Domain Of P: Solve For X In (4x²-4x+1)/((x+1)(2x-1)²)
Alright, guys, let's dive into finding the values of x for which the expression P = (4x² - 4x + 1) / ((x + 1)(2x - 1)²) exists. This is all about figuring out the domain of the function, which means identifying any values of x that would make the function undefined. In simpler terms, we need to avoid any situations where we're dividing by zero, as that's a big no-no in the math world. So, our mission is to pinpoint the values of x that would cause the denominator of our fraction to equal zero.
Identifying Potential Issues
The expression we're dealing with is P = (4x² - 4x + 1) / ((x + 1)(2x - 1)²). The existence of P hinges on its denominator, (x + 1)(2x - 1)², not being equal to zero. If the denominator equals zero, P is undefined. Our goal is to find those x values that make the denominator zero and exclude them from the domain of P.
Setting the Denominator to Zero
To find the problematic x values, we set the denominator equal to zero: (x + 1)(2x - 1)² = 0. This equation is satisfied if either (x + 1) = 0 or (2x - 1)² = 0. Let's solve each of these.
Solving (x + 1) = 0
Solving (x + 1) = 0 is straightforward. Subtract 1 from both sides to get x = -1. So, x = -1 is one value that makes the denominator zero, and therefore, P is undefined at x = -1.
Solving (2x - 1)² = 0
For (2x - 1)² = 0, we take the square root of both sides, which gives us 2x - 1 = 0. Adding 1 to both sides, we get 2x = 1. Dividing by 2, we find x = 1/2. Thus, x = 1/2 is another value that makes the denominator zero, and P is undefined at x = 1/2.
Determining the Domain of P
We've identified that P is undefined when x = -1 and x = 1/2. Therefore, the domain of P includes all real numbers except -1 and 1/2. We can express this in several ways:
- Set Notation: {x ∈ ℝ | x ≠ -1, x ≠ 1/2}
- Interval Notation: (-∞, -1) ∪ (-1, 1/2) ∪ (1/2, ∞)
Conclusion
In conclusion, the values of x for which P = (4x² - 4x + 1) / ((x + 1)(2x - 1)²) exists are all real numbers except -1 and 1/2. These are the values that would make the denominator zero, rendering P undefined. So, we've successfully navigated through the math to find the domain of P! Remember, always watch out for those pesky denominators!
Let's elaborate further to ensure we have a robust understanding. When we're dealing with rational functions (functions that are fractions with polynomials), the domain is a critical concept. It tells us all the possible input values (x-values) that we can plug into the function without causing any mathematical errors, like dividing by zero or taking the square root of a negative number (in the realm of real numbers, of course!). In our specific case, the function P = (4x² - 4x + 1) / ((x + 1)(2x - 1)²) is a rational function, and our main concern is the denominator.
Why is the denominator so important?
The denominator is the key because division by zero is undefined in mathematics. Think about it: if you have a pizza and try to divide it among zero people, how much does each person get? It doesn't make sense! Similarly, in mathematical terms, any number divided by zero is undefined. Therefore, to find the domain of P, we need to identify any values of x that would make the denominator, (x + 1)(2x - 1)², equal to zero.
Step-by-Step Analysis
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Identify the Denominator:
The denominator of our function is (x + 1)(2x - 1)². This is the expression we need to focus on.
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Set the Denominator Equal to Zero:
To find the values of x that make the denominator zero, we set the entire expression equal to zero: (x + 1)(2x - 1)² = 0.
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Solve for x:
This equation is satisfied if either of the factors is equal to zero. So, we have two cases to consider:
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Case 1: (x + 1) = 0
Solving for x, we get x = -1. This means that if we plug x = -1 into the denominator, it will become zero, and the function P will be undefined.
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Case 2: (2x - 1)² = 0
To solve this, we first take the square root of both sides: √(2x - 1)² = √0, which simplifies to 2x - 1 = 0. Adding 1 to both sides gives us 2x = 1, and then dividing by 2, we get x = 1/2. This means that if we plug x = 1/2 into the denominator, it will also become zero, making the function P undefined.
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Exclude the Problematic Values from the Domain:
We've found that x = -1 and x = 1/2 make the denominator zero. Therefore, we must exclude these values from the domain of P. The domain of P is all real numbers except -1 and 1/2.
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Express the Domain in Different Notations:
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Set Notation: {x ∈ ℝ | x ≠ -1, x ≠ 1/2}
This notation reads as "the set of all x that belong to the real numbers such that x is not equal to -1 and x is not equal to 1/2."
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Interval Notation: (-∞, -1) ∪ (-1, 1/2) ∪ (1/2, ∞)
This notation uses intervals to represent the domain. It includes all numbers from negative infinity up to -1 (but not including -1), then all numbers from -1 up to 1/2 (but not including 1/2), and finally all numbers from 1/2 to positive infinity. The ∪ symbol means "union," which combines these intervals.
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Additional Considerations
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The Numerator:
In this particular problem, the numerator (4x² - 4x + 1) doesn't affect the domain because it's a polynomial, and polynomials are defined for all real numbers. However, in other problems, the numerator might contain expressions that could further restrict the domain (e.g., square roots or logarithms).
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Holes vs. Vertical Asymptotes:
It's also worth noting that when a factor cancels out from both the numerator and the denominator, it creates a "hole" in the graph of the function rather than a vertical asymptote. In our case, the numerator can be factored as (2x - 1)², so P can be simplified to (2x-1)²/((x+1)(2x-1)²). The term (2x-1)² can be cancelled. This means there is a hole in x = 1/2.
Practical Examples
Let's consider some practical examples to solidify our understanding. Suppose we want to evaluate P at x = 0. Plugging in x = 0, we get P = (4(0)² - 4(0) + 1) / ((0 + 1)(2(0) - 1)²) = 1 / (1 * 1) = 1. So, P is defined at x = 0.
Now, let's try x = -2. Plugging in x = -2, we get P = (4(-2)² - 4(-2) + 1) / ((-2 + 1)(2(-2) - 1)²) = (16 + 8 + 1) / ((-1)(-5)²) = 25 / (-1 * 25) = -1. So, P is also defined at x = -2.
However, if we try to evaluate P at x = -1 or x = 1/2, we'll find that the denominator becomes zero, and P is undefined.
Final Thoughts
Understanding the domain of a function is crucial in mathematics. It ensures that we're working with valid inputs and outputs. In the case of rational functions, we must always be mindful of the denominator and exclude any values of x that would make it zero. By following a systematic approach, we can confidently determine the domain of any rational function and avoid mathematical pitfalls. So keep an eye on those denominators, and happy calculating!