Let `f (x)= max (x,x ^(2) x ^(3)) in -2 le x le 2.` Then:

Options

A. f (x) is continous in `-2 le x le 2`
B. `f (x)` is not differentiable at `x=1`
C. `f (-1)+f ((3)/(2))=(35)/(8)`
D. `f (-1)f' ((3)/(2)) =(-35)/(4)`

Answer

To solve the problem, we need to analyze the function \( f(x) = \max(x, x^2, x^3) \) over the interval \([-2, 2]\).

### Step-by-Step Solution:

1. **Identify the Functions**:
The function \( f(x) \) is defined as the maximum of three functions: \( x \), \( x^2 \), and \( x^3 \). We will analyze these functions over the interval \([-2, 2]\).

2. **Evaluate the Functions at Key Points**:
We will evaluate \( f(x) \) at some key points within the interval:
- At \( x = -2 \):
\[ f(-2) = \max(-2, (-2)^2, (-2)^3) = \max(-2, 4, -8) = 4 \]
- At \( x = -1 \):
\[ f(-1) = \max(-1, (-1)^2, (-1)^3) = \max(-1, 1, -1) = 1 \]
- At \( x = 0 \):
\[ f(0) = \max(0, 0^2, 0^3) = \max(0, 0, 0) = 0 \]
- At \( x = 1 \):
\[ f(1) = \max(1, 1^2, 1^3) = \max(1, 1, 1) = 1 \]
- At \( x = 2 \):
\[ f(2) = \max(2, 2^2, 2^3) = \max(2, 4, 8) = 8 \]

3. **Determine the Maximum Function Over the Interval**:
Now we need to determine which function is the maximum in different segments of the interval \([-2, 2]\):
- For \( x < 0 \) (specifically between -2 and 0), \( x^2 \) is the maximum function since it is always non-negative and greater than \( x \) and \( x^3 \).
- For \( x = 0 \) to \( x = 1 \), \( x^2 \) and \( x \) are equal at \( x = 1 \), but \( x^2 \) is greater than \( x \) for \( x < 1 \).
- For \( x > 1 \) (specifically between 1 and 2), \( x^3 \) becomes the maximum function since it grows faster than \( x^2 \) and \( x \).

4. **Piecewise Definition of \( f(x) \)**:
Based on the analysis:
\[ f(x) = \begin{cases} x^2 & \text{for } -2 \leq x < 0 \\ x & \text{for } 0 \leq x \leq 1 \\ x^3 & \text{for } 1 < x \leq 2 \end{cases} \]

5. **Check Continuity**:
To check if \( f(x) \) is continuous at the transition points \( x = 0 \) and \( x = 1 \):
- At \( x = 0 \):
\[ \lim_{x \to 0^-} f(x) = 0^2 = 0, \quad f(0) = 0 \]
- At \( x = 1 \):
\[ \lim_{x \to 1^-} f(x) = 1^2 = 1, \quad f(1) = 1 \]
Both limits match the function values, hence \( f(x) \) is continuous.

6. **Check Differentiability**:
To check differentiability at \( x = 1 \):
- The derivative from the left is:
\[ f'(x) = 2x \quad \text{for } x < 1 \implies f'(1^-) = 2(1) = 2 \]
- The derivative from the right is:
\[ f'(x) = 3x^2 \quad \text{for } x > 1 \implies f'(1^+) = 3(1^2) = 3 \]
Since \( f'(1^-) \neq f'(1^+) \), \( f(x) \) is not differentiable at \( x = 1 \).

### Final Results:
- \( f(x) \) is continuous on \([-2, 2]\).
- \( f(x) \) is not differentiable at \( x = 1 \).