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
∴