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Question
Test the continuity and differentiability of f(x) `{:(= 3 x + 2, "if" x > 2),(= 12 - x^2, "if" x ≤ 2):}}` at x = 2
Solution
Differentiability at x = 2
f(x) = 12 – x2 if x ≤ 2
∴ f(2) = 12 – (2)2 = 8
Lf'(2) = `lim_("h" -> 0^-) ("f"(2 + "h") - "f"(2))/"h"`
= `lim_("h" -> 0) ([12 - (2 + "h")^2] - 8)/"h"` ...[∵ f(x) = 12 − x2, if x ≤ 2]
= `lim_("h" -> 0) (12 - 4 - 4"h" - "h"^2 - 8)/"h"`
= `lim_("h" -> 0) (-4"h" - "h"^2)/"h"`
= `lim_("h" -> 0) ("h"[-4 - "h"])/"h"`
= `lim_("h" -> 0) [-4 - "h"]` ...[∵ h → 0 ∴ h → 0]
= – 4 – 0
= – 4
Rf'(2) = `lim_("h" -> 0^+) ("f"(2 + "h") - "f"(2))/"h"`
= `lim_("h" -> 0) ([3(2 + "h") + 2] - 8)/"h"` ...[∵ f(x) = 3x + 2, if x > 2]
= `lim_("h" -> 0) (6 + 3"h" + 2 - 8)/"h"`
= `lim_("h" -> 0) (3"h")/"h"`
= `lim_("h" -> 0) (3)` ...[∵ h → 0 ∴ h ≠ 0]
= 3
∴ Lf'(2) ≠ Rf'(2)
∴ f is not differentiable at x = 2
Continuity at x = 2
`lim_(x -> 2^-) "f"(x) = lim_(x -> 2) (12 - x^2)` = 12 – 4 = 8
f(2) = 12 – 4 = 8
`lim_(x -> 2^+) "f"(x) = lim_(x -> 2) (3x + 2)` = 3(2) + 2 = 8
∴ f(2) = `lim_(x - 2^+) "f"(x) = lim_(x -> 2^-) "f"(x)`
∴ f is continuous at x = 2
Hence, f is continuous at x = 2 but not differentiable at x = 2.
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