If `f(x)={{:(1+x",",-1lexlt0),(x^(2)-1",",0ltxlt2),(2x",",2lex):}`
Find `f(3),f(-2),f(0),f((1)/(2))f(2-h),f(-1+h)`, where `hgt0` is very small.

Answer

To solve the problem, we need to evaluate the function \( f(x) \) at several specific points based on the piecewise definition provided. The function is defined as follows:

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

Now, let's evaluate \( f(3) \), \( f(-2) \), \( f(0) \), \( f\left(\frac{1}{2}\right) \), \( f(2-h) \), and \( f(-1+h) \) where \( h > 0 \) is very small.

### Step 1: Calculate \( f(3) \)

Since \( 3 \geq 2 \), we use the third case of the function:

\[ f(3) = 2 \cdot 3 = 6 \]

### Step 2: Calculate \( f(-2) \)

Since \( -2 < -1 \), the function is not defined for \( x < -1 \):

\[ f(-2) \text{ is not defined.} \]

### Step 3: Calculate \( f(0) \)

Since \( 0 \) is not included in the interval \( -1 \leq x < 0 \) and \( 0 < x < 2 \), the function is not defined at \( x = 0 \):

\[ f(0) \text{ is not defined.} \]

### Step 4: Calculate \( f\left(\frac{1}{2}\right) \)

Since \( \frac{1}{2} \) is in the interval \( 0 < x < 2 \), we use the second case of the function:

\[ f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - 1 = \frac{1}{4} - 1 = -\frac{3}{4} \]

### Step 5: Calculate \( f(2-h) \)

Since \( 2-h < 2 \) (because \( h > 0 \)), we use the second case of the function:

\[ f(2-h) = (2-h)^2 - 1 = 4 - 4h + h^2 - 1 = 3 - 4h + h^2 \]

### Step 6: Calculate \( f(-1+h) \)

Since \( -1 + h \) is in the interval \( -1 \leq x < 0 \) (as \( h \) is very small and positive), we use the first case of the function:

\[ f(-1+h) = 1 + (-1 + h) = h \]

### Summary of Results

- \( f(3) = 6 \)
- \( f(-2) \) is not defined.
- \( f(0) \) is not defined.
- \( f\left(\frac{1}{2}\right) = -\frac{3}{4} \)
- \( f(2-h) = 3 - 4h + h^2 \)
- \( f(-1+h) = h \)