The function f(x)=2|x|+|x+2|-||x+2|-2|x|| has | vedclass

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(D) Given the function $f(x) = 2|x| + |x+2| - ||x+2| - 2|x||$.We analyze the function in different intervals:$1$. For $x \leq -2$: $f(x) = 2(-x) + (-(

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$(A, B)$

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(D) Given the function $f(x) = 2|x| + |x+2| - ||x+2| - 2|x||$.We analyze the function in different intervals:$1$. For $x \leq -2$: $f(x) = 2(-x) + (-(x+2)) - |-(x+2) - 2(-x)| = -2x - x - 2 - |-x - 2 + 2x| = -3x - 2 - |x - 2| = -3x - 2 - (2 - x) = -2x - 4$.$2$. For $-2 $3$. For $-2/3 $4$. For $0 $5$. For $x > 2$: $f(x) = 2x + (x+2) - |(x+2) - 2x| = 3x + 2 - (x - 2) = 2x + 4$.Summarizing the function:$f(x) = \begin{cases} -2x-4, & x \leq -2 \ 2x+4, & -2  2 \end{cases}$From the graph and the piecewise definition:Local minima occur at $x = -2$ and $x = 0$.Local maxima occur at $x = -2/3$.Comparing with the options,the points $x = -2$ and $x = -2/3$ correspond to option $(A)$ and $(B)$. Thus,the correct option is $(D)$.

The function $f(x)=2|x|+|x+2|-||x+2|-2|x||$ has a local minimum or a local maximum at $x=$

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