Problem 8 For the function $$ g(x)=\fr... [FREE SOLUTION] (2024)

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Chapter 4: Problem 8

For the function $$ g(x)=\frac{x-2}{x+4} $$ solve each of the following. $$g(x) \leq 0$$

Short Answer

Expert verified

$x \in (-4, 2]$

Step by step solution

Set Up the Inequality

Write the given inequality using the function expression: \[\frac{x-2}{x+4} \leq 0\]

Find the Critical Points

Set the numerator and the denominator equal to zero to find the critical points because these points will determine the intervals to test. 1. Numerator: \[x-2=0 \implies x=2\]2. Denominator: \[x+4=0 \implies x=-4\]

Identify Intervals

The critical points from Step 2 divide the number line into three intervals:1. $( -\infty, -4 )$2. $(-4, 2)$3. $( 2, \infty )$

Test the Intervals

Choose a test point from each interval and determine whether the inequality $\frac{x-2}{x+4} \leq 0$ holds:1. For $ x = -5 $ in interval $( -\infty, -4 )$, \[\frac{-5-2}{-5+4} = \frac{-7}{-1} = 7 \implies 7 ot\leq 0 \]2. For $ x = 0 $ in interval $(-4, 2)$, \[\frac{0-2}{0+4} = \frac{-2}{4} = -\frac{1}{2} \implies -\frac{1}{2} \leq 0\]3. For $ x = 3 $ in interval $( 2, \infty )$, \[\frac{3-2}{3+4} = \frac{1}{7} \implies \frac{1}{7} ot\leq 0\]

Include Critical Points with Zero Numerator

The function is zero at $ x = 2 $, so this point is included in the solution set: \[\frac{2-2}{2+4} = \frac{0}{6} = 0 \implies 0 \leq 0 \]

Exclude Points with Zero Denominator

The function is undefined at $ x = -4 $, so this point is excluded from the solution set.

Combine the Results

Combine the valid intervals and points:$(-4, 2] $

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

critical points

Critical points are vital in solving rational inequalities. They are the values of the variable where the function changes its sign or becomes undefined. To find these points, set the numerator and denominator of the function to zero separately.
For the function \ g(x)=\frac{x-2}{x+4} \, set both parts to zero:

Numerator: $ x-2=0 $ leads to $ x=2 $
Denominator: $ x+4=0 $ leads to $ x=-4 $

Critical points $ x=2 $ and $ x=-4 $ help us form the intervals on the number line which are crucial for further steps.

function intervals

Function intervals occur between the critical points identified earlier. They help visualize how the function behaves on different segments of the number line.
For $ g(x)=\frac{x-2}{x+4} $, the critical points $ x=2 $ and $ x=-4 $ divide the number line into three intervals:

$( -\infty, -4 )$
$(-4, 2)$
$( 2, \infty )$

Understanding these intervals is necessary to see where the function is positive, negative, or zero.

test points

Test points are selected from each function interval to determine where the inequality holds true. You pick a value from each interval and plug it into the function. Let's test the function in each interval:

For $ x = -5 $ in $ ( -\infty, -4 ) $: $ \frac{-5-2}{-5+4} = \frac{-7}{-1} = 7 $, which is not $ \leq 0 $.
For $ x = 0 $ in $ (-4, 2) $: $ \frac{0-2}{0+4} = \frac{-2}{4} = -\frac{1}{2} $, which is $ \leq 0 $.
For $ x = 3 $ in $ ( 2, \infty ) $: $ \frac{3-2}{3+4} = \frac{1}{7} $, which is not $ \leq 0 $.

By using these test points, you can identify which intervals satisfy the inequality.

numerator and denominator analysis

Numerator and denominator analysis is essential when solving rational inequalities. This step helps understand how the function behaves around the critical points and within intervals.
Set the numerator $ x-2 $ and denominator $ x+4 $ to zero:

$ x-2=0 \rightarrow x=2 $
$ x+4=0 \rightarrow x=-4 $

We see the function is zero at $ x = 2 $ and undefined at $ x = -4 $.
Note the behavior of the function:

Numerator zero: The function itself is zero.
Denominator zero: The function is undefined.

These insights, combined with interval testing, allow for a complete and accurate solution.

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$Problem 8 For the function $$ g(x)=\fr... [FREE SOLUTION] (3)$

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