The derivative of f(x)=3|2+x| at x=-3 is

Options

A. 3
B. `-3`
C. 0
D. does not exist

Answer

To find the derivative of the function \( f(x) = 3|2+x| \) at \( x = -3 \), we will follow these steps:

### Step 1: Identify the function and the point of interest
We have the function:
\[ f(x) = 3|2+x| \]
We need to find the derivative at \( x = -3 \).

### Step 2: Calculate \( f(-3) \)
First, we need to evaluate \( f(-3) \):
\[ f(-3) = 3|2 + (-3)| = 3|2 - 3| = 3|-1| = 3 \times 1 = 3 \]

### Step 3: Set up the limit definition of the derivative
The derivative of a function at a point \( c \) is defined as:
\[ f'(c) = \lim_{h \to 0} \frac{f(c + h) - f(c)}{h} \]
In our case, \( c = -3 \):
\[ f'(-3) = \lim_{h \to 0} \frac{f(-3 + h) - f(-3)}{h} \]

### Step 4: Calculate \( f(-3 + h) \)
Now we need to evaluate \( f(-3 + h) \):
\[ f(-3 + h) = 3|2 + (-3 + h)| = 3|2 - 3 + h| = 3|-1 + h| \]

### Step 5: Analyze the absolute value
The expression \( |-1 + h| \) depends on the value of \( h \):
- If \( h < 1 \), then \( -1 + h < 0 \) and \( |-1 + h| = -(-1 + h) = 1 - h \).
- If \( h \geq 1 \), then \( -1 + h \geq 0 \) and \( |-1 + h| = -1 + h \).

However, since we are taking the limit as \( h \) approaches 0, we will consider the case where \( h \) is small and negative (which is the most relevant for our limit).

Thus, we can write:
\[ f(-3 + h) = 3(1 - h) = 3 - 3h \]

### Step 6: Substitute into the limit
Now we substitute back into the limit:
\[ f'(-3) = \lim_{h \to 0} \frac{(3 - 3h) - 3}{h} = \lim_{h \to 0} \frac{-3h}{h} \]

### Step 7: Simplify the expression
This simplifies to:
\[ f'(-3) = \lim_{h \to 0} -3 = -3 \]

### Conclusion
Thus, the derivative of \( f(x) = 3|2+x| \) at \( x = -3 \) is:
\[ \boxed{-3} \]