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

Options

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

Answer

To solve the problem, we need to analyze the function:

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

We will check the continuity of the function at \(x = 0\) and find its maximum value on the interval \([-2, 2]\).

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

To determine if \(f(x)\) is continuous at \(x = 0\), we need to check if:

\[ \lim_{x \to 0} f(x) = f(0) \]

Since \(f(0) = 0\), we need to find \(\lim_{x \to 0} f(x)\).

### Step 2: Calculate \(\lim_{x \to 0^-} f(x)\)

For \(x < 0\):

\[ f(x) = (x + 1)e^{-\left(\frac{1}{|x|} + \frac{1}{x}\right)} = (x + 1)e^{-\left(-\frac{1}{x} + \frac{1}{x}\right)} = (x + 1)e^{0} = (x + 1) \]

Now, we compute the limit:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x + 1) = 1 \]

### Step 3: Calculate \(\lim_{x \to 0^+} f(x)\)

For \(x > 0\):

\[ f(x) = (x + 1)e^{-\left(\frac{1}{x} + \frac{1}{x}\right)} = (x + 1)e^{-2/x} \]

As \(x \to 0^+\), \(e^{-2/x} \to 0\):

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x + 1)e^{-2/x} = 0 \]

### Step 4: Compare Limits

Now we compare the two limits:

\[ \lim_{x \to 0^-} f(x) = 1 \quad \text{and} \quad \lim_{x \to 0^+} f(x) = 0 \]

Since the left-hand limit and right-hand limit are not equal, we conclude that:

\[ \lim_{x \to 0} f(x) \neq f(0) \]

Thus, \(f(x)\) is not continuous at \(x = 0\).

### Step 5: Analyze Continuity on the Interval \([-2, 2]\)

The function is defined for all \(x \in [-2, 2]\) except at \(x = 0\). Therefore, it is continuous on \([-2, 0)\) and \((0, 2]\).

### Step 6: Find Maximum Value on \([-2, 2]\)

To find the maximum value of \(f(x)\) on the interval, we will evaluate \(f(x)\) at the endpoints and check for critical points in the intervals.

1. **Evaluate at endpoints:**
- \(f(-2) = (-2 + 1)e^{-\left(\frac{1}{2} + \frac{1}{-2}\right)} = -1 \cdot e^{0} = -1\)
- \(f(2) = (2 + 1)e^{-\left(\frac{1}{2} + \frac{1}{2}\right)} = 3e^{-1} = \frac{3}{e}\)

2. **Check critical points in \((-2, 0)\) and \((0, 2)\)**:
- Since \(f(x)\) is continuous in these intervals, we can use calculus to find critical points by taking the derivative and setting it to zero. However, for simplicity, we can evaluate \(f(x)\) at a few points:
- \(f(-1) = 0\)
- \(f(1) = 2e^{-2}\)

### Step 7: Compare Values

Now we compare the values:

- \(f(-2) = -1\)
- \(f(-1) = 0\)
- \(f(1) = 2e^{-2} \approx 0.27\)
- \(f(2) = \frac{3}{e} \approx 1.10\)

The maximum value occurs at \(x = 2\):

\[ \text{Maximum value} = \frac{3}{e} \]

### Conclusion

1. \(f(x)\) is not continuous at \(x = 0\).
2. The maximum value of \(f(x)\) on the interval \([-2, 2]\) is \(\frac{3}{e}\).