Question

The global minimum value of
f(x) = |x - 1| + |x - 2|²
on \(\R\) is equal to _________. (rounded off to two decimal places)

Answer 1

Correct Answer: 0.74 - 0.76

Finding the Global Minimum of \( f(x) = |x - 1| + |x - 2|^2 \) 

To find the global minimum of the function \( f(x) = |x - 1| + |x - 2|^2 \) on \( \mathbb{R} \), we first analyze the behavior of the function in different regions defined by the absolute values: \( x < 1 \), \( 1 \leq x < 2 \), and \( x \geq 2 \).

1. Region \( x < 1 \):

For \( x < 1 \), \( |x - 1| = 1 - x \). Thus, the function becomes:

\[ f(x) = 1 - x + (x - 2)^2 = 1 - x + x^2 - 4x + 4 = x^2 - 5x + 5 \]

The derivative of \( f(x) \) is:

\[ f'(x) = 2x - 5 \]

The critical point is found by setting \( f'(x) = 0 \), which gives \( x = 2.5 \), but this point does not lie in the region \( x < 1 \).

2. Region \( 1 \leq x < 2 \):

For \( 1 \leq x < 2 \), \( |x - 1| = x - 1 \). Thus, the function becomes:

\[ f(x) = x - 1 + (x - 2)^2 = x - 1 + x^2 - 4x + 4 = x^2 - 3x + 3 \]

The derivative of \( f(x) \) is:

\[ f'(x) = 2x - 3 \]

The critical point is found by setting \( f'(x) = 0 \), which gives \( x = 1.5 \). This point lies within the interval \( 1 \leq x < 2 \). Evaluating \( f(1.5) \), we get:

\[ f(1.5) = (1.5)^2 - 3(1.5) + 3 = 2.25 - 4.5 + 3 = 0.75 \]

3. Region \( x \geq 2 \):

For \( x \geq 2 \), \( |x - 1| = x - 1 \). The function is the same as in the previous region:

\[ f(x) = x^2 - 3x + 3 \]

The derivative is the same as before, \( f'(x) = 2x - 3 \), with no new minima.

Critical Points and Function Values:

The critical points and their corresponding function values are:

• At \( x = 0 \), \( f(0) = 5 \)
• At \( x = 1.5 \), \( f(1.5) = 0.75 \)
• At \( x = 2.5 \), \( f(2.5) = 2.25 \)

Conclusion:

The minimum value occurs at \( x = 1.5 \), where \( f(1.5) = 0.75 \). This value is within the given range \( [0.74, 0.76] \). Therefore, the global minimum value is:

0.75