Unique Solution: Finding M For |x-1| + |x+1| = Mx + 1
Hey guys! Let's dive into a cool math problem today: finding the values of m that make the equation |x-1| + |x+1| = mx + 1 have only one real solution. This might sound tricky, but we'll break it down step by step so it's super easy to understand. We're going to explore how absolute values work and how they interact with linear equations. So, buckle up and let's get started!
Understanding Absolute Value Equations
To tackle this problem effectively, we first need to understand the behavior of absolute value functions. Remember, the absolute value of a number is its distance from zero, so |x| is x if x is positive or zero, and -x if x is negative. This means that when we have absolute value expressions in an equation, we often need to consider different cases depending on the value of x. Specifically, for our equation |x-1| + |x+1| = mx + 1, the critical points are x = 1 and x = -1 because these are the points where the expressions inside the absolute values change sign.
- Critical Points and Cases: Identifying the critical points is crucial. These points divide the number line into intervals where the absolute value expressions have different behaviors. In our case, the critical points x = -1 and x = 1 split the number line into three intervals: x < -1, -1 ≤ x ≤ 1, and x > 1. We'll analyze the equation separately in each of these intervals to remove the absolute value signs and solve the resulting linear equations. Understanding these cases is fundamental to solving absolute value equations, so make sure you're comfortable with this concept. By breaking down the problem into manageable chunks, we can systematically find the solutions in each interval and then combine them to find the overall solution. Let's look at the first case and see how it unfolds. Remember, the key is to carefully consider the sign of the expressions inside the absolute values within each interval.
Case 1: x < -1
In this interval, both (x - 1) and (x + 1) are negative. This means that |x - 1| = -(x - 1) = 1 - x and |x + 1| = -(x + 1) = -x - 1. So, our equation becomes:
1 - x + (-x - 1) = mx + 1
Simplifying this, we get:
-2x = mx + 1
Rearranging the terms, we have:
(-2 - m)x = 1
Now, if (-2 - m) ≠0, then we can solve for x:
x = 1 / (-2 - m)
For this solution to be valid, it must satisfy the condition x < -1. So,
1 / (-2 - m) < -1
This inequality will give us a range of values for m for which there is a solution in this interval. We need to be careful when dealing with inequalities and fractions, especially when the denominator can be negative. To solve this inequality, we can consider two sub-cases: when -2 - m is positive and when it's negative. This ensures we don't accidentally flip the inequality sign when multiplying or dividing.
Case 2: -1 ≤ x ≤ 1
Here, (x - 1) is negative or zero, so |x - 1| = 1 - x, and (x + 1) is non-negative, so |x + 1| = x + 1. Our equation becomes:
1 - x + x + 1 = mx + 1
Simplifying, we get:
2 = mx + 1
Rearranging, we have:
mx = 1
If m ≠0, then:
x = 1 / m
For this solution to be valid, it must satisfy the condition -1 ≤ x ≤ 1. Thus,
-1 ≤ 1 / m ≤ 1
This inequality will give us another range of values for m for which there is a solution in this interval. Remember, the sign of m plays a critical role here. If m is positive, the inequality behaves differently than if m is negative. We need to analyze both situations to get the correct range of m values. Solving inequalities involving fractions requires careful attention to detail, but it's a crucial skill for this type of problem.
Case 3: x > 1
In this interval, both (x - 1) and (x + 1) are positive. So, |x - 1| = x - 1 and |x + 1| = x + 1. Our equation becomes:
x - 1 + x + 1 = mx + 1
Simplifying, we get:
2x = mx + 1
Rearranging the terms, we have:
(2 - m)x = 1
If (2 - m) ≠0, then:
x = 1 / (2 - m)
For this solution to be valid, it must satisfy the condition x > 1. So,
1 / (2 - m) > 1
This inequality will give us the range of m values for which there is a solution in this interval. Again, we need to consider the sign of (2 - m) when solving this inequality. Just like in the first case, understanding how the sign of the denominator affects the inequality is essential to finding the correct solution. Let's move on to the next step and see how we combine these findings to get the final answer.
Analyzing for a Unique Solution
Now that we have solutions (or conditions for solutions) in each interval, the real challenge is to find the values of m for which the equation has exactly one solution. This means we need to analyze the solutions from each case and see when they overlap or when only one of them is valid.
To ensure a unique solution, we need to consider the following:
- Overlapping Solutions: If we find solutions in multiple intervals, we need to check if they are the same. If they are, that counts as one solution. If they are different, then we have more than one solution, which we don't want.
