Let f(x) be defined as follows `f(x)={(x^(6),x^(2)gt1),(x^(3),x^(2)le1):}` Then f(x) is

Options

A. continous everywhere
B. differentiable everywhere
C. discontinuous at `x=-1`
D. not differentiable at x=1

Answer

To solve the problem, we need to analyze the function \( f(x) \) defined as follows:

\[ f(x) = \begin{cases} x^6 & \text{if } x^2 > 1 \\ x^3 & \text{if } x^2 \leq 1 \end{cases} \]

### Step 1: Identify the intervals
The condition \( x^2 > 1 \) implies \( x < -1 \) or \( x > 1 \). The condition \( x^2 \leq 1 \) implies \( -1 \leq x \leq 1 \). Therefore, we can rewrite the function as:

\[ f(x) = \begin{cases} x^6 & \text{if } x < -1 \text{ or } x > 1 \\ x^3 & \text{if } -1 \leq x \leq 1 \end{cases} \]

### Step 2: Check continuity at the points of interest
We need to check the continuity of \( f(x) \) at the points \( x = -1 \) and \( x = 1 \).

**At \( x = -1 \):**
- Left-hand limit:
\[ \lim_{x \to -1^-} f(x) = (-1)^6 = 1 \]
- Right-hand limit:
\[ \lim_{x \to -1^+} f(x) = (-1)^3 = -1 \]
Since the left-hand limit (1) is not equal to the right-hand limit (-1), \( f(x) \) is not continuous at \( x = -1 \).

**At \( x = 1 \):**
- Left-hand limit:
\[ \lim_{x \to 1^-} f(x) = (1)^3 = 1 \]
- Right-hand limit:
\[ \lim_{x \to 1^+} f(x) = (1)^6 = 1 \]
Since both limits are equal, \( f(x) \) is continuous at \( x = 1 \).

### Step 3: Check differentiability at the points of interest
Next, we check the differentiability of \( f(x) \) at \( x = -1 \) and \( x = 1 \).

**At \( x = -1 \):**
- Left-hand derivative:
\[ f'(x) = 6x^5 \quad \text{for } x < -1 \Rightarrow f'(-1) = 6(-1)^5 = -6 \]
- Right-hand derivative:
\[ f'(x) = 3x^2 \quad \text{for } -1 \leq x \leq 1 \Rightarrow f'(-1) = 3(-1)^2 = 3 \]
Since the left-hand derivative (-6) is not equal to the right-hand derivative (3), \( f(x) \) is not differentiable at \( x = -1 \).

**At \( x = 1 \):**
- Left-hand derivative:
\[ f'(x) = 3x^2 \quad \text{for } -1 \leq x \leq 1 \Rightarrow f'(1) = 3(1)^2 = 3 \]
- Right-hand derivative:
\[ f'(x) = 6x^5 \quad \text{for } x > 1 \Rightarrow f'(1) = 6(1)^5 = 6 \]
Since the left-hand derivative (3) is not equal to the right-hand derivative (6), \( f(x) \) is not differentiable at \( x = 1 \).

### Conclusion
- \( f(x) \) is not continuous at \( x = -1 \).
- \( f(x) \) is not differentiable at \( x = -1 \) and \( x = 1 \).

Thus, the correct options are that \( f(x) \) is discontinuous at \( x = -1 \) and not differentiable at \( x = 1 \).

### Final Answer
The function \( f(x) \) is not continuous at \( x = -1 \) and not differentiable at \( x = 1 \).

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