Problem 9 For the function $$ g(x)=\fr... [FREE SOLUTION] (2024)

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Chapter 4: Problem 9

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

Short Answer

Expert verified

The solution is $ x > 2 $.

Step by step solution

Determine where the function is positive

Identify the values of x for which the function $ g(x)=\frac{x-2}{x+4} $ is greater than 0. This occurs when the numerator and denominator have the same sign.

Solve inequalities for numerator and denominator

Set the numerator greater than 0: $ x-2 > 0 $. Solve to get x > 2. Set the denominator greater than 0: $ x+4 > 0 $. Solve to get x > -4.

Analyze sign changes

Evaluate the signs of the fractions in each region determined by the critical points x = -4 and x = 2. Consider the regions: x < -4, -4 < x < 2, and x > 2.

Determine intervals where the function is positive

For x > -4: $ x+4 > 0 $; and for x > 2: $ x-2 > 0 $. Both must be positive, so we find that for x > 2, the function $ g(x) $ is positive.

Include or exclude boundaries

Check boundaries: At x = -4 and x = 2, $ g(x) $ is undefined and equals 0 respectively. Therefore, exclude x = -4 and x = 2 from the solution.

Key Concepts

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

Rational Functions

Rational functions are expressions that can be represented as the quotient of two polynomials. In this exercise, we have the function:
\[g(x) = \frac{x-2}{x+4}\] Here, $x-2$ is the numerator and $x+4$ is the denominator. It's essential to understand that rational functions can have values that are undefined when the denominator equals zero. For this function, $x+4 = 0$ when $x = -4$, meaning $g(x)$ is undefined at $x = -4$. Rational functions often exhibit interesting behavior around these critical points.
Understanding rational functions' properties is crucial for analyzing where they are positive or negative, which brings us to the next concept.

Inequalities

Inequalities involve finding the values of the variable for which a function meets a certain condition, such as being greater than zero. In this problem, we need to solve:
\[g(x) > 0\]
to determine where the function $g(x)$ is positive. Inequalities require us to look at the numerator and the denominator separately:

The numerator $x-2$
The denominator $x+4$

We solve these inequalities independently and then combine the results to find where the function meets the specified condition.

Sign Analysis

Sign analysis involves figuring out the signs (positive or negative) of a function's components over different intervals. For the rational function:
\[g(x) = \frac{x-2}{x+4}\],
we determine the critical points where the zeros of the numerator and the undefined points of the denominator impact its sign.
In this case:

The numerator $x-2$ is zero at $x=2$
The denominator $x+4$ is zero at $x=-4$

We consider the regions:

$x < -4$
$-4 < x < 2$
$x > 2$

By substituting values from these intervals into both the numerator and the denominator, we can determine the overall sign of $g(x)$ in each region.

Critical Points

Critical points are where the function changes its behavior, such as going from positive to negative or becoming undefined. For the function $g(x)=\frac{x-2}{x+4}$, the critical points are:

$x=-4$: where the denominator is zero, making the function undefined
$x=2$: where the numerator is zero, making the function equal to zero

We use these critical points to divide the number line into intervals and analyze the sign changes in each one.
In the exercise solution:

For $x > 2$, both the numerator and denominator are positive, making $g(x) > 0$
For $-4 < x < 2$, the numerator is negative, and the denominator is positive, making $g(x) < 0$
$x < -4$ is not important here, since it does not meet our inequality condition $g(x) > 0$

Hence, the solution to $g(x) > 0$ is $x > 2$, excluding the critical point.

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$Problem 9 For the function $$ g(x)=\fr... [FREE SOLUTION] (3)$

Most popular questions from this chapter

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