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

Answer 1

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.