Question

If \[ f(x)=|x^2-3x+2|, \] then \[ \frac{df}{dx}= \]

• A. \(2x-3,\; \text{when } 1<x<2\)
• B. \(3-2x,\; \text{when } x>2\)
• C. \(2x-3,\; \text{when } x>2\)
• D. \(3+2x,\; \text{when } 1<x<2\)

Hint:

For functions of the form \(|g(x)|\), first determine where \(g(x)\) is positive and negative. Remove the modulus accordingly and then differentiate piecewise.

Answer 1

The Correct Option is C

Step 1: Factor the quadratic inside the absolute value.
f(x) = |x² - 3x + 2| = |(x - 1)(x - 2)|.

Step 2: Determine the sign of the quadratic in each interval.
For x<1: both factors negative → product positive. For 1<x<2: one positive, one negative → product negative. For x>2: both positive → product positive.

Step 3: Remove the modulus piecewise.
f(x) = x² - 3x + 2 for x<1 and x>2. f(x) = -x² + 3x - 2 for 1<x<2.

Step 4: Differentiate in each region.
For x<1 and x>2: df/dx = 2x - 3. For 1<x<2: df/dx = 3 - 2x.

Step 5: Identify the correct option.
The statement df/dx = 2x - 3 when x>2 is correct, matching option (3).

Step 6: Final conclusion.
The derivative is 2x - 3 for x>2.