Question

Let $ f(x) = |x^2 - 4x + 3| + |x^2 - 5x + 6| $. The minimum value of $ f(x) $ is:

• A. 2
• B. 3
• C. 1
• D. 0

Hint:

To minimize functions with modulus, break into intervals and evaluate at critical points (like roots of expressions inside modulus).

Answer 1

The Correct Option is D

To find the minimum value of \( f(x) = |x^2 - 4x + 3| + |x^2 - 5x + 6| \), we first consider the expressions inside the absolute value functions. 
\(x^2 - 4x + 3\) can be factored as \((x-1)(x-3)\), and \(x^2 - 5x + 6\) can be factored as \((x-2)(x-3)\). 
Analysis of these factors reveals the critical points: \(x = 1, 2, 3\). These points indicate where the expression inside the absolute values changes sign. We will analyze \(f(x)\) piecewise over the intervals determined by these critical points: 

• For \(x < 1\): Both expressions are positive, so \(f(x) = (x-1)(x-3) + (x-2)(x-3) = 2x^2 - 10x + 15\).
• For \(1 \leq x < 2\): \((x-1)(x-3)\) is non-negative, but \((x-2)(x-3)\) is negative: \(f(x) = (x-1)(x-3) - (x-2)(x-3) = 2(x-3)\).
• For \(2 \leq x < 3\): Both expressions are negative: \(f(x) = -(x-1)(x-3) - (x-2)(x-3) = 10 - 2x^2\).
• For \(x = 3\): \(f(x) = 0\) since both expressions are zero.
• For \(x > 3\): Both expressions are non-negative, so \(f(x) = (x-1)(x-3) + (x-2)(x-3) = 2x^2 - 10x + 15\).

Evaluating these, we find that for \(x = 3\), \(f(x) = 0\). This is the minimum value since the other evaluated values are greater than zero.
Thus, the minimum value of \( f(x) \) is 0.

Answer 2

Factor both expressions: \[ f(x) = |(x - 1)(x - 3)| + |(x - 2)(x - 3)| \] The critical points are at \( x = 1, 2, 3 \). We analyze piecewise:

• Case 1: \( x<1 \) Both expressions are positive:
  \[ f(x) = (1 - x)(3 - x) + (2 - x)(3 - x) \]
  
• Case 2: \( 1 \leq x<2 \) \[ f(x) = (x - 1)(3 - x) + (2 - x)(3 - x) \]
  
• Case 3: \( 2 \leq x<3 \) \[ f(x) = (x - 1)(x - 3) + (x - 2)(x - 3) \]
  
• Case 4: \( x>3 \)
  Both are positive again: \[ f(x) = (x - 1)(x - 3) + (x - 2)(x - 3) \]
  

Now test values at critical points:
- At \( x = 1 \): \( f(1) = |1 - 4 + 3| + |1 - 5 + 6| = |0| + |2| = 2 \)
- At \( x = 2 \): \( f(2) = |4 - 8 + 3| + |4 - 10 + 6| = | -1 | + |0| = 1 \)
- At \( x = 3 \): \( f(3) = |9 - 12 + 3| + |9 - 15 + 6| = |0| + |0| = 0 \)