Let f: R rightarrow R be given by f(x) = |x^2 | vedclass

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(C) The function is defined as $f(x) = |x^2 - 1|$.We can analyze the behavior of the function by looking at its graph or by examining its critical poi

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$f$ has a local minima at $x = \pm 1$ and a local maxima at $x = 0$

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(C) The function is defined as $f(x) = |x^2 - 1|$.We can analyze the behavior of the function by looking at its graph or by examining its critical points.$1$. The function $f(x) = |x^2 - 1|$ is non-negative for all $x \in R$.$2$. At $x = 1$ and $x = -1$,$f(x) = |1^2 - 1| = 0$. Since $f(x) \ge 0$ for all $x$,the points $x = 1$ and $x = -1$ are points of local minima where the minimum value is $0$.$3$. At $x = 0$,$f(0) = |0^2 - 1| = |-1| = 1$. For values of $x$ in a small neighborhood around $0$ (e.g.,$x \in (-1, 1)$),$f(x) = |x^2 - 1| = 1 - x^2$. Since $1 - x^2 Thus,$f$ has local minima at $x = \pm 1$ and a local maxima at $x = 0$.

Let $f: R \rightarrow R$ be given by $f(x) = |x^2 - 1|$,then

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