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If \(f (x) = |\; x\; |^3\), show that \(f ″(x)\) exists for all real \(x\) and find it.

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Toolbox:
  • $\large\frac{d}{dx}$$(x^{\large n})=nx^{\large n-1}$
Step 1:
When $x\geq 0$ then $f(x)=\mid x\mid^3=x^3$
$f'(x)=3x^2$
$f''(x)=6x$ which exists for all real values of $x$
Step 2:
When $x\leq 0$ then $f(x)=\mid x\mid^3=(-x)^3=-x^3$
$f'(x)=-3x^2$
$f''(x)=-6x$,which exists for all real value of $x$.
Step 3:
Hence $f''(x)=\left\{\begin{array}{1 1}6x&if\;x\geq 0\\-6x&if\;x<0\end{array}\right.$
answered May 15, 2013 by sreemathi.v
 

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