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 the function \( f(x) \) at three different points: \( f(-1) \), \( f(2) \), and \( f(4) \). The function \( f \) is defined piecewise as follows:

- \( f(x) = 2x \) for \( x > 3 \)
- \( f(x) = x^2 \) for \( 1 < x \leq 3 \)
- \( f(x) = 3x \) for \( x \leq 1 \)

Now, let's evaluate each of the required values step by step.

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

Since \( -1 \leq 1 \), we use the third piece of the function:
\[ f(-1) = 3(-1) = -3 \]

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

Since \( 2 \) lies between \( 1 \) and \( 3 \) (i.e., \( 1 < 2 \leq 3 \)), we use the second piece of the function:
\[ f(2) = 2^2 = 4 \]

### Step 3: Calculate \( f(4) \)

Since \( 4 > 3 \), we use the first piece of the function:
\[ f(4) = 2(4) = 8 \]

### Step 4: Sum the values

Now we sum the results from the previous steps:
\[ f(-1) + f(2) + f(4) = -3 + 4 + 8 \]

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

### Final Result

Thus, the value of \( f(-1) + f(2) + f(4) \) is:
\[ \boxed{9} \]

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