Q.Let $f(x) = \begin{cases} -x - 1 & x \le -1 \ x^2 - 1 & -1 < x < 1 \ x + 1 & x \ge 1 \end{cases}$ . Then $f$ is

continuous and differentiable everywhere

continuous everywhere but not differentiable at

x=\pm1

differentiable everywhere but not continuous at

x=\pm1

neither continuous nor differentiable at

x=\pm1

The correct solution involves applying the fundamental concept to derive the final value step by step...

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Q.Let f(x)={−x−1x≤−1 x2−1−1<x<1 x+1x≥1f(x) = \begin{cases} -x - 1 & x \le -1 \ x^2 - 1 & -1 < x < 1 \ x + 1 & x \ge 1 \end{cases}f(x)={−x−1​x≤−1 x2−1​−1<x<1 x+1​x≥1​. Then fff is