Question

Discuss whether the function f(x) = |x + 1| + |x  – 1| is differentiable ∀ x ∈ R

Answer 1

f(x) = |x + 1| + |x – 1|

= – (1 + x) + (1 – x),   x < – 1

= 1 + x + 1 – x,       –1 ≤ x < 1

= x + 1 + x – 1,         x ≥ 1

i.e., f(x) `{:(= - 2x",", x < -1),(= 2",", -1 ≤ x < 1),(= 2x",", x ≥ 1):}`

Differentiability at x = – 1:

Lf'(– 1) = `lim_("h" -> 0^-) ("f"(- 1 + "h") - "f"(- 1))/"h"`

= `lim_("h" -> 0^-) (-2(-1 + "h") - (2))/"h"`

= `lim_("h" -> 0^-) ((-2"h")/"h") = -2`

 Rf'(– 1) = `lim_("h" -> 0^+) ("f"(-1 + "h") - "f"(-1))/"h"`

= `lim_("h" -> 0^+) (2 - 2)/"h"` = 0

∵ Lf'(– 1) ≠ Rf'(– 1)

∴ f is not differentiable at x = – 1

Differentiability at x = 1:

Lf'(1) = `lim_("h" -> 0^-) ("f"(1 + "h") - "f"(1))/"h"`

= `lim_("h" -> 0^-) (2 - 2)/"h"` = 0

Rf'(1) = `lim_("h" -> 0^+) ("f"(1 + "h") - "f"(1))/"h"`

= `lim_("h" -> 0^+) (2(1 + "h") - (2))/"h"`

= `lim_("h" -> 0^-) ((2"h")/"h")` = 2

∵ Lf'(1) ≠ Rf'(1)

∴ f is not differentiable at x = 1.

∴ f is not differentiable at x = – 1 and x = 1

∴ and not differentiable  ∀ x ∈ R.