Let `f (x) =|2x-9|+|2x+9|.` Which of the following are true ?

Options

A. f (x) is not differerntiable at `x=9/2`
B. f (x) is not differentiable at `x = (-9)/(2)`
C. f (x) is differentiable at `x=0`
D. f (x) is differentiable at `x = (-9)/(2), 0, (9)/(2)`

Answer

To solve the problem, we need to analyze the function \( f(x) = |2x - 9| + |2x + 9| \) and determine its differentiability at specific points.

### Step 1: Identify critical points
The critical points for the absolute value function occur where the expressions inside the absolute values equal zero. Thus, we solve:
1. \( 2x - 9 = 0 \) → \( x = \frac{9}{2} \)
2. \( 2x + 9 = 0 \) → \( x = -\frac{9}{2} \)

These points, \( x = \frac{9}{2} \) and \( x = -\frac{9}{2} \), are where the function may not be differentiable.

### Step 2: Define the function piecewise
We can express \( f(x) \) in a piecewise manner based on the critical points:

- For \( x < -\frac{9}{2} \):
\[ f(x) = -(2x - 9) - (2x + 9) = -2x + 9 - 2x - 9 = -4x \]

- For \( -\frac{9}{2} \leq x \leq \frac{9}{2} \):
\[ f(x) = -(2x - 9) + (2x + 9) = -2x + 9 + 2x + 9 = 18 \]

- For \( x > \frac{9}{2} \):
\[ f(x) = (2x - 9) + (2x + 9) = 2x - 9 + 2x + 9 = 4x \]

Thus, we have:
\[ f(x) = \begin{cases} -4x & \text{if } x < -\frac{9}{2} \\ 18 & \text{if } -\frac{9}{2} \leq x \leq \frac{9}{2} \\ 4x & \text{if } x > \frac{9}{2} \end{cases} \]

### Step 3: Check differentiability at \( x = -\frac{9}{2} \)
To check differentiability at \( x = -\frac{9}{2} \), we find the left-hand and right-hand derivatives.

**Left-hand derivative:**
\[ L'(-\frac{9}{2}) = \lim_{h \to 0} \frac{f(-\frac{9}{2} - h) - f(-\frac{9}{2})}{-h} \]
Since \( -\frac{9}{2} - h < -\frac{9}{2} \) for small \( h \):
\[ = \lim_{h \to 0} \frac{-4(-\frac{9}{2}) - 18}{-h} = \lim_{h \to 0} \frac{18 + 4h - 18}{-h} = \lim_{h \to 0} \frac{4h}{-h} = -4 \]

**Right-hand derivative:**
\[ R'(-\frac{9}{2}) = \lim_{h \to 0} \frac{f(-\frac{9}{2} + h) - f(-\frac{9}{2})}{h} \]
Since \( -\frac{9}{2} + h \) is in the interval \( -\frac{9}{2} \leq x \leq \frac{9}{2} \):
\[ = \lim_{h \to 0} \frac{18 - 18}{h} = 0 \]

Since \( L'(-\frac{9}{2}) \neq R'(-\frac{9}{2}) \), \( f(x) \) is not differentiable at \( x = -\frac{9}{2} \).

### Step 4: Check differentiability at \( x = \frac{9}{2} \)
**Left-hand derivative:**
\[ L'(\frac{9}{2}) = \lim_{h \to 0} \frac{f(\frac{9}{2} - h) - f(\frac{9}{2})}{-h} = \lim_{h \to 0} \frac{18 - 18}{-h} = 0 \]

**Right-hand derivative:**
\[ R'(\frac{9}{2}) = \lim_{h \to 0} \frac{f(\frac{9}{2} + h) - f(\frac{9}{2})}{h} = \lim_{h \to 0} \frac{4(\frac{9}{2} + h) - 18}{h} = \lim_{h \to 0} \frac{18 + 4h - 18}{h} = 4 \]

Since \( L'(\frac{9}{2}) \neq R'(\frac{9}{2}) \), \( f(x) \) is not differentiable at \( x = \frac{9}{2} \).

### Step 5: Check differentiability at \( x = 0 \)
**Left-hand derivative:**
\[ L'(0) = \lim_{h \to 0} \frac{f(0 - h) - f(0)}{-h} = \lim_{h \to 0} \frac{18 - 18}{-h} = 0 \]

**Right-hand derivative:**
\[ R'(0) = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0} \frac{18 - 18}{h} = 0 \]

Since \( L'(0) = R'(0) = 0 \), \( f(x) \) is differentiable at \( x = 0 \).

### Conclusion
1. \( f(x) \) is not differentiable at \( x = -\frac{9}{2} \) (True).
2. \( f(x) \) is not differentiable at \( x = \frac{9}{2} \) (True).
3. \( f(x) \) is differentiable at \( x = 0 \) (True).
4. \( f(x) \) is not differentiable at \( x = -\frac{9}{2}, 0, \frac{9}{2} \) (False).

### Final Answer
The true statements are:
- 1. True
- 2. True
- 3. True
- 4. False