If a function `f:RtoR` is defined by
`f(x){{:(2x,xgt3),(x^(2),1lexle3),(3x,xlt1):}`
Then the value of `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 function \( f(x) \) at three specific points: \( f(-1) \), \( f(2) \), and \( f(4) \). The function is defined piecewise as follows:

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

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

Since \(-1 < 1\), we use the third piece of the function:

\[ f(-1) = 3(-1) = -3 \]

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

Since \(2\) is between \(1\) and \(3\), 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: Add the values

Now, we can add the three values we calculated:

\[ f(-1) + f(2) + f(4) = -3 + 4 + 8 \]

Calculating this gives:

\[ -3 + 4 = 1 \\ 1 + 8 = 9 \]

Thus, the final result is:

\[ f(-1) + f(2) + f(4) = 9 \]

### Final Answer:

The value of \( f(-1) + f(2) + f(4) \) is \( \boxed{9} \).

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