Let g(x) = ||x + 2| - 3|. If a denotes the nu | vedclass

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(D) The function is $g(x) = ||x + 2| - 3|$.To find the relative minima and maxima,we analyze the graph of $g(x)$.The inner function $f(x) = |x + 2| -

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$9$

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(D) The function is $g(x) = ||x + 2| - 3|$.To find the relative minima and maxima,we analyze the graph of $g(x)$.The inner function $f(x) = |x + 2| - 3$ has a minimum at $x = -2$ with value $-3$.The absolute value function $g(x) = |f(x)|$ reflects the negative part of $f(x)$ across the $x$-axis.Thus,$g(x)$ has relative minima at $x = -2$ (where $g(-2) = 3$) and at the roots of $|x + 2| - 3 = 0$,which are $x = 1$ and $x = -5$ (where $g(x) = 0$).So,the relative minima are at $x = -5, -2, 1$. Thus,$a = 3$.There is a relative maximum at $x = -2$ for the inner function,but for $g(x)$,the local maximum occurs at $x = -2$ where $g(-2) = 3$. Wait,checking the graph: $g(x)$ has a local maximum at $x = -2$ with value $3$. Thus,$b = 1$.The zeroes of $g(x)$ occur when $|x + 2| - 3 = 0$,so $x + 2 = 3$ or $x + 2 = -3$. This gives $x = 1$ and $x = -5$.The product of the zeroes is $c = 1 	imes (-5) = -5$.Calculating the expression: $a + 2b - c = 3 + 2(1) - (-5) = 3 + 2 + 5 = 10$. Re-evaluating: The relative minima are at $x = -5$ and $x = 1$ (value $0$),and the relative maximum is at $x = -2$ (value $3$). So $a = 2$ and $b = 1$. Then $a + 2b - c = 2 + 2(1) - (-5) = 2 + 2 + 5 = 9$.

Let $g(x) = ||x + 2| - 3|$. If $a$ denotes the number of relative minima,$b$ denotes the number of relative maxima,and $c$ denotes the product of the zeroes of $g(x)$,then the value of $(a + 2b - c)$ is:

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