Question

The minimum value of the function \( \max\{x, x^2\} \) is equal to:

• A. \( 0 \)
• B. \( 1 \)
• C. \( 2 \)
• D. \( \frac{1}{2} \)
• E. \( \frac{3}{2} \)

Hint:

Visualizing this with a graph helps tremendously. Draw \( y=x \) and \( y=x^2 \) on the same axes. Trace only the parts of the lines that are higher than the other. The lowest point on this "traced" path is your minimum.

Answer 1

The Correct Option is A

Concept: To find the minimum of \( \max\{f(x), g(x)\} \), we analyze where the two functions intersect and which one is "on top" in different intervals. The intersection points of \( y=x \) and \( y=x^2 \) occur at \( x=0 \) and \( x=1 \).

Step 1: Analyze the function behavior in intervals.

• For \( x < 0 \): \( x^2 > x \), so \( \max\{x, x^2\} = x^2 \). As \( x \rightarrow 0 \), \( x^2 \rightarrow 0 \).
• For \( 0 \leq x \leq 1 \): \( x \geq x^2 \), so \( \max\{x, x^2\} = x \).
• For \( x > 1 \): \( x^2 > x \), so \( \max\{x, x^2\} = x^2 \).

Step 2: Find the minimum value of the resulting piece-wise function.
The function \( H(x) = \max\{x, x^2\} \) behaves as follows: - It decreases from \( \infty \) to \( 0 \) on the interval \( (-\infty, 0) \). - It increases from \( 0 \) to \( 1 \) on the interval \( [0, 1] \). - It increases from \( 1 \) to \( \infty \) on the interval \( (1, \infty) \). The overall minimum is clearly at \( x = 0 \), where the value is \( 0 \).