Question

Show that the function `f(x)=|x-3|,x in R` is continuous but not differentiable at x = 3.

Answer 1

`f(x)=|x-3|={(3-x, x<3),(x-3,x>=3) :}`

Let c be a real number.

Case I: c < 3. Then f (c) = 3 − c.

`lim_(x->c)f(x)=lim_(x->c)(3-x)=3-c`

Since `lim_(x->c)f(x)=f(c)` f is continuous at all negative real numbers.

Case II: c = 3. Then f (c) = 3 − 3 = 0

`lim_(x->c)f(x)=lim_(x->c)(x-3)=3-3=0`

Since , `lim_(x->3)f(x)=f(3)` ,f is continuous at x = 3.

Case III: c > 3. Then f (c) = c − 3.

`lim_(x->c)f(x)=lim_(x->c)(x-3)=c-3`

Since `lim_(x->c)f(x)=f(c)`  f is continuous at all positive real numbers.

Therefore, f is continuous function.

We will now show that f(x)=|x-3|,x in R is not differentiable at x = 3.

Consider the left hand limit of f at x = 3

`lim_(h->0^-)(f(3+h)-f(3))/h=lim_(h->0^-)(|3+h-3|-|3-3|)/h=lim_(h->0^-)(|h|-0)/h=lim_(h->0^-)-h/h=-1`

consider the right hand limit of f at x=3

`lim_(h->0^-)(f(3+h)-f(3))/h=lim_(h->0^-)(|3+h-3|-|3-3|)/h=lim_(h->0^-)(|h|-0)/h=lim_(h->0^-)h/h=-1`

Since the left and right hand limits are not equal, f is not differentiable at x = 3.