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In the interval $(7, \infty) f(x)=|x-5|+2|x-7|$ is
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The correct answer is:
increasing function
$|x-5|=$ $\begin{cases}x-5, & x \geq 5 \\ -(x-5), & x < 5\end{cases}$
$x>7$
$f(x)=x-5+2(x-7)$
$=3 x-5-14$
$\begin{aligned} & f(x)=3 x-19 . \\ & f^{\prime}(x)=3>0\end{aligned}$
$x>7$
$f(x)=x-5+2(x-7)$
$=3 x-5-14$
$\begin{aligned} & f(x)=3 x-19 . \\ & f^{\prime}(x)=3>0\end{aligned}$
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