Question

Test the continuity and differentiability of f(x) `{:(= 3 x + 2, "if"  x > 2),(= 12 - x^2, "if"  x ≤ 2):}}` at x = 2

Answer 1

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.