Question:
Absolute minimum value of \( f(x) = (x - 2)^2 + 5 \) in the interval \( [-3, 2] \) is :
Absolute minimum value of \( f(x) = (x - 2)^2 + 5 \) in the interval \( [-3, 2] \) is :
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For quadratic functions in the form \( (x-h)^2 + k \), the vertex is at \( (h, k) \). Since it's a "upward" parabola, the minimum value is simply \( k \) if \( h \) falls within your interval.
Updated On: Sep 10, 2026
- \( -3 \)
- \( 2 \)
- \( 5 \)
- \( 30 \)
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The Correct Option is C
Solution and Explanation
Concept:
• For a continuous function on a closed interval, the absolute minimum occurs either at the critical points (where \( f'(x) = 0 \)) or at the endpoints of the interval.
Step 1: Find the critical points of the function
Differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx} ((x - 2)^2 + 5) = 2(x - 2) \]
Set the derivative to zero to find critical points:
\[ 2(x - 2) = 0 \implies x = 2 \]
The critical point \( x = 2 \) is within the given interval \( [-3, 2] \).
Step 2: Evaluate the function at the critical point and endpoints
The critical point and one endpoint are the same (\( x = 2 \)).
1. At \( x = 2 \):
\[ f(2) = (2 - 2)^2 + 5 = 0 + 5 = 5 \]
2. At \( x = -3 \):
\[ f(-3) = (-3 - 2)^2 + 5 = (-5)^2 + 5 = 25 + 5 = 30 \]
Step 3: Identify the absolute minimum value
Comparing the values \( \{5, 30\} \), the smallest value is 5.
Therefore, the absolute minimum value is 5.
• For a continuous function on a closed interval, the absolute minimum occurs either at the critical points (where \( f'(x) = 0 \)) or at the endpoints of the interval.
Step 1: Find the critical points of the function
Differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx} ((x - 2)^2 + 5) = 2(x - 2) \]
Set the derivative to zero to find critical points:
\[ 2(x - 2) = 0 \implies x = 2 \]
The critical point \( x = 2 \) is within the given interval \( [-3, 2] \).
Step 2: Evaluate the function at the critical point and endpoints
The critical point and one endpoint are the same (\( x = 2 \)).
1. At \( x = 2 \):
\[ f(2) = (2 - 2)^2 + 5 = 0 + 5 = 5 \]
2. At \( x = -3 \):
\[ f(-3) = (-3 - 2)^2 + 5 = (-5)^2 + 5 = 25 + 5 = 30 \]
Step 3: Identify the absolute minimum value
Comparing the values \( \{5, 30\} \), the smallest value is 5.
Therefore, the absolute minimum value is 5.
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