Test whether the function f(x) `{:(= x^2 + 1",", "for" x ≥ 2),(= 2x + 1",", "for" x < 2):}` is differentiable at x = 2
Solution
f(x) = x2 + 1, for x ≥ 2
∴ f(2) = 22 + 1 = 5
Now, Lf'(2) = `lim_("h" -> 0^-) ("f"(2 + "h") - "f"(2))/"h"`
= `lim_("h" -> 0) ([2(2 + "h") + 1] - 5)/"h"` ...[∵ f(x) = 2x + 1, for x < 2]
= `lim_("h" -> 0) (4 + 2"h" + 1 - 5)/"h"`
= `lim_("h" -> 0) (2"h")/"h"`
= `lim_("h" -> 0) 2` ...[∵ h → 0, ∴ h ≠ 0]
= 2
Rf'(2) = `lim_("h" -> 0^+) ("f"(2 + "h") - "f"(2))/"h"`
= `lim_("h" -> 0) ([(2 + "h")^2 + 1] - 5)/"h"` ...[∵ f(x) = x2 + 1, for x ≥ 2]
= `lim_("h" -> 0) (4 + 4"h" + "h"^2 + 1 - 5)/"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
∴ Lf'(2) ≠ Rf'(2)
∴ f is not differentiable at x = 2.
