Question

Examine the differentiability of the function f defined by
f(x) = `{{:(2x + 3",",  "if"  -3 ≤ x < - 2),(x + 1",",  "if"  -2 ≤ x < 0),(x + 2",",  "if"  0 ≤ x ≤ 1):}`

Answer 1

The only doubtful points for differentiability of f(x) are x = – 2 and x = 0.

Differentiability at x = – 2.

Now Lf'(–2) = `lim_("h" -> 0) ("f"(-2 + "h")  "f"(-2))/"h"`

= `lim_("h" -> 0^-) (2(-2 + "h") + 3 - (-2 + 1))/"h"`

= `lim_("h" -> 0^-)  (2"h")/"h"`

= `lim_("h" -> 0^-) 2`

= 2

And Rf'(–2) = `lim_("h" -> 0^+) ("f"(-2 + "h") - "f"(-2))/"h"`

= `lim_("h" ->0^+) (-2 + "h" + 1 - (-2 + 1))/"h"`

= `lim_("h" ->0^+) ("h" - 1 - (-1))/"h"`

= `lim_("h" -> 0^+) "h"/"h"`

= 1

Thus R f′(–2) ≠ Lf′(–2).

Therefore f is not differentiable at x = – 2.

Similarly, for differentiability at x = 0, we have

Lf'(0) = `lim_("h" -> 0^-) ("f"(0 + "h") - "f"(0))/"h"`

= `lim_("h" -> 0^-) (0 + "h" + 1 - (0 + 2))/"h"`

= `lim_("h" -> 0^-) ("h" - 1)/"h"`

= `lim_("h" ->0^-) (1 - 1/"h")`

Which does not exist.

Hence f is not differentiable at x = 0.