Question

Let \( f(x) = x^2 \), \( x \in [-1, 1] \). Then which of the following are correct?

• A. \( f \) has a minimum at \( x = 0 \).
• B. \( f \) has the maximum at \( x = 1 \).
• C. \( f \) is continuous on \( [-1, 1] \).
• D. \( f \) is bounded on \( [-1, 1] \).

Hint:

Whenever the domain is a closed interval and the function is continuous, by the \textbf{Extreme Value Theorem}, the function must attain both maximum and minimum values, and must be bounded.

Answer 1

The Correct Option is B

Step 1: Function Analysis: \( f(x) = x^2 \) on \( [-1,1] \).
\n\nThe function \( f(x) = x^2 \) represents an upward-opening parabola.
\n\nAt \( x=1 \), \( f(1) = 1 \).
\nAt \( x=-1 \), \( f(-1) = 1 \).
\nTherefore, the maximum value of \( 1 \) is attained at both \( x=1 \) and \( x=-1 \).
\n\n
Step 2: Option Evaluation.
\n\begin{itemize}\n \item (A) \( f \) has a minimum at \( x=0 \). \n \(\quad\) Incorrect. While the minimum is at \( x=0 \), the statement focuses on the maximum value.\n \item (B) \( f \) has the maximum at \( x=1 \). \n \(\quad\) Correct, since \( f(1) = 1 \).\n \item (C) \( f \) is continuous on \( [-1, 1] \). \n \(\quad\) Correct. \( f(x) = x^2 \) is continuous everywhere.\n \item (D) \( f \) is bounded on \( [-1, 1] \). \n \(\quad\) Correct. The function is bounded between 0 and 1.\n\end{itemize}