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Question: The domain of the derivative of the function \(f(x) = \left\{ \begin{matrix} \tan^{- 1}x & , & |x| ...

The domain of the derivative of the function

f(x)={tan⁡−1x,∣x∣≤112(∣x∣−1),∣x∣>1 f(x) = \left\{ \begin{matrix} \tan^{- 1}x & , & |x| \leq 1 \\ \frac{1}{2}(|x| - 1) & , & |x| > 1 \end{matrix} \right.\ is

A

R−{0}R - \{ 0\}

B

R−{1}R - \{ 1\}

C

R−{−1}R - \{ - 1\}

D

R−{−1,1}R - \{ - 1,1\}

Answer

R−{−1}R - \{ - 1\}

Explanation

Solution

f(x)={12(−x−1),x<−1tan⁡−1x,−1≤x≤112(x+1),x>1 f(x) = \left\{ \begin{aligned} & \frac{1}{2}( - x - 1),x < - 1 \\ & \tan^{- 1}x, - 1 \leq x \leq 1 \\ & \frac{1}{2}(x + 1),x > 1 \end{aligned} \right.\ ⇒

& - \frac{1}{2},x < - 1 \\ & \frac{1}{1 + x^{2}}, - 1 < x < 1 \\ & \frac{1}{2},x > 1 \end{aligned} \right.\ $$ $f ^ { \prime } ( - 1 - 0 ) = - \frac { 1 } { 2 } ; f ^ { \prime } ( - 1 + 0 ) = \frac { 1 } { 1 + ( - 1 + 0 ) ^ { 2 } } = \frac { 1 } { 2 }$ $f^{'}(1 - 0) = \frac{1}{1 + (1 - 0)^{2}} = \frac{1}{2};f^{'}(1 + 0) = \frac{1}{2}$ ∴$f^{'}( - 1)$ does not exist. ∴domain of $f^{'}(x) = R - \{ - 1\}$.