If `f: R to R` be defined by `f(x)={(2x:xgt3),(x^(2):1lt x le 3),(3x:x le 1):}`
Then,`f(-1)+f(2)+f(4)` is

Options

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

Explanation

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

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

### 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 \(3\) (i.e., \(1 < 2 \leq 3\)), 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 add the values we calculated:
\[ f(-1) + f(2) + f(4) = -3 + 4 + 8 \]

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

### Final Answer
Thus, the final result is:
\[ f(-1) + f(2) + f(4) = 9 \]