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

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(C) Given $f(x) = |x^2 - 3x + 2|$.We can factorize the quadratic expression as $f(x) = |(x - 1)(x - 2)|$.The expression inside the absolute value is p

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$2x - 3$,when $x > 2$

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(C) Given $f(x) = |x^2 - 3x + 2|$.We can factorize the quadratic expression as $f(x) = |(x - 1)(x - 2)|$.The expression inside the absolute value is positive for $x  2$,and negative for $1 Thus,$f(x) = \begin{cases} x^2 - 3x + 2 & 	ext{if } x \leq 1 	ext{ or } x \geq 2 \ -(x^2 - 3x + 2) & 	ext{if } 1 Differentiating with respect to $x$:$f'(x) = \begin{cases} 2x - 3 & 	ext{if } x  2 \ -2x + 3 & 	ext{if } 1 Comparing this with the given options,for $x > 2$,$f'(x) = 2x - 3$,which matches option $C$.

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

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