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