If `f(x)=|x^2-5x+6|,t h e nf^(prime)(x)e q u a l s`

Options

A. `2x-5" for "2ltxlt3`
B. `5-2x" for "2ltxlt3`
C. `2x-5" for "2lexle3`
D. `5-2x" for "2lexle3`

Answer

To find the derivative \( f'(x) \) of the function \( f(x) = |x^2 - 5x + 6| \), we will first analyze the expression inside the absolute value and then differentiate accordingly.

### Step 1: Factor the quadratic expression
The expression inside the absolute value is \( x^2 - 5x + 6 \). We can factor this as follows:

\[ x^2 - 5x + 6 = (x - 2)(x - 3) \]

### Step 2: Determine the intervals for the absolute value
Next, we need to find the points where the expression changes sign. The roots of the equation \( x^2 - 5x + 6 = 0 \) are \( x = 2 \) and \( x = 3 \). We will analyze the sign of \( (x - 2)(x - 3) \) in the intervals defined by these roots:

- For \( x < 2 \): Both \( (x - 2) < 0 \) and \( (x - 3) < 0 \) → Product is positive.
- For \( 2 < x < 3 \): \( (x - 2) > 0 \) and \( (x - 3) < 0 \) → Product is negative.
- For \( x > 3 \): Both \( (x - 2) > 0 \) and \( (x - 3) > 0 \) → Product is positive.

Thus, we can express \( f(x) \) as:

\[ f(x) = \begin{cases} x^2 - 5x + 6 & \text{if } x < 2 \\ -(x^2 - 5x + 6) & \text{if } 2 \leq x < 3 \\ x^2 - 5x + 6 & \text{if } x \geq 3 \end{cases} \]

### Step 3: Write the piecewise function
Now we can write the piecewise definition of \( f(x) \):

\[ f(x) = \begin{cases} x^2 - 5x + 6 & \text{if } x < 2 \\ - (x^2 - 5x + 6) & \text{if } 2 \leq x < 3 \\ x^2 - 5x + 6 & \text{if } x \geq 3 \end{cases} \]

### Step 4: Differentiate each piece
Now we differentiate each piece of the function:

1. For \( x < 2 \):
\[ f'(x) = \frac{d}{dx}(x^2 - 5x + 6) = 2x - 5 \]

2. For \( 2 \leq x < 3 \):
\[ f'(x) = \frac{d}{dx}(- (x^2 - 5x + 6)) = - (2x - 5) = 5 - 2x \]

3. For \( x \geq 3 \):
\[ f'(x) = \frac{d}{dx}(x^2 - 5x + 6) = 2x - 5 \]

### Step 5: Combine the results
Thus, the derivative \( f'(x) \) can be expressed as:

\[ f'(x) = \begin{cases} 2x - 5 & \text{if } x < 2 \\ 5 - 2x & \text{if } 2 \leq x < 3 \\ 2x - 5 & \text{if } x \geq 3 \end{cases} \]

### Summary of the solution
The final answer for \( f'(x) \) is:

\[ f'(x) = \begin{cases} 2x - 5 & \text{if } x < 2 \\ 5 - 2x & \text{if } 2 \leq x < 3 \\ 2x - 5 & \text{if } x \geq 3 \end{cases} \]