Let `f:RrarrR` be defined by :
`f(x)={(2x,,xgt3),(x^(2),,1ltxlt3),(3x,,xle1):}`
Then, `f(-1)+f(2)+f(4)` is :

Options

A. 9
B. 14
C. 5
D. None of these

Answer

To solve the problem, we need to evaluate the expression \( f(-1) + f(2) + f(4) \) using the piecewise function defined as:

\[ f(x) = \begin{cases} 2x & \text{if } x > 3 \\ x^2 & \text{if } 1 < x < 3 \\ 3x & \text{if } x \leq 1 \end{cases} \]

### Step 1: Calculate \( f(-1) \)
Since \(-1 \leq 1\), we use the third case of the piecewise function:
\[ f(-1) = 3(-1) = -3 \]

### Step 2: Calculate \( f(2) \)
Since \(2\) is between \(1\) and \(3\), we use the second case of the piecewise function:
\[ f(2) = 2^2 = 4 \]

### Step 3: Calculate \( f(4) \)
Since \(4 > 3\), we use the first case of the piecewise function:
\[ f(4) = 2(4) = 8 \]

### Step 4: Combine the results
Now we can combine the results from the previous steps:
\[ f(-1) + f(2) + f(4) = -3 + 4 + 8 \]

Calculating this gives:
\[ -3 + 4 = 1 \]
\[ 1 + 8 = 9 \]

Thus, the final answer is:
\[ f(-1) + f(2) + f(4) = 9 \]

### Final Answer:
\[ \boxed{9} \]