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: Factorize the expression inside the modulus.
\[ f(x)=|x^2-3x+2|. \] Factorizing, \[ x^2-3x+2=(x-1)(x-2). \] Hence, \[ f(x)=|(x-1)(x-2)|. \]

Step 2: Determine the sign of \(x^2-3x+2\).
The critical points are \[ x=1,\qquad x=2. \] For \[ x<1, \] both factors are negative, so \[ x^2-3x+2>0. \] For \[ 1<x<2, \] one factor is positive and the other is negative, so \[ x^2-3x+2<0. \] For \[ x>2, \] both factors are positive, so \[ x^2-3x+2>0. \]

Step 3: Remove the modulus in different intervals.
Therefore, \[ f(x)=x^2-3x+2, \qquad x2. \] and \[ f(x)=-(x^2-3x+2), \qquad 1<x<2. \]

Step 4: Differentiate.
For \[ x2, \] \[ \frac{df}{dx} = \frac{d}{dx}(x^2-3x+2) = 2x-3. \] For \[ 1<x<2, \] \[ \frac{df}{dx} = \frac{d}{dx}(-x^2+3x-2) = 3-2x. \]

Step 5: Compare with the options.
Option (3) states \[ \frac{df}{dx}=2x-3, \qquad x>2, \] which is correct.

Step 6: Final conclusion.
Therefore, \[ \boxed{\frac{df}{dx}=2x-3 \text{ when } x>2} \] and the correct option is \[ \boxed{(3)}. \]