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Question
Show that f(x) = |x – 5| is continuous but not differentiable at x = 5.
Solution
We have f(x) = |x – 5|
⇒ f(x) = `{{:(-(x - 5)",", "if" x - 5 < 0 or x < 5),(x - 5",", "if" x - 5 > 0 or x > 5):}`
For continuity at x = 5
L.H.L. `lim_("h" -> 5^-) "f"(x)` = – (x – 5)
= `lim_("h" -> 0) - (5 - "h" - 5)`
= `lim_("h" -> 0) "h" = 0`
R.H.L. `lim_(x -> 5^+) "f"(x)` = x – 5
= `lim_("h" -> 0) (5 + "h" - 5)`
= `lim_("h" -> 0) "h"` = 0
L.H.L. = R.H.L.
So, f(x) is continuous at x = 5
Now, for differentiability
Lf'(5) = `lim_("h" -> 0) ("f"(5 - "h") - "f"(5))/(-"h")`
= `lim_("h" -> 0) (-(5 - "h" - 5) - (5 - 5))/(-"h")`
= `lim_("h" -> 0) "h"/(-"h")`
= – 1
Rf'(5) = `lim_("h" -> 0) ("f"(5 + "h") - "f"(5))/"h"`
= `lim_("h" -> 0) ((5 + "h" - 5) - (5 - 5))/"h"`
= `lim_("h" -> 0) ("h" - 0)/"h"`
= 1
∵ Lf'(5) ≠ Rd'(5)
Hence, f(x) is not differentiable at x = 5.
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