Let `h(x) = "min" {x,x^(2)}` , for every real number of x. Then

Options

A. h continuous for all x
B. h is differentiable for all x
C. `h'(x) =1` for all `x gt 1`
D. h is not differentiable at two values of x

Answer

To solve the problem, we need to analyze the function \( h(x) = \min \{ x, x^2 \} \) for all real numbers \( x \). We will determine the behavior of this function in different intervals and check its continuity and differentiability.

### Step 1: Define the function in different intervals

1. **For \( x < 0 \)**:
- Here, \( x^2 \) is always positive and greater than \( x \). Therefore, \( h(x) = \min \{ x, x^2 \} = x \).

2. **For \( 0 \leq x < 1 \)**:
- In this interval, \( x^2 < x \) (since \( x^2 \) is less than \( x \) for \( 0 < x < 1 \)). Thus, \( h(x) = \min \{ x, x^2 \} = x^2 \).

3. **For \( x \geq 1 \)**:
- Here, \( x^2 \) is greater than \( x \). Therefore, \( h(x) = \min \{ x, x^2 \} = x \).

Combining these, we can summarize the function as:
\[ h(x) = \begin{cases} x & \text{if } x < 0 \\ x^2 & \text{if } 0 \leq x < 1 \\ x & \text{if } x \geq 1 \end{cases} \]

### Step 2: Check for continuity

To check if \( h(x) \) is continuous, we need to check the limits at the points where the definition of the function changes, specifically at \( x = 0 \) and \( x = 1 \).

1. **At \( x = 0 \)**:
- \( \lim_{x \to 0^-} h(x) = 0 \) (from the left, \( h(x) = x \))
- \( \lim_{x \to 0^+} h(x) = 0 \) (from the right, \( h(x) = x^2 \))
- Since both limits equal \( h(0) = 0 \), \( h(x) \) is continuous at \( x = 0 \).

2. **At \( x = 1 \)**:
- \( \lim_{x \to 1^-} h(x) = 1 \) (from the left, \( h(x) = x^2 \))
- \( \lim_{x \to 1^+} h(x) = 1 \) (from the right, \( h(x) = x \))
- Since both limits equal \( h(1) = 1 \), \( h(x) \) is continuous at \( x = 1 \).

Since \( h(x) \) is continuous at both points \( x = 0 \) and \( x = 1 \), we conclude that \( h(x) \) is continuous for all \( x \).

### Step 3: Check for differentiability

To check for differentiability, we need to look at the points where the function changes its definition, specifically at \( x = 0 \) and \( x = 1 \).

1. **At \( x = 0 \)**:
- The left-hand derivative: \( h'(0^-) = 1 \) (derivative of \( h(x) = x \))
- The right-hand derivative: \( h'(0^+) = 0 \) (derivative of \( h(x) = x^2 \))
- Since the left-hand and right-hand derivatives are not equal, \( h(x) \) is not differentiable at \( x = 0 \).

2. **At \( x = 1 \)**:
- The left-hand derivative: \( h'(1^-) = 2 \) (derivative of \( h(x) = x^2 \))
- The right-hand derivative: \( h'(1^+) = 1 \) (derivative of \( h(x) = x \))
- Since the left-hand and right-hand derivatives are not equal, \( h(x) \) is not differentiable at \( x = 1 \).

### Conclusion

- \( h(x) \) is continuous for all \( x \).
- \( h(x) \) is not differentiable at \( x = 0 \) and \( x = 1 \).

### Final Answer

1. \( h \) is continuous for all \( x \) (True).
2. \( h \) is differentiable for all \( x \) (False).
3. \( h \) is not differentiable at 2 values of \( x \) (True).
4. \( h' \) for \( x > 1 \) is \( 1 \) (True).