Question

If f : R → R is defined by f(x) = |x|³, show that f''(x) exists for all real x and find it.

Answer 1

We have `f(x) = |x|^3,  {:{(x^3", if"  x ≥ 0),((-x)^3 = -x^3", if"  x < 0):}`

Now, (L.H.D. at x = 0) =`lim_(x->0^-)(f(x) - f(0))/(x - 0)`

= `lim_(x->0^-)((-x^3 - 0)/(x))`

= `lim_(x->0^-)(-x^2)`

= 0

(RHD at x = 0) `lim_(x->0^+)(f(x) - f(0))/(x - 0)`

= `lim_(x->0^+)((x^3 - 0)/(x))`

= `lim_(x->0)(-x^2)`

= 0

∴ (LHD of f(x) at x = 0) = (RHD of f(x) at x = 0)

So, f(x) is differentiable at x = 0 and the derivative of f(x) is given by

f'(x) = `{:{(3x^2", if"  x ≥ 0),(-3x^2", if"  x < 0):}`

Now, (LHD of f'(x) at x = 0) = `lim_(x->0^-)(f^'(x) - f^'(0))/(x - 0)`

= `lim_(x->0^-)((-3x^2 - 0)/x)`

= `lim_(x->0^-)(-3x)`

= 0

(RHD of f'(x) at x = 0) = `lim_(x->0^+)(f^'(x) - f^'(0))/(x - 0)`

= `lim_(x->0^+)((3x^2 - 0)/(x - 0))`

= `lim_(x->0^+)(3x)`

= 0

∴ (LHD of f'(x) at x = 0) = (RHD of f'(x) at x = 0) 

So, f'(x) is differentiable at x = 0.

Hence, f''(x) = `{:{(6x", if"  x ≥ 0),(-6x", if"  x < 0):}`.