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Question
Examine the differentiability of f, where f is defined by
f(x) = `{{:(x[x]",", "if" 0 ≤ x < 2),((x - 1)x",", "if" 2 ≤ x < 3):}` at x = 2
Solution
We know that a function f is differentiable at a point ‘a’ in its domain if
Lf'(x) = Rf'(c)
where Lf'(c) = `lim_("h" -> 0) ("f"("a" - "h") - "f"("a"))/(-"h")` and Rf'(c) = `lim_("h" -> 0) ("f"("a" + "h") - "f"("a"))/"h"`
Here, f(x) = `{{:(x[x]",", "if" 0 ≤ x < 2),((x - 1)x",", "if" 2 ≤ x < 3):}` at x = 2
Lf'(c) = `lim_("h" -> 0) ("f"(2 - "h") - "f"(2))/(-"h")`
= `lim_("h" -> 0) ((2 - "h")[2 - "h"] - (2 - 1)2)/(-"h")`
= `lim_("h" -> 0) ((2 - "h") * 1 - 2)/(-"h")` ....[∵ [2 – h] = 1]
= `lim_("h" -> 0) (2 - "h" - 2)/(-"h")`
= 1
Rf'(c) = `lim_("h" -> 0) ("f"(2 + "h") - "f"(2))/"h"`
= `lim_("h" -> 0) ((2 + "h" - 1)(2 + "h") - (2 - 1)*2)/"h"`
= `lim_("h" -> 0) ((1 + "h")(2 + "h") - 2)/"h"`
= `lim_("h" -> 0) (2 - "h" + 2"h" + "h"^2 - 2)/"h"`
= `lim_("h" -> 0) (3"h" + "h"^2)/"h"`
= `lim_("h" -> 0) ("h"(3 + "h"))/"h"`
= 3
Lf"(2) ≠ Rf'(2)
Hence, f(x) is not disserentiable at x = 2.
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