Consider the following for the next two (02) items that follow : Consider the function f(x) = |x - 2| + |3 - x| + |4 - x|, where x ∈ R.
What is the minimum value of the function ?
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Understanding the Absolute Value Function
The given function is \(f(x) = |x - 2| + |3 - x| + |4 - x|\), where \(x \in R\). This function involves absolute values. Recall that the absolute value \(|a|\) of a real number \(a\) is its distance from zero on the number line. Mathematically, it's defined as:
\[|a| = \begin{cases} a & \text{if } a \ge 0 \\ -a & \text{if } a < 0 \end{cases}\]A function that is a sum of absolute values is piecewise linear and is convex. The minimum value of such a function occurs at one of the "critical points" where the expressions inside the absolute values become zero.
Identifying Critical Points
For the function \(f(x) = |x - 2| + |3 - x| + |4 - x|\), the expressions inside the absolute values are \(x - 2\), \(3 - x\), and \(4 - x\). The critical points are the values of \(x\) for which these expressions are equal to zero:
- \(x - 2 = 0 \implies x = 2\)
- \(3 - x = 0 \implies x = 3\)
- \(4 - x = 0 \implies x = 4\)
So, the critical points are \(x = 2\), \(x = 3\), and \(x = 4\). These points divide the number line into intervals, and the function definition changes in each interval.
Analyzing the Function in Intervals
We can rewrite the function by considering the sign of the expressions \(x-2\), \(3-x\), and \(4-x\) in the intervals defined by the critical points:
- Interval 1: \(x < 2\)
In this interval, \(x - 2 < 0\), \(3 - x > 0\), and \(4 - x > 0\).
\(f(x) = -(x - 2) + (3 - x) + (4 - x) = -x + 2 + 3 - x + 4 - x = -3x + 9\) - Interval 2: \(2 \le x < 3\)
In this interval, \(x - 2 \ge 0\), \(3 - x > 0\), and \(4 - x > 0\).
\(f(x) = (x - 2) + (3 - x) + (4 - x) = x - 2 + 3 - x + 4 - x = -x + 5\) - Interval 3: \(3 \le x < 4\)
In this interval, \(x - 2 \ge 0\), \(3 - x \le 0\), and \(4 - x > 0\).
\(f(x) = (x - 2) - (3 - x) + (4 - x) = x - 2 - 3 + x + 4 - x = x - 1\) - Interval 4: \(x \ge 4\)
In this interval, \(x - 2 \ge 0\), \(3 - x < 0\), and \(4 - x \le 0\).
\(f(x) = (x - 2) - (3 - x) - (4 - x) = x - 2 - 3 + x - 4 + x = 3x - 9\)
Evaluating the Function at Critical Points
The minimum value of a sum of absolute values usually occurs at one of the critical points. Let's evaluate \(f(x)\) at \(x = 2\), \(x = 3\), and \(x = 4\).
- At \(x = 2\):
\(f(2) = |2 - 2| + |3 - 2| + |4 - 2| = |0| + |1| + |2| = 0 + 1 + 2 = 3\) - At \(x = 3\):
\(f(3) = |3 - 2| + |3 - 3| + |4 - 3| = |1| + |0| + |1| = 1 + 0 + 1 = 2\) - At \(x = 4\):
\(f(4) = |4 - 2| + |3 - 4| + |4 - 4| = |2| + |-1| + |0| = 2 + 1 + 0 = 3\)
Comparing the values at the critical points: \(f(2) = 3\), \(f(3) = 2\), and \(f(4) = 3\). The minimum value among these is 2.
Determining the Minimum Value
Based on the analysis of intervals and the evaluation at critical points, the function \(f(x)\) behaves as follows:
- For \(x < 2\), \(f(x) = -3x + 9\) (decreasing). As \(x \to 2^-\), \(f(x) \to 3\).
- For \(2 \le x < 3\), \(f(x) = -x + 5\) (decreasing). At \(x=2\), \(f(2)=3\). As \(x \to 3^-\), \(f(x) \to 2\).
- For \(3 \le x < 4\), \(f(x) = x - 1\) (increasing). At \(x=3\), \(f(3)=2\). As \(x \to 4^-\), \(f(x) \to 3\).
- For \(x \ge 4\), \(f(x) = 3x - 9\) (increasing). At \(x=4\), \(f(4)=3\).
The function decreases until \(x = 3\) and then increases. This confirms that the minimum value occurs at \(x = 3\).
The minimum value of the function \(f(x) = |x - 2| + |3 - x| + |4 - x|\) is \(f(3) = 2\).
Revision Table: Key Concepts for Minimum Value of Absolute Functions
| Concept | Description | Relevance to the Problem |
|---|---|---|
| Absolute Value | \(|a|\) is the distance of \(a\) from zero; \(|a|=a\) if \(a \ge 0\), \(|a|=-a\) if \(a < 0\). | Used to define the function \(f(x)\). Understanding its definition is key to analyzing the function in intervals. |
| Critical Points | Values of the variable where the expression inside an absolute value is zero. For \(|ax+b|\), the critical point is \(x = -b/a\). | These points (\(x=2, 3, 4\)) divide the domain into intervals where the function can be written without absolute values. The minimum value often occurs at one of these points. |
| Piecewise Function | A function defined by multiple sub-functions, each applying to a certain interval of the domain. | The given function \(f(x)\) is a piecewise linear function defined by different expressions in the intervals \(x < 2\), \(2 \le x < 3\), \(3 \le x < 4\), and \(x \ge 4\). |
| Minimum Value | The lowest value that a function attains in its domain. | The objective of the problem is to find this value for the given function \(f(x)\). For sums of absolute values, checking critical points is an efficient way to find the minimum. |
Additional Information: Properties of Sums of Absolute Values
Functions formed by the sum of absolute values of linear expressions, like \(f(x) = \sum_{i=1}^n |x - a_i|\), where \(a_1 < a_2 < ... < a_n\), have interesting properties:
- They are continuous everywhere.
- They are convex functions.
- The graph is piecewise linear.
- The minimum value occurs at the median of the values \(a_i\).
- If \(n\) is odd, the minimum occurs uniquely at the median \(a_{(n+1)/2}\).
- If \(n\) is even, the minimum occurs at any point in the interval \([a_{n/2}, a_{(n/2)+1}]\).
In our case, the function is \(f(x) = |x - 2| + |x - 3| + |x - 4|\) (since \(|3-x|=|x-3|\) and \(|4-x|=|x-4|\)). The values \(a_i\) are 2, 3, and 4. There are \(n=3\) values, which is odd. The median is the middle value, which is 3. According to the property, the minimum should occur at \(x=3\), which matches our calculation. The minimum value is \(f(3) = 2\).
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