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


When, x ≥ 0,


f(x) = |x|3 = x3


So, f’(x) = 3x2


And f’’(x) = d(f’(x))/dx = 6x


∴ f’’(x) = 6x


When x < 0,


f(x) = |x|3 = ( – x)3 = – x3


f’(x) = – 3x2


f’’(x) = – 6x


∴


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