On the interval `I=[-2,2]`, the function `f(x)={{:(,(x+1)e^(-((1)/(|x|)+(1)/(x))),x ne 0),(,0,x=0):}`

Options

A. f(x) is continuous for all values of `x in I`
B. f(x) is continuous for `x in I - {0}`
C. f(x) assumes all intermediate values from f(-2) to f(2)
D. f(x) has a maximum value equal to 3/e

Answer

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

\[ f(x) = \begin{cases} (x + 1)e^{-\left(\frac{1}{|x|} + \frac{1}{x}\right)} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

We will check the continuity of \( f(x) \) on the interval \( I = [-2, 2] \).

### Step 1: Check continuity at \( x = 0 \)

To check continuity at \( x = 0 \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches 0 from both sides and compare it to \( f(0) \).

1. **Limit as \( x \to 0^- \)** (approaching from the left):
\[ f(x) = (x + 1)e^{-\left(\frac{1}{|x|} + \frac{1}{x}\right)} = (x + 1)e^{-\left(\frac{1}{-x} + \frac{1}{x}\right)} = (x + 1)e^{-\left(-\frac{1}{x} + \frac{1}{x}\right)} = (x + 1)e^{0} = x + 1 \]
Thus,
\[ \lim_{x \to 0^-} f(x) = 1. \]

2. **Limit as \( x \to 0^+ \)** (approaching from the right):
\[ f(x) = (x + 1)e^{-\left(\frac{1}{|x|} + \frac{1}{x}\right)} = (x + 1)e^{-\left(\frac{1}{x} + \frac{1}{x}\right)} = (x + 1)e^{-\frac{2}{x}}. \]
As \( x \to 0^+ \), \( e^{-\frac{2}{x}} \) approaches 0 very quickly, hence:
\[ \lim_{x \to 0^+} f(x) = (0 + 1) \cdot 0 = 0. \]

3. **Evaluate \( f(0) \)**:
\[ f(0) = 0. \]

Since \( \lim_{x \to 0^-} f(x) = 1 \) and \( \lim_{x \to 0^+} f(x) = 0 \), we see that:
\[ \lim_{x \to 0} f(x) \text{ does not exist.} \]
Thus, \( f(x) \) is not continuous at \( x = 0 \).

### Step 2: Check continuity for \( x \neq 0 \)

For \( x \in [-2, 2] \) and \( x \neq 0 \), the function \( f(x) = (x + 1)e^{-\left(\frac{1}{|x|} + \frac{1}{x}\right)} \) is continuous since it is a product of continuous functions.

### Conclusion

The function \( f(x) \) is continuous on the interval \( I = [-2, 2] \) except at \( x = 0 \). Therefore, we conclude that \( f(x) \) is continuous on \( I \setminus \{0\} \).

### Final Answer

The function \( f(x) \) is continuous for \( x \in I \setminus \{0\} \).

---