To solve this problem, we need to analyze the function \( h(x) = \min(x, x^2) \). This function takes the minimum of two expressions: \( x \) and \( x^2 \). Let's break this down:
Consider the function \( h(x) = \min(x, x^2) \).
- For \( x > 1 \), \( x^2 > x \). Therefore, \( h(x) = x \).
- For \( 0 \le x \le 1 \), \( x^2 \le x \). Therefore, \( h(x) = x^2 \).
- For \( x < 0 \), \( x^2 > x \). Therefore, \( h(x) = x \).
The next step is to check the continuity of \( h(x) \).
- The functions \( x \) and \( x^2 \) are continuous everywhere.
- The function \( h(x) \) is piecewise continuous using \( x \) and \( x^2 \), so \( h(x) \) is continuous for all \( x \).
Examine differentiability:
For \( x > 1 \) or \( x < 0 \), \( h(x) = x \), which is differentiable with \( h'(x) = 1 \).
For \( 0 < x < 1 \), \( h(x) = x^2 \), which is differentiable with \( h'(x) = 2x \).
Check at key points \( x = 0 \) and \( x = 1 \) where the form of the function changes.
- At \( x = 0 \), from \( x^2 \) (approach from the right) and \( x \) (approach from the left), the derivative is \( 0 \) (from right) and \( 1 \) (from left). The derivatives do not agree, so \( h(x) \) is not differentiable at \( x = 0 \).
- At \( x = 1 \), from \( x \) (approach from the right) and \( x^2 \) (approach from the left), the derivative is \( 1 \) (from right) and \( 2 \times 1 = 2 \) (from left). The derivatives do not agree, so \( h(x) \) is not differentiable at \( x = 1 \).
Thus, the correct answer is: \( h \) is not differentiable at two values of \( x \) (which are \( x = 0 \) and \( x = 1 \)).