as real number.

Answer to Problem 99E

  $x\in \left[-\frac{4}{3},\frac{4}{3}\right]$ .

Explanation of Solution

Given:

The given expression is $\sqrt{16-9x^2}$ .

Concept used:

Guidelines for solving nonlinear inequality:

1. Move all terms to one side.
2. Factor the non-zero side of the inequality.
3. Find the value for which each factor is zero. The number will divide the real lines into interval. List the interval determined by these numbers.
4. Make a table or diagram by using test values of the signs of each factor on each interval. In the last row of the table determining the sign of the product of these factors.
5. Determine the solution of the inequality from the last row of the sign table.

Calculation:

The domain of $16-9x^2$ is greater than or equal to 0.

Thus, the factor is

  $\begin{array}{l}16-9x^2\ge 0\\ \left(4-3x\right)\left(4+3x\right)\ge 0\end{array}$

First to find the zeros of the expression in the numerator and demniminator, then

  $\begin{array}{l}\left(4-3x\right)=0\Rightarrow x=\frac{3}{4}\\ \left(4+3x\right)=0\Rightarrow x=\frac{-3}{4}\end{array}$

From the three zeros above, it extracts the following intervals:

  $\left(-\infty ,-\frac{4}{3}\right)\left(-\frac{4}{3},\frac{4}{3}\right)\left(\frac{4}{3},\infty \right)$

Now, make a table by using test values of the signs of each factor on each interval.

Invertal	$\left(-\infty ,-\frac{4}{3}\right)$	$\left(-\frac{4}{3},\frac{4}{3}\right)$	$\left(\frac{4}{3},\infty \right)$
$\left(4-3x\right)$	+	+	$-$
$\left(x-3\right)$	$-$	+	+
$\left(4+3x\right)$	$-$	+	$-$

As it is seen that the less the equivalent form of the solution is $\left[-\frac{4}{3}\le x\le \frac{4}{3}\right]$

Hence, the solution set is $x\in \left[-\frac{4}{3},\frac{4}{3}\right]$ .