A function `f :R to R` is given by `f (x) = {{:(x ^(4) (2+ sin ""(1)/(x)), x ne0),(0, x=0):},` then

Options

A. f has a continous derivative `AA x in R`
B. f is a bounded function
C. f has an global minimum at `x=0`
D. f" is continous `AA x in R`

Answer

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

\[ f(x) = \begin{cases} x^4(2 + \sin(1/x)) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

We will find the first derivative \( f'(x) \) and the second derivative \( f''(x) \), and then analyze the continuity and boundedness of the function and its derivatives.

### Step 1: Find the first derivative \( f'(x) \)

For \( x \neq 0 \), we can use the product rule to differentiate \( f(x) \):

\[ f'(x) = \frac{d}{dx}[x^4(2 + \sin(1/x))] \]

Using the product rule \( (u \cdot v)' = u'v + uv' \):

- Let \( u = x^4 \) and \( v = 2 + \sin(1/x) \).
- Then \( u' = 4x^3 \) and \( v' = \cos(1/x) \cdot (-1/x^2) \).

Thus,

\[ f'(x) = 4x^3(2 + \sin(1/x)) + x^4\left(-\frac{\cos(1/x)}{x^2}\right) \]

Simplifying this gives:

\[ f'(x) = 4x^3(2 + \sin(1/x)) - x^2 \cos(1/x) \]

For \( x = 0 \), we need to find the limit of \( f'(x) \) as \( x \) approaches 0:

\[ f'(0) = \lim_{x \to 0} f'(x) = \lim_{x \to 0} \left(4x^3(2 + \sin(1/x)) - x^2 \cos(1/x)\right) \]

As \( x \to 0 \), \( \sin(1/x) \) oscillates between -1 and 1, and \( \cos(1/x) \) also oscillates between -1 and 1. However, both terms \( 4x^3(2 + \sin(1/x)) \) and \( -x^2 \cos(1/x) \) approach 0. Therefore:

\[ f'(0) = 0 \]

### Step 2: Find the second derivative \( f''(x) \)

Now we differentiate \( f'(x) \):

\[ f'(x) = 4x^3(2 + \sin(1/x)) - x^2 \cos(1/x) \]

Using the product rule again:

\[ f''(x) = \frac{d}{dx}[4x^3(2 + \sin(1/x))] - \frac{d}{dx}[x^2 \cos(1/x)] \]

Calculating each part:

1. For \( 4x^3(2 + \sin(1/x)) \):
- Differentiate using product rule:
\[ f''(x) = 12x^2(2 + \sin(1/x)) + 4x^3\left(\frac{\cos(1/x)}{x^2}\right) \]

2. For \( -x^2 \cos(1/x) \):
- Differentiate using product rule:
\[ \frac{d}{dx}[-x^2 \cos(1/x)] = -2x \cos(1/x) + x^2 \sin(1/x) \cdot \frac{1}{x^2} \]

Combining these results gives us:

\[ f''(x) = 12x^2(2 + \sin(1/x)) + 4x \cos(1/x) + 2x \cos(1/x) - x \sin(1/x) \]

### Step 3: Analyze continuity and boundedness

1. **Continuity of \( f'(x) \)**:
We need to check \( f'(0) \) and the limits as \( x \to 0 \). Since both limits approach 0, \( f'(x) \) is continuous at \( x = 0 \).

2. **Boundedness of \( f(x) \)**:
As \( x \to \pm \infty \), \( f(x) \) approaches \( +\infty \) because \( x^4 \) dominates. Thus, \( f(x) \) is unbounded.

3. **Global minimum**:
Since \( f'(x) \geq 0 \) for all \( x \) and \( f'(0) = 0 \), \( f(x) \) has a global minimum at \( x = 0 \).

4. **Continuity of \( f''(x) \)**:
The oscillatory nature of \( \sin(1/x) \) and \( \cos(1/x) \) means that \( f''(x) \) does not have a limit as \( x \to 0 \), indicating that \( f''(x) \) is not continuous at \( x = 0 \).

### Conclusion

- \( f(x) \) has a continuous first derivative for all \( x \).
- \( f(x) \) is unbounded.
- \( f(x) \) has a global minimum at \( x = 0 \).
- \( f''(x) \) is not continuous at \( x = 0 \).