Question

If \[ f(x)=|x-2|+|x+1|, \] then the minimum value of \(f(x)\) is:

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

For expressions of the form \[ |x-a|+|x-b|, \] the minimum value is always \[ |a-b|. \] This fact can save considerable time in competitive examinations.

Answer 1

The Correct Option is C

Concept: Modulus expressions often have a geometric interpretation. The quantity \[ |x-a| \] represents the distance of the point \(x\) from the fixed point \(a\) on the number line. Similarly, \[ |x-2|+|x+1| \] represents the sum of the distances of the point \(x\) from the points \(2\) and \(-1\). A fundamental result from coordinate geometry states that the minimum sum of distances from two fixed points occurs when the moving point lies between those two points. In such a case, the minimum value becomes equal to the distance between the fixed points themselves. This interpretation provides a very elegant way to solve modulus problems.

Step 1: Identify the fixed points. The given expression is \[ |x-2|+|x+1|. \] The fixed points are \[ 2 \] and \[ -1. \]

Step 2: Determine the interval where the minimum occurs. The minimum value occurs when \(x\) lies between the two fixed points. Therefore, \[ -1\le x\le 2. \]

Step 3: Remove the modulus signs in this interval. For \[ -1\le x\le 2, \] we have \[ |x-2|=2-x \] and \[ |x+1|=x+1. \] Hence \[ f(x) = (2-x)+(x+1). \]

Step 4: Simplify the expression. Combining like terms, \[ f(x) = 2-x+x+1. \] \[ = 3. \] Thus throughout the interval \[ [-1,2], \] the function has the constant value \[ 3. \]

Step 5: Verify geometrically. The distance between the fixed points is \[ |2-(-1)| = 3. \] Since the sum of distances can never be less than the direct distance between the points, \(3\) is indeed the minimum possible value.

Step 6: Final Conclusion. Therefore, \[ \boxed{\min f(x)=3} \] Hence the correct answer is \[ \boxed{\text{Option (C)}}. \]