Question

Absolute minimum value of \( f(x) = (x - 2)^2 + 5 \) in the interval \( [-3, 2] \) is :

• A. \( -3 \)
• B. \( 2 \)
• C. \( 5 \)
• D. \( 30 \)

Hint:

Since a squared expression \( (x - 2)^2 \) is always non-negative (\( \ge 0 \)), its minimum possible value is \( 0 \), which occurs at \( x = 2 \). Thus, the absolute minimum value is simply \( 0 + 5 = 5 \).

Answer 1

The Correct Option is C

Concept: To determine the absolute maximum or absolute minimum value of a continuous function on a closed interval \( [a, b] \), we must find the values of the function at all its critical points located inside the interval, as well as evaluate it directly at the endpoints \( x = a \) and \( x = b \).

Step 1: Locate critical points by differentiating \( f(x) \). The function given is: \[ f(x) = (x - 2)^2 + 5 \] Differentiating with respect to \( x \): \[ f'(x) = 2(x - 2) \cdot \frac{d}{dx}(x - 2) = 2(x - 2)(1) = 2(x - 2) \] To find the critical points, set the first derivative equal to zero: \[ 2(x - 2) = 0 \implies x - 2 = 0 \implies x = 2 \] Notice that \( x = 2 \) is a critical point that serves simultaneously as the right endpoint of our given interval \( [-3, 2] \).

Step 2: Evaluate the function at the boundary endpoints. We need to test the function's values at \( x = -3 \) and at \( x = 2 \):
• At \( x = -3 \): \[ f(-3) = (-3 - 2)^2 + 5 = (-5)^2 + 5 = 25 + 5 = 30 \]
• At \( x = 2 \): \[ f(2) = (2 - 2)^2 + 5 = (0)^2 + 5 = 0 + 5 = 5 \]

Step 3: Compare values to find the absolute minimum. Comparing the computed values: \[ f(-3) = 30, \quad f(2) = 5 \] The smallest value among these is \( 5 \). Hence, the absolute minimum value of the function over the closed interval is \( 5 \).