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 determine the minimum value of \( f(x) = |x^2 - 4x + 3| + |x^2 - 5x + 6| \), we first factor the expressions within the absolute value functions. \(x^2 - 4x + 3\) factors to \((x-1)(x-3)\), and \(x^2 - 5x + 6\) factors to \((x-2)(x-3)\).
The critical points where these expressions change sign are \(x = 1, 2, 3\). We analyze \(f(x)\) piecewise based on these critical points: 

• For \(x < 1\): Both \( (x-1)(x-3) \) and \( (x-2)(x-3) \) are positive. Thus, \(f(x) = (x-1)(x-3) + (x-2)(x-3) = 2x^2 - 10x + 15\).
• For \(1 \leq x < 2\): \( (x-1)(x-3) \) is non-negative, while \( (x-2)(x-3) \) is negative. Thus, \(f(x) = (x-1)(x-3) - (x-2)(x-3) = 2(x-3)\).
• For \(2 \leq x < 3\): Both \( (x-1)(x-3) \) and \( (x-2)(x-3) \) are negative. Thus, \(f(x) = -(x-1)(x-3) - (x-2)(x-3) = 10 - 2x^2\).
• For \(x = 3\): Both expressions are zero, resulting in \(f(x) = 0\).
• For \(x > 3\): Both \( (x-1)(x-3) \) and \( (x-2)(x-3) \) are non-negative. Thus, \(f(x) = (x-1)(x-3) + (x-2)(x-3) = 2x^2 - 10x + 15\).

Evaluating \(f(x)\) at \(x = 3\) yields \(f(3) = 0\). Since other evaluated values are greater than zero, this is the minimum value.
Therefore, the minimum value of \( f(x) \) is 0.