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  • Let . The set of points where <img alt="\mathrm{f(x) }" s

Let \mathrm{f(x)=x \mid x }. The set of points where \mathrm{f(x) } is twice differentiable, is :

Option: 1

\mathrm{ R-\{0\} }


Option: 2

\mathrm{ R-\{0,1\} }


Option: 3

\mathrm{ R }


Option: 4

\mathrm{R-\{1\}}


Answers (1)

best_answer

\mathrm{f(x)=x|x|=\left\{\begin{array}{l} -x^2, x<0 \\ x^2, x \geq 0 \end{array}\right.}
\mathrm{\Rightarrow f^{\prime}(x)=\left\{\begin{array}{l} -2 x ; x<0 \\ 2 x ; x \geq 0 \end{array} \Rightarrow f^{\prime \prime}(x)=\left\{\begin{array}{l} -2 ; x<0 \\ 2 ; x \geq 0 \end{array}\right.\right..}
Clearly,\mathrm{ f^{\prime \prime}(x).} exists at every point except at \mathrm{x=0}. Thus, \mathrm{f(x)} is twice differentiable on \mathrm{R-\{0\}.}

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