If f(x) = begin{cases} tan^{-1} x, & text{whe | vedclass

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(A) The function $f(x)$ is defined as:$f(x) = \begin{cases} \frac{1}{2}(-x-1), & 	ext{if } x < -1 \ 	an^{-1} x, & 	ext{if } -1 \leq x \leq 1 \ \f

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$R - \{-1, 1\}$

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(A) The function $f(x)$ is defined as:$f(x) = \begin{cases} \frac{1}{2}(-x-1), & 	ext{if } x  1 \end{cases}$Check continuity at $x = -1$:$\lim_{x 	o -1^-} f(x) = \frac{1}{2}(1-1) = 0$$f(-1) = 	an^{-1}(-1) = -\frac{\pi}{4}$Since $\lim_{x 	o -1^-} f(x) 
eq f(-1)$,$f(x)$ is discontinuous at $x = -1$.Check continuity at $x = 1$:$\lim_{x 	o 1^+} f(x) = \frac{1}{2}(1-1) = 0$$f(1) = 	an^{-1}(1) = \frac{\pi}{4}$Since $\lim_{x 	o 1^+} f(x) 
eq f(1)$,$f(x)$ is discontinuous at $x = 1$.$A$ function must be continuous to be differentiable. Since $f(x)$ is discontinuous at $x = -1$ and $x = 1$,it is not differentiable at these points.Thus,the domain of $f'(x)$ is $R - \{-1, 1\}$.

If $f(x) = \begin{cases} \tan^{-1} x, & \text{when } |x| \leq 1 \\ \frac{1}{2}(|x|-1), & \text{when } |x| > 1 \end{cases}$,then the domain of $\frac{d}{dx} f(x)$ is

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