If `f:RtoR` defined by `f(x)={(4x-1" for " xgt4),(x^(2)-2" for "-2lexlt4),(3x+4" for "xlt-2):}`
find `f(5)+f(0)+f(-5)`

Answer

To solve the problem, we need to evaluate the function \( f(x) \) at three different points: \( f(5) \), \( f(0) \), and \( f(-5) \). The function \( f \) is defined piecewise as follows:

\[ f(x) = \begin{cases} 4x - 1 & \text{if } x > 4 \\ x^2 - 2 & \text{if } -2 \leq x < 4 \\ 3x + 4 & \text{if } x < -2 \end{cases} \]

Now, let's calculate each of these values step by step.

### Step 1: Calculate \( f(5) \)

Since \( 5 > 4 \), we use the first piece of the function:

\[ f(5) = 4(5) - 1 \]

Calculating this gives:

\[ f(5) = 20 - 1 = 19 \]

### Step 2: Calculate \( f(0) \)

Since \( 0 \) lies in the interval \( -2 \leq x < 4 \), we use the second piece of the function:

\[ f(0) = 0^2 - 2 \]

Calculating this gives:

\[ f(0) = 0 - 2 = -2 \]

### Step 3: Calculate \( f(-5) \)

Since \( -5 < -2 \), we use the third piece of the function:

\[ f(-5) = 3(-5) + 4 \]

Calculating this gives:

\[ f(-5) = -15 + 4 = -11 \]

### Step 4: Combine the results

Now, we need to find \( f(5) + f(0) + f(-5) \):

\[ f(5) + f(0) + f(-5) = 19 + (-2) + (-11) \]

Calculating this gives:

\[ 19 - 2 - 11 = 19 - 13 = 6 \]

### Final Answer

Thus, the final answer is:

\[ \boxed{6} \]