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If $f(x)=\left|x^2-3 x+2\right|$, then $\frac{d f}{d x}=$
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The correct answer is:
$2 x-3$, when $x>2$
$f(x)=\left|x^2-3 x+2\right|$
$\begin{aligned} & =|(x-1)(x-2)| \\ & =\left\{\begin{array}{l}(x-1)(x-2) ; x < 1 \text { and } x \geq 2 \\ -(x-1)(x-2) ; 1 \leq x < 2\end{array}\right.\end{aligned}$
$f(x)=\left\{\begin{array}{l}x^2-3 x+2 ; x < 1 \text { and } x \geq 2 \\ -x^2+3 x-2 ; 1 \leq x < 2\end{array}\right.$
$\Rightarrow f^{\prime}(x)=\left\{\begin{array}{l}2 x-3 ; x < 1 \text { and } x \geq 2 \\ -2 x+3 ; 1 \leq x < 2\end{array}\right.$
$\begin{aligned} & =|(x-1)(x-2)| \\ & =\left\{\begin{array}{l}(x-1)(x-2) ; x < 1 \text { and } x \geq 2 \\ -(x-1)(x-2) ; 1 \leq x < 2\end{array}\right.\end{aligned}$
$f(x)=\left\{\begin{array}{l}x^2-3 x+2 ; x < 1 \text { and } x \geq 2 \\ -x^2+3 x-2 ; 1 \leq x < 2\end{array}\right.$
$\Rightarrow f^{\prime}(x)=\left\{\begin{array}{l}2 x-3 ; x < 1 \text { and } x \geq 2 \\ -2 x+3 ; 1 \leq x < 2\end{array}\right.$
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