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 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 will check the behavior of the function at the critical points and the endpoints of the intervals to determine if there are any maxima or minima.

### Step 1: Evaluate the function at the critical point \(x = 0\)

We need to find the left-hand limit and right-hand limit of \(f(x)\) as \(x\) approaches \(0\).

- For \(x\) approaching \(0\) from the left (\(x \to 0^-\)):
\[ f(0^-) = 0 + 2 = 2 \]

- For \(x\) approaching \(0\) from the right (\(x \to 0^+\)):
\[ f(0^+) = \frac{0}{2} = 0 \]

- At \(x = 0\):
\[ f(0) = 1 \]

### Step 2: Compare the values

Now we compare the values of \(f(x)\) around \(x = 0\):
- \(f(0^-) = 2\)
- \(f(0) = 1\)
- \(f(0^+) = 0\)

### Step 3: Determine maxima and minima

To determine whether \(x = 0\) is a point of maxima or minima, we analyze the behavior of \(f(x)\):
- As \(x\) approaches \(0\) from the left, \(f(x)\) is decreasing from \(2\) to \(1\).
- As \(x\) approaches \(0\) from the right, \(f(x)\) is decreasing from \(1\) to \(0\).

Since \(f(0^-) > f(0) > f(0^+)\), we can conclude that:
- \(f(0)\) is less than the value of \(f(x)\) just before it and greater than the value of \(f(x)\) just after it.

### Conclusion

Since \(f(0)\) is lower than both \(f(0^-)\) and \(f(0^+)\), it indicates that \(x = 0\) is a point of local maxima. However, since \(f(0^+) < f(0)\), it is not a global maximum or minimum.

Thus, we conclude that \(x = 0\) is neither a point of maxima nor minima.

### Final Answer
The point \(x = 0\) is neither a point of maxima nor minima.

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