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Let $f:R\rightarrow R$ be a function defined by $f(x)=max(x,x^3)$.The set of all points where $f(x)$ is not differentiable

$(a)\;\{-1,1\}\qquad(b)\;\{-1,0\}\qquad(c)\{0,1\}\qquad(d)\;\{-1,0,1\}$

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$f(x)=max\{x,x^3\}$
$\;\;\;\;\;\;\;=\left\{\begin{array}{1 1}x;&x< -1\\x^3;&-1\leq x\leq 0\\x;&0\leq x\leq 1\\x^3;&x\geq 1\end{array}\right.$
$\therefore f'(x)=\left\{\begin{array}{1 1}1;&x< -1\\3x^2;&-1\leq x\leq 0\\1;&0\leq x\leq 1\\3x^2;&x\geq 1\end{array}\right.$
$\Rightarrow f$ is not differentiable at $-1,0$ and $1$
Hence (d) is the correct answer.
answered Dec 20, 2013 by sreemathi.v
 

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