Let `f(x) = a|x|^(2) + b|x| +c` where a,b,c are real constants. Then f'(x) exists at x=0 if:

Options

A. a=0
B. b=0
C. c=0
D. a=b

Answer

To determine the conditions under which the derivative \( f'(x) \) exists at \( x = 0 \) for the function \( f(x) = a|x|^2 + b|x| + c \), we need to analyze the left-hand derivative and the right-hand derivative at that point.

### Step-by-Step Solution:

1. **Define the Function**:
\[ f(x) = a|x|^2 + b|x| + c \]

2. **Find \( f(0) \)**:
\[ f(0) = a|0|^2 + b|0| + c = c \]

3. **Calculate the Left-Hand Derivative**:
The left-hand derivative at \( x = 0 \) is given by:
\[ f'_-(0) = \lim_{h \to 0^-} \frac{f(0 + h) - f(0)}{h} \]
For \( h < 0 \), \( |h| = -h \):
\[ f(0 + h) = f(h) = a|h|^2 + b|h| + c = a(-h)^2 + b(-h) + c = ah^2 - bh + c \]
Therefore:
\[ f'_-(0) = \lim_{h \to 0^-} \frac{(ah^2 - bh + c) - c}{h} = \lim_{h \to 0^-} \frac{ah^2 - bh}{h} \]
Simplifying:
\[ f'_-(0) = \lim_{h \to 0^-} (ah - b) = -b \quad \text{(as } h \to 0^- \text{)} \]

4. **Calculate the Right-Hand Derivative**:
The right-hand derivative at \( x = 0 \) is given by:
\[ f'_+(0) = \lim_{h \to 0^+} \frac{f(0 + h) - f(0)}{h} \]
For \( h > 0 \), \( |h| = h \):
\[ f(0 + h) = f(h) = a|h|^2 + b|h| + c = ah^2 + bh + c \]
Therefore:
\[ f'_+(0) = \lim_{h \to 0^+} \frac{(ah^2 + bh + c) - c}{h} = \lim_{h \to 0^+} \frac{ah^2 + bh}{h} \]
Simplifying:
\[ f'_+(0) = \lim_{h \to 0^+} (ah + b) = b \quad \text{(as } h \to 0^+ \text{)} \]

5. **Set Left-Hand Derivative Equal to Right-Hand Derivative**:
For \( f'(0) \) to exist, we need:
\[ f'_-(0) = f'_+(0) \]
Therefore:
\[ -b = b \]
This implies:
\[ 2b = 0 \quad \Rightarrow \quad b = 0 \]

6. **Conclusion**:
The derivative \( f'(x) \) exists at \( x = 0 \) if:
\[ b = 0 \]