Solving Inequalities And Analyzing Rational Functions

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Hey guys! Today, we're diving into the fascinating world of functions and inequalities. We'll be tackling a specific function and inequality, breaking down each step so you can follow along and master these concepts. Let's get started!

The Function and the Inequality

Our mission, should we choose to accept it, is to analyze the function:

f(x) = - rac{7(2x - 1)}{(x - 5)(x + 4)}

And, we need to solve the inequality:

- rac{7(2x - 1)}{(x - 5)(x + 4)} ge 0

Sounds a bit intimidating, right? Don't worry, we'll break it down into manageable chunks. First, let's figure out how many critical values this function has.

Part 1: Unveiling the Critical Values

When we talk about critical values, we're essentially looking for the points where the function's behavior might change. These points are crucial for solving inequalities and understanding the function's graph. Critical values come in two main flavors:

  1. x-intercepts: These are the points where the function crosses the x-axis, meaning f(x) = 0.
  2. Vertical Asymptotes: These are the vertical lines that the function approaches but never quite touches. They occur where the denominator of a rational function equals zero.

Finding the x-intercepts

To find the x-intercepts, we need to set the function equal to zero and solve for x:

- rac{7(2x - 1)}{(x - 5)(x + 4)} = 0

A fraction is equal to zero only when its numerator is zero. So, we focus on the numerator:

7(2x−1)=07(2x - 1) = 0

Divide both sides by -7:

2x−1=02x - 1 = 0

Add 1 to both sides:

2x=12x = 1

Divide by 2:

x = rac{1}{2}

So, we have one x-intercept at x = 1/2. This is a key point where the function might change its sign (from positive to negative or vice versa).

Locating the Vertical Asymptotes

Vertical asymptotes pop up when the denominator of our rational function equals zero. This is because division by zero is a big no-no in the math world, leading to undefined function values. So, we need to find the values of x that make the denominator zero:

(x−5)(x+4)=0(x - 5)(x + 4) = 0

This gives us two possibilities:

  1. x - 5 = 0 => x = 5
  2. x + 4 = 0 => x = -4

Therefore, we have vertical asymptotes at x = 5 and x = -4. These lines act as barriers, guiding the function's behavior as it gets closer and closer but never actually touching them. They also mark potential points where the function's sign might flip.

Counting the Critical Values

Now, let's count 'em up! We've found:

  • One x-intercept: x = 1/2
  • Two vertical asymptotes: x = 5 and x = -4

Adding them together, we have a grand total of three critical values. These three values will be our guide as we navigate the inequality.

Visualizing Critical Values

Think of these critical values as dividing the number line into different intervals. The function's behavior (whether it's positive or negative) will remain consistent within each interval. These values are where the function can change signs. Our number line is now split into the following intervals by these critical values:

  • (-∞, -4)
  • (-4, 1/2)
  • (1/2, 5)
  • (5, ∞)

Understanding these intervals is the key to cracking the inequality!

Part 2: Solving the Inequality

Now that we've identified the critical values, we can finally solve the inequality:

- rac{7(2x - 1)}{(x - 5)(x + 4)} ge 0

The goal here is to find the values of x that make this inequality true. We'll use a sign chart to help us organize our thoughts and arrive at the solution.

Crafting the Sign Chart

A sign chart is a fantastic tool for visualizing how the sign of an expression changes across different intervals. Here's how we'll build ours:

  1. Draw a number line: Mark our critical values (-4, 1/2, and 5) on the number line. These values divide the line into the intervals we identified earlier.
  2. List the factors: Identify the factors in our expression: (2x - 1), (x - 5), and (x + 4). We'll analyze the sign of each factor in each interval.
  3. Determine the signs: For each factor and each interval, determine whether the factor is positive or negative. We do this by picking a test value within the interval and plugging it into the factor. For example, in the interval (-∞, -4), we could pick x = -5.
  4. Calculate the expression's sign: Once we know the signs of the individual factors, we can determine the sign of the entire expression by multiplying (or dividing) the signs together. Remember: a negative times a negative is a positive, and so on.

Here's what our sign chart might look like:

Interval Test Value 2x - 1 x - 5 x + 4 -7(2x-1) / ((x-5)(x+4))
(-∞, -4) x = -5 - - - -
(-4, 1/2) x = 0 - - + +
(1/2, 5) x = 1 + - + -
(5, ∞) x = 6 + + + +

Let's walk through filling in the chart for the interval (-∞, -4) with the test value x = -5:

  • 2x - 1: 2(-5) - 1 = -11 (negative)
  • x - 5: -5 - 5 = -10 (negative)
  • x + 4: -5 + 4 = -1 (negative)
  • Expression: A negative divided by the product of two negatives is a negative. However, the leading negative sign in the function flips the sign, resulting in a final negative sign for the entire expression in this interval.

We repeat this process for each interval, carefully tracking the signs.

Reading the Sign Chart

The sign chart is our map to the solution! We're looking for the intervals where the expression is greater than or equal to zero (≥ 0). This means we want the intervals where the sign is positive or zero.

From our sign chart (the example above), we can see that the expression is positive in the intervals:

  • (-4, 1/2)
  • (5, ∞)

We also need to consider the points where the expression is equal to zero. This occurs at the x-intercept, x = 1/2. So, we include this point in our solution.

However, we must be careful about the vertical asymptotes. The function is undefined at x = -4 and x = 5, so we cannot include these values in our solution. The expression doesn't exist there, so it can't be greater than or equal to zero.

The Grand Solution

Putting it all together, the solution to the inequality is:

x elongs to (-4, rac{1}{2}] igcup (5, ext{∞})

This means that the inequality holds true for all values of x in the interval between -4 and 1/2 (including 1/2), as well as all values of x greater than 5. We use a parenthesis for -4 and 5 to indicate that these values are not included in the solution (due to the asymptotes), and a bracket for 1/2 to indicate that it is included (because the expression equals zero at this point).

Key Takeaways

Let's recap what we've learned today:

  • Critical values (x-intercepts and vertical asymptotes) are crucial for analyzing functions and solving inequalities.
  • x-intercepts occur where the function equals zero.
  • Vertical asymptotes occur where the denominator of a rational function equals zero.
  • A sign chart is a powerful tool for visualizing the sign of an expression across different intervals.
  • When solving inequalities, remember to consider both the intervals where the expression is positive/negative and the points where it equals zero.
  • Always exclude vertical asymptotes from the solution set.

Solving inequalities might seem tricky at first, but with a systematic approach and a trusty sign chart, you'll be conquering them in no time! Keep practicing, and don't hesitate to revisit these steps as needed.

Part 3: ShowDiscussion Category

For the "ShowDiscussion" category, let's dive deeper into the mathematics behind these concepts. We can discuss things like:

  • The behavior of rational functions near vertical asymptotes.
  • How the multiplicity of roots affects the sign changes of a function.
  • The relationship between the graph of a function and the solution to an inequality.

Feel free to ask questions, share your insights, and let's learn together!