- Validity within Intervals: Each solution must fall within the interval it was derived from. If a solution doesn't satisfy the interval condition, it's not a valid solution.
- No Solution: It's also possible that for some values of m, there might be no solution in one or more intervals. This can help us narrow down the possible values of m for a unique solution.
Let's take a look at the conditions we derived in each case and see how they play out.
Combining the Conditions
From Case 1 (x < -1), we have x = 1 / (-2 - m) and the condition 1 / (-2 - m) < -1. Solving this inequality involves considering the sign of -2 - m. If -2 - m > 0 (i.e., m < -2), then 1 < -1*(-2 - m*), which simplifies to 1 < 2 + m, or m > -1. This contradicts our initial assumption of m < -2, so there are no solutions in this sub-case. If -2 - m < 0 (i.e., m > -2), then 1 > -1*(-2 - m*), which simplifies to 1 > 2 + m, or m < -1. Combining this with m > -2, we have -2 < m < -1. So, in this interval, we might have a solution if m is within this range.
From Case 2 (-1 ≤ x ≤ 1), we have x = 1 / m and the condition -1 ≤ 1 / m ≤ 1. This gives us two inequalities to consider: -1 ≤ 1 / m and 1 / m ≤ 1. For -1 ≤ 1 / m, if m > 0, then -m ≤ 1, so m ≥ -1 (which is always true since we assumed m > 0). If m < 0, then -m ≥ 1, so m ≤ -1. For 1 / m ≤ 1, if m > 0, then 1 ≤ m, so m ≥ 1. If m < 0, then 1 ≥ m, which is always true since we assumed m < 0. Combining these, we see that m ≥ 1 or m ≤ -1.
From Case 3 (x > 1), we have x = 1 / (2 - m) and the condition 1 / (2 - m) > 1. Solving this inequality involves considering the sign of 2 - m. If 2 - m > 0 (i.e., m < 2), then 1 > 2 - m, which simplifies to m > 1. So, we have 1 < m < 2. If 2 - m < 0 (i.e., m > 2), then 1 < 2 - m, which simplifies to m < 1. This contradicts our assumption of m > 2, so there are no solutions in this sub-case. Therefore, in this interval, we might have a solution if 1 < m < 2.
Finding the Unique Solution
To have a unique solution, we need to find the values of m where only one of these cases gives us a valid solution. Let's analyze the intervals:
- Case 1 (-2 < m < -1): We have a potential solution here.
- Case 2 (m ≤ -1 or m ≥ 1): We have potential solutions here.
- Case 3 (1 < m < 2): We have a potential solution here.
Notice that if m = -1, from Case 2, x = -1, which is the boundary value. Plugging m = -1 into Case 1, we have x = 1 / (-2 - (-1)) = -1, which is the same solution. So, m = -1 gives us a unique solution x = -1.
If m = 1, from Case 2, x = 1, which is the boundary value. Plugging m = 1 into Case 3, we have x = 1 / (2 - 1) = 1, which is the same solution. So, m = 1 gives us a unique solution x = 1.
Now, let's consider the intervals:
- If -2 < m < -1, we have a solution in Case 1. We need to make sure there's no solution in Cases 2 and 3. For Case 2, m ≤ -1, so this interval doesn't overlap with -2 < m < -1. For Case 3, 1 < m < 2, which also doesn't overlap. So, these values of m give a unique solution.
- If 1 < m < 2, we have a solution in Case 3. We need to make sure there's no solution in Cases 1 and 2. Case 1 doesn't overlap. For Case 2, m ≥ 1, so there's a potential overlap. However, we need -1 ≤ 1 / m ≤ 1, which means 1 ≤ m. So, these values of m give a unique solution.
The Final Set M
So, guys, after all this analysis, we've found that the set M of values for m for which the equation has a unique real solution is:
M = {-1} ∪ {1} ∪ (-2, -1) ∪ (1, 2)
Which simplifies to:
M = (-2, -1] ∪ [1, 2)
This means that for any m in this set, the original equation |x-1| + |x+1| = mx + 1 will have exactly one real solution. How cool is that? We took a seemingly complex problem and broke it down using careful analysis and a bit of algebraic manipulation. Remember, the key to solving problems like this is to understand the fundamental concepts and take it one step at a time. Keep practicing, and you'll become a math whiz in no time!