Following questions are based on the given information for the following functions f (x)
`f(x) =2bx +f(-x)," if " x lt 0`
`f(x)=a, " if "x =0`
`f(x)=b+c-2cx +f(x-1)," if "x gt 0`
f(-19) equals:

Options

A. `a-19b+361c `
B. `a+8(b-8c)`
C. `8(a+b-c)`
D. None of these

Answer

To find the value of \( f(-19) \) using the given piecewise function definitions, we will follow these steps:

### Step 1: Identify the function for \( f(x) \) when \( x < 0 \)
Since \( -19 < 0 \), we will use the first definition of the function:
\[ f(x) = 2bx + f(-x) \quad \text{for } x < 0 \]

### Step 2: Substitute \( x = -19 \) into the function
We substitute \( x = -19 \):
\[ f(-19) = 2b(-19) + f(19) = -38b + f(19) \]

### Step 3: Identify the function for \( f(x) \) when \( x > 0 \)
Now we need to find \( f(19) \). Since \( 19 > 0 \), we will use the third definition of the function:
\[ f(x) = b + c - 2cx + f(x-1) \quad \text{for } x > 0 \]

### Step 4: Substitute \( x = 19 \) into the function
We substitute \( x = 19 \):
\[ f(19) = b + c - 2c(19) + f(18) = b + c - 38c + f(18) = b - 37c + f(18) \]

### Step 5: Find \( f(18) \)
Since \( 18 > 0 \), we again use the third definition:
\[ f(18) = b + c - 2c(18) + f(17) = b + c - 36c + f(17) = b - 35c + f(17) \]

### Step 6: Continue this process for \( f(17), f(16), \ldots, f(1) \)
Following the same pattern, we can express \( f(17), f(16), \ldots, f(1) \):
- \( f(17) = b - 34c + f(16) \)
- \( f(16) = b - 33c + f(15) \)
- ...
- \( f(1) = b - 2c + f(0) \)

### Step 7: Find \( f(0) \)
From the definition, we know:
\[ f(0) = a \]

### Step 8: Substitute back to find \( f(1) \)
Substituting \( f(0) \) into the equation for \( f(1) \):
\[ f(1) = b - 2c + a \]

### Step 9: Substitute \( f(1) \) back to find \( f(2) \), \( f(3) \), ..., \( f(19) \)
Continuing this substitution:
- \( f(2) = b - 4c + f(1) = b - 4c + (b - 2c + a) = 2b - 6c + a \)
- \( f(3) = b - 6c + f(2) = b - 6c + (2b - 6c + a) = 3b - 12c + a \)

Continuing this process, we find:
\[ f(n) = nb - (n(n+1)/2) \cdot 2c + a \]

### Step 10: Find \( f(19) \)
Substituting \( n = 19 \):
\[ f(19) = 19b - 19 \cdot 20c + a = 19b - 380c + a \]

### Step 11: Substitute \( f(19) \) back into \( f(-19) \)
Now we substitute back into the equation for \( f(-19) \):
\[ f(-19) = -38b + (19b - 380c + a) = -19b - 380c + a \]

### Final Result
Thus, the value of \( f(-19) \) is:
\[ f(-19) = -19b - 380c + a \]

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