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.
At what value of x does the function attain minimum value ?
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Finding the Minimum Value of an Absolute Value Function
The given function is \(f(x) = |x - 2| + |3 - x| + |4 - x|\), where \(x\) belongs to the set of real numbers, \(\mathbb{R}\). We are asked to find the value of \(x\) at which this function attains its minimum value.
This function is a sum of absolute value terms. Each term \(|x - a|\) represents the distance between \(x\) and the point \(a\) on the number line. So, \(f(x)\) represents the sum of the distances from \(x\) to the points 2, 3, and 4.
Understanding Absolute Value Functions
The absolute value function \(|u|\) is defined as:
- \(|u| = u\), if \(u \ge 0\)
- \(|u| = -u\), if \(u < 0\)
To analyze the function \(f(x)\), we need to consider the points where the expressions inside the absolute values become zero. These are called critical points. For \(f(x) = |x - 2| + |3 - x| + |4 - x|\), the critical points are:
- \(x - 2 = 0 \implies x = 2\)
- \(3 - x = 0 \implies x = 3\)
- \(4 - x = 0 \implies x = 4\)
These critical points (2, 3, and 4) divide the number line into intervals. We can analyze the function in each interval by removing the absolute value signs based on the sign of the expression inside.
Analyzing the Function in Intervals
Let's look at the function's form in different intervals:
- Interval 1: \(x < 2\)
In this interval, \(x-2 < 0\), \(3-x > 0\), and \(4-x > 0\).
\(f(x) = -(x - 2) + (3 - x) + (4 - x) = -x + 2 + 3 - x + 4 - x = 9 - 3x\)
In this interval, the function is decreasing.
- Interval 2: \(2 \le x < 3\)
In this interval, \(x-2 \ge 0\), \(3-x > 0\), and \(4-x > 0\).
\(f(x) = (x - 2) + (3 - x) + (4 - x) = x - 2 + 3 - x + 4 - x = 5 - x\)
In this interval, the function is decreasing.
- Interval 3: \(3 \le x < 4\)
In this interval, \(x-2 > 0\), \(3-x \le 0\), and \(4-x > 0\).
\(f(x) = (x - 2) - (3 - x) + (4 - x) = x - 2 - 3 + x + 4 - x = x - 1\)
In this interval, the function is increasing.
- Interval 4: \(x \ge 4\)
In this interval, \(x-2 \ge 0\), \(3-x \le 0\), and \(4-x \le 0\).
\(f(x) = (x - 2) - (3 - x) - (4 - x) = x - 2 - 3 + x - 4 + x = 3x - 9\)
In this interval, the function is increasing.
We can see that the function decreases up to \(x=3\) and increases from \(x=3\) onwards. This indicates that the minimum value occurs at \(x = 3\). Let's calculate the value of the function at the critical points to confirm.
- At \(x = 2\): \(f(2) = |2 - 2| + |3 - 2| + |4 - 2| = 0 + 1 + 2 = 3\)
- At \(x = 3\): \(f(3) = |3 - 2| + |3 - 3| + |4 - 3| = 1 + 0 + 1 = 2\)
- At \(x = 4\): \(f(4) = |4 - 2| + |3 - 4| + |4 - 4| = 2 + |-1| + 0 = 2 + 1 + 0 = 3\)
Comparing the values, \(f(3) = 2\) is the smallest value among \(f(2)=3\), \(f(3)=2\), and \(f(4)=3\). Also, observing the function's behavior in intervals confirms that the minimum is at \(x=3\).
The Median Rule for Minimum Value
For a function of the form \(f(x) = \sum_{i=1}^n |x - a_i|\), where \(a_1 < a_2 < \dots < a_n\), the minimum value is attained at the median of the values \(a_1, a_2, \dots, a_n\).
In this problem, the points are 2, 3, and 4. When sorted, they are 2, 3, 4. The number of points is \(n=3\), which is odd. The median is the middle value, which is the \((3+1)/2 = 2\)nd value, which is 3.
According to the median rule, the function \(f(x) = |x - 2| + |3 - x| + |4 - x|\) attains its minimum value at \(x = 3\).
The function attains its minimum value at \(x=3\).
| Value of \(x\) | \(|x - 2|\) | \(|3 - x|\) | \(|4 - x|\) | \(f(x)\) |
|---|---|---|---|---|
| 2 | \(|2 - 2| = 0\) | \(|3 - 2| = 1\) | \(|4 - 2| = 2\) | \(0 + 1 + 2 = 3\) |
| 3 | \(|3 - 2| = 1\) | \(|3 - 3| = 0\) | \(|4 - 3| = 1\) | \(1 + 0 + 1 = 2\) |
| 4 | \(|4 - 2| = 2\) | \(|3 - 4| = |-1| = 1\) | \(|4 - 4| = 0\) | \(2 + 1 + 0 = 3\) |
The table also shows that the minimum value of \(f(x)\) is 2, which is attained at \(x = 3\).
Revision Table: Minimum of Absolute Value Functions
| Concept | Description |
|---|---|
| Absolute Value \(|a|\) | Distance of \(a\) from 0 on the number line. Defined as \(a\) if \(a \ge 0\) and \(-a\) if \(a < 0\). |
| \(|x - a|\) | Distance between \(x\) and \(a\) on the number line. |
| Function \(f(x) = \sum |x - a_i|\) | Sum of distances from \(x\) to multiple points \(a_i\). |
| Critical Points | Values of \(x\) where the expression inside an absolute value becomes zero. These are the points \(a_i\). |
| Finding Minimum | The minimum value of \(f(x) = \sum_{i=1}^n |x - a_i|\) is attained at the median of the points \(a_1, a_2, \dots, a_n\). |
Additional Information: Properties of Absolute Value Functions
Functions that are sums of absolute values, like the one discussed, are convex functions. A convex function has a single minimum (or a range of points where the minimum is attained). For the sum of distances function \(f(x) = \sum_{i=1}^n |x - a_i|\), the critical points \(a_i\) are where the function changes its slope. The slope of \(f(x)\) increases by 2 each time \(x\) crosses a point \(a_i\) from left to right.
Let's check the slopes in our example:
- \(x < 2\): \(f(x) = 9 - 3x\), slope is -3.
- \(2 < x < 3\): \(f(x) = 5 - x\), slope is -1. (Slope increased by -1 - (-3) = 2 at x=2)
- \(3 < x < 4\): \(f(x) = x - 1\), slope is 1. (Slope increased by 1 - (-1) = 2 at x=3)
- \(x > 4\): \(f(x) = 3x - 9\), slope is 3. (Slope increased by 3 - 1 = 2 at x=4)
The minimum of such a function occurs at the point where the slope changes from negative to positive. This transition point is precisely the median of the \(a_i\) values. If there is a single median (for an odd number of points), the minimum is unique. If there is an interval of medians (for an even number of points), the minimum is attained at any point within that interval.
Important Questions from Special Functions
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