Let `f(x)={x+2,-1lelt0`
`1,x=0
(x)/(2),0ltxle1`

Options

A. a point of minima
B. a point of maxima
C. both points of minima and maxima
D. neither a pointof minimanor that of maxima

Explanation

To solve the given problem, we need to analyze the piecewise function defined as:

\[ f(x) = \begin{cases} x + 2 & \text{for } -1 \leq x < 0 \\ 1 & \text{for } x = 0 \\ \frac{x}{2} & \text{for } 0 < x \leq 1 \end{cases} \]

We are tasked with finding the points of maximum and minimum for this function.

### Step 1: Evaluate the function at the boundary points
We need to evaluate the function at the boundary points, specifically at \(x = 0\) and the limits approaching \(0\) from both sides.

1. **Calculate \(f(0^-)\)** (as \(x\) approaches \(0\) from the left):
\[ f(0^-) = 0 + 2 = 2 \]

2. **Calculate \(f(0)\)**:
\[ f(0) = 1 \]

3. **Calculate \(f(0^+)\)** (as \(x\) approaches \(0\) from the right):
\[ f(0^+) = \frac{0}{2} = 0 \]

### Step 2: Analyze the values
Now we have:
- \(f(0^-) = 2\)
- \(f(0) = 1\)
- \(f(0^+) = 0\)

### Step 3: Determine the nature of \(x = 0\)
To determine if \(x = 0\) is a point of maximum or minimum, we compare the values:

- As \(x\) approaches \(0\) from the left, \(f(x)\) approaches \(2\).
- At \(x = 0\), \(f(0) = 1\).
- As \(x\) approaches \(0\) from the right, \(f(x)\) approaches \(0\).

Since \(f(0) = 1\) is less than \(f(0^-)\) and greater than \(f(0^+)\), we can conclude that:

- \(f(0)\) is a local minimum since the function value decreases when moving to the right and increases when moving to the left.

### Step 4: Check the intervals
Next, we check the intervals defined by the piecewise function:

- For \(x < 0\) (specifically in the interval \([-1, 0)\)), the function \(f(x) = x + 2\) is increasing.
- For \(0 < x \leq 1\), the function \(f(x) = \frac{x}{2}\) is also increasing.

### Conclusion
Since \(f(x)\) is increasing on both sides of \(x = 0\) and \(f(0)\) is a local minimum, we conclude that:

- The point \(x = 0\) is a point of minimum for the function \(f(x)\).

### Final Answer
Thus, the answer is:
- **Point of minimum at \(x = 0\)**.

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