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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):}`
Solution
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.
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