Let f: R rightarrow R be given by f(x) = |x^{ | 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 testing points.At $x = \

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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 testing points.At $x = \pm 1$, $f(x) = |(\pm 1)^{2} - 1| = |1 - 1| = 0$. Since the absolute value function is always non-negative, $f(x) \geq 0$ for all $x \in R$. Thus, $f(x) = 0$ at $x = \pm 1$ represents the absolute minimum value of the function, which is also a local minimum.At $x = 0$, $f(0) = |0^{2} - 1| = |-1| = 1$. For values of $x$ close to $0$, say $x = 0.1$ or $x = -0.1$, $f(0.1) = |(0.1)^{2} - 1| = |0.01 - 1| = 0.99$. Since $f(0) = 1 > 0.99$, $x = 0$ is a point of local maximum.Therefore, $f$ has local minima at $x = \pm 1$ and a local maximum at $x = 0$.

Let $f: R \rightarrow R$ be given by $f(x) = |x^{2} - 1|$, $x \in R$. Then:

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