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(D) The function is given by $f(x) = ||x+2|-2|x||$. To find the local extrema,we analyze the behavior of the function by considering the critical poin

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$4$

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(D) The function is given by $f(x) = ||x+2|-2|x||$. To find the local extrema,we analyze the behavior of the function by considering the critical points where the expression inside the absolute values changes sign,which are $x = -2$,$x = 0$,and the points where $|x+2| = 2|x|$. Solving $|x+2| = 2|x|$: Case $1$: $x \geq 0 \implies x+2 = 2x \implies x = 2$. Case $2$: $-2 \leq x Case $3$: $x Thus,the critical points are $x = -2, -2/3, 0, 2$. By plotting the graph or analyzing the sign changes of $f(x)$,we observe: - At $x = -2/3$,$f(x) = 0$,which is a local minimum. - At $x = 0$,$f(x) = 2$,which is a local maximum. - At $x = 2$,$f(x) = 0$,which is a local minimum. - At $x = -2$,$f(x) = 4$,which is a local maximum. Therefore,the points of local minima are $x = -2/3$ and $x = 2$,so $m = 2$. The points of local maxima are $x = -2$ and $x = 0$,so $n = 2$. Thus,$m+n = 2+2 = 4$.

Let $f : R \rightarrow R$ be a function defined by $f(x) = ||x+2|-2|x||$. If $m$ is the number of points of local minima and $n$ is the number of points of local maxima of $f$,then $m+n$ is

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