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ID:
Class: 12
Subject: Math
Topic: Application of Derivative
Type: Short
Question:
Find the absolute maximum and minimum values of the function \( f(x) = (x - 1)^2 + 3 \) in the interval \( [-3, 1] \).
Official Solution
Explanation:
We have \( f(x) = (x - 1)^2 + 3 \)
\[
f'(x) = 2(x - 1) \Rightarrow f'(x) = 0 \Rightarrow x = 1
\]
Now evaluate \( f \) at endpoints and critical point:
\[
f(-3) = (-3 - 1)^2 + 3 = 16 + 3 = 19 \\
f(1) = (1 - 1)^2 + 3 = 0 + 3 = 3
\]
\(\textbf{Absolute maximum}\) is \( 19 \) at \( x = -3 \) and \(\textbf{Absolute minimum}\) is \( 3 \) at \( x = 1 \)
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