If f(x) = x5sgn(sgn(x)), where sgn(.) denotes signum function, then which is/are correct

f(x) is discontinuous at x = 0

f(x) is continuous but not differentiable at x = 0

f '(0+) = 0

f(x) is continuous as well as differentiable at x = 0

Given the function \( f(x) = x^5 \cdot \text{sgn}(\text{sgn}(x)) \), where \(\text{sgn}(.)\) denotes the signum function, let's analyze the properties of \( f(x) \).

### Step 1: Simplify \( f(x) \)
First, recall the definition of the signum function:
\[
\text{sgn}(x) = \begin{cases} 
1 & \text{if } x > 0 \\
0 & \text{if } x = 0 \\
-1 & \text{if } x < 0 
\end{cases}
\]

Then, \(\text{sgn}(\text{sgn}(x))\) can be analyzed as:
\[
\text{sgn}(\text{sgn}(x)) = \begin{cases} 
1 & \text{if } x \neq 0 \\
0 & \text{if } x = 0 
\end{cases}
\]

So, the function \( f(x) \) simplifies to:
\[
f(x) = \begin{cases} 
x^5 & \text{if } x \neq 0 \\
0 & \text{if } x = 0 
\end{cases}
\]

### Step 2: Continuity at \( x = 0 \)
To check the continuity of \( f(x) \) at \( x = 0 \), we need to verify that:
\[
\lim_{x \to 0} f(x) = f(0)
\]

Since \( f(0) = 0 \), we need to compute the limit:
\[
\lim_{x \to 0} f(x) = \lim_{x \to 0} x^5 = 0
\]

Thus,
\[
\lim_{x \to 0} f(x) = f(0) = 0
\]
Therefore, \( f(x) \) is continuous at \( x = 0 \).

### Step 3: Differentiability at \( x = 0 \)
To check differentiability, we need to evaluate:
\[
f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h^5 - 0}{h} = \lim_{h \to 0} h^4 = 0
\]

So, \( f'(0) = 0 \).

Next, we check the differentiability from both sides. For \( x > 0 \):
\[
f'(x) = \frac{d}{dx}(x^5) = 5x^4
\]
For \( x < 0 \):
\[
f'(x) = \frac{d}{dx}(-x^5) = -5x^4
\]

At \( x = 0 \), the derivative from the left-hand side and right-hand side is:
\[
f'(0^+) = \lim_{x \to 0^+} 5x^4 = 0
\]
\[
f'(0^-) = \lim_{x \to 0^-} -5x^4 = 0
\]

Therefore, the derivative exists and is 0 at \( x = 0 \). 

### Step 4: Summary
Given the analysis, we can conclude that:
1. \( f(x) \) is continuous at \( x = 0 \).
2. \( f(x) \) is differentiable at \( x = 0 \) with \( f'(0) = 0 \).

Hence, the correct statements are:
- \( f(x) \) is continuous as well as differentiable at \( x = 0 \).

### Final Answer
The correct statement is:
\[ \text{f(x) is continuous as well as differentiable at } x = 0. \]

Would you like more details or have any questions?

Here are 8 related questions:
1. How is the signum function defined mathematically?
2. What is the significance of differentiability in a function?
3. Can you give an example of a function that is continuous but not differentiable at a point?
4. How do you determine the continuity of a piecewise function?
5. What are the steps to check the differentiability of a function at a point?
6. Why is the limit definition used to find the derivative?
7. Can a function be differentiable at a point but have different left-hand and right-hand derivatives?
8. How would \( f(x) \) change if \( f(0) \neq 0 \)?

**Tip:** Always verify the continuity of a function at a point before checking for differentiability, as differentiability implies continuity.

Explore the function f(x) = x^5sgn(sgn(x)) and analyze its continuity and differentiability properties at x = 0. This detailed analysis covers the signum function, limit definitions, and derivative calculations, suitable for advanced high school students and beyond.

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