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: Analyze the function \( f(x) = x^2 \) on the interval \( [-1,1] \).
The function \( f(x) = x^2 \) is a parabola opening upwards.
At \( x=1 \), \( f(1) = 1 \).
At \( x=-1 \), \( f(-1) = 1 \).
Thus, the maximum value \( 1 \) occurs at both \( x=1 \) and \( x=-1 \).

Step 2: Check each option.
\begin{itemize} \item (A) \( f \) has a minimum at \( x=0 \). \(\quad\) False, minimum is at \( x=0 \) but the statement about maximum needed. \item (B) \( f \) has the maximum at \( x=1 \). \(\quad\) True, \( f(1) = 1 \). \item (C) \( f \) is continuous on \( [-1, 1] \). \(\quad\) True, \( f(x) = x^2 \) is continuous everywhere. \item (D) \( f \) is bounded on \( [-1, 1] \). \(\quad\) True, bounded between 0 and 1. \end{itemize}