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(C) The function is given by $f(x) = |x| + |x^2 - 1|$.We analyze the function in different intervals:$1$. For $x < -1$: $f(x) = -x + x^2 - 1 = x^2 - x

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$5$

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(C) The function is given by $f(x) = |x| + |x^2 - 1|$.We analyze the function in different intervals:$1$. For $x $2$. For $-1 \leq x $3$. At $x = -1$: $f(-1) = 1$. Since $f(x)$ decreases to $1$ from the left and increases to $5/4$ from the right,$x = -1$ is a local minimum.$4$. At $x = 0$: $f(0) = 1$. Since $f(x)$ decreases to $1$ from the left and increases to $1$ from the right,$x = 0$ is a local minimum.$5$. For $0 $6$. At $x = 1$: $f(1) = 1$. Since $f(x)$ decreases to $1$ from the left and increases to $1$ from the right,$x = 1$ is a local minimum.Summarizing the points:- Local minima at $x = -1, 0, 1$ ($3$ points).- Local maxima at $x = -1/2, 1/2$ ($2$ points).Total number of points = $3 + 2 = 5$.

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be defined as $f(x)=|x|+|x^2-1|$. The total number of points at which $f$ attains either a local maximum or a local minimum is

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