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Let $\mathrm{f(x)=(x+|x|)|x|}$ , Then for all x
Option: 1
f is continuous

Option: 2
$\mathrm{f ~is~ differentiable~ for~ some~ x }$

Option: 3
$\mathrm{f^{\prime}}$ is continuous

Option: 4
$\mathrm{f^{\prime \prime}~ is~ continuous}$

Answers (1)

best_answer

$\mathrm{{ We ~have, } f(x)=\left\{\begin{array}{l} (x-x)(-x)=0, x<0 \\ (x+x) x=2 x^2, x \geq 0 \end{array}\right. \text {. }}$

As is evident from the graph of f(x) that it is continuous and differentiable for all x. Also,
$\mathrm{ f^{\prime}(x)= \begin{cases}0, & x<0 \\ 4 x, & x \geq 0^{\circ}\end{cases} }$
Clearly, $\mathrm{ f^{\prime}(x) }$ is continuous for all x but it is not differentiable at x=0.

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