Question

Show that f(x) = |x – 5| is continuous but not differentiable at x = 5.

Answer 1

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.