Let `f(x)={{:(-|x-2|, ,x le 3 ),(x^(2)-2x-4,, x gt 3 ):}`
Then the number of critical points on the graph of the function is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the number of critical points for the given piecewise function

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

we will follow these steps:

### Step 1: Break down the absolute value function
For \( x \leq 3 \), we need to express \(-|x-2|\) without the absolute value. The expression \(|x-2|\) can be rewritten as:
- When \( x < 2 \), \(|x-2| = 2-x\) (since \( x-2 < 0 \))
- When \( 2 \leq x \leq 3\), \(|x-2| = x-2\) (since \( x-2 \geq 0 \))

Thus, we can rewrite \( f(x) \) as:
\[ f(x) = \begin{cases} 2 - x & \text{if } x < 2 \\ -(x - 2) = 2 - x & \text{if } 2 \leq x \leq 3 \\ x^2 - 2x - 4 & \text{if } x > 3 \end{cases} \]

### Step 2: Find the first derivative \( f'(x) \)
Now, we will find the derivative of each piece of the function.

1. For \( x < 2 \):
\[ f'(x) = -1 \]

2. For \( 2 \leq x \leq 3 \):
\[ f'(x) = -1 \]

3. For \( x > 3 \):
\[ f'(x) = 2x - 2 \]

### Step 3: Set the derivative equal to zero
To find critical points, we set \( f'(x) = 0 \):

1. For \( x < 2 \):
\[ -1 \neq 0 \quad \text{(no critical points)} \]

2. For \( 2 \leq x \leq 3 \):
\[ -1 \neq 0 \quad \text{(no critical points)} \]

3. For \( x > 3 \):
\[ 2x - 2 = 0 \implies x = 1 \quad \text{(not in this interval)} \]

### Step 4: Check points where the derivative is undefined
Next, we check the points where the derivative might be undefined, specifically at the boundaries of the piecewise function:

1. At \( x = 2 \):
- Left-hand derivative: \( f'(2^-) = -1 \)
- Right-hand derivative: \( f'(2^+) = -1 \)
- Since both derivatives are equal, there is no critical point here.

2. At \( x = 3 \):
- Left-hand derivative: \( f'(3^-) = -1 \)
- Right-hand derivative: \( f'(3^+) = 2(3) - 2 = 4 \)
- Since the left-hand derivative does not equal the right-hand derivative, \( x = 3 \) is a critical point.

### Conclusion
The only critical point we found is at \( x = 3 \). Therefore, the total number of critical points on the graph of the function is:

\[ \text{Number of critical points} = 1 \]