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Consider the function \mathrm{f(x)=\left|x^3\right|}, where x is real. then the function \mathrm{f(x) \, \, at \, \, x=0} is

Option: 1

continuous but not differentiable
 


Option: 2

once differentiable but not twice
 


Option: 3

twice differentiable but not thrice
 


Option: 4

three differentiable


Answers (1)

best_answer

                            \mathrm{f(x)=|x|^3=\left\{\begin{array}{ccc} x^3 & \text { if } & x>0 \\ -x^3 & \text { if } & x<0 \\ 0 & \text { if } & x=0 \end{array}\right.}

\mathrm{f(x)} is continuous at \mathrm{x=0}

\mathrm{(\because LHS = RHS =f(0)=0)}

Now,                    \mathrm{ f^{\prime}(x)=\left\{\begin{array}{ccc} 3 x^2 & \text { if } & x>0 \\ -3 x^2 & \text { if } & x<0 \\ 0 & \text { if } & x=0 \end{array}\right. }

\mathrm{\because L H D=0=R H D \text { so } f(x) \text { is once differentiable }}

Again                           \mathrm{f^{\prime \prime}(x)=\left\{\begin{array}{ccc} 6 x & \text { if } & x>0 \\ -6 x & \text { if } & x<0 \\ 0 & \text { if } & x=0 \end{array}\right.}

\mathrm{\because \text { LHD }=0=\text { RHD so } f(x) \text { is twice differentiable }}

Again                        \mathrm{f^{\prime \prime}(x)=\left\{\begin{array}{ccc} 6 & \text { if } & x>0 \\ -6 & \text { if } & x=0 \\ \text { Does not if } x=0 \end{array}\right.}

\mathrm{\because \mathrm{LHD}=-6\, \, \& \, \, \mathrm{RHD}=6 \text { ie LHD } \neq \text { RHD }}

So  \mathrm{f(x) } is not thrice differentiable.

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