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