$f(X)$ is differentiable at $x=a$ but it is discontinuous

Question

I have defined a function such that:

$$f(x)=\begin{cases}mx,&(x\le a), \\ mx+c,&(x>a).\end{cases}$$

Here according to the derivative definition :

$f '(a) = \lim_{x\to a} [f (x) - f (a) ]/ [x - a] $

we can show that this limit exits by taking the LHS and RHS limits , and showing that they are equal. Since the gradient is the same I think it is trivial .

Can anyone please explain ?

Thank you !

Look carefully at how you have defined $f(a)$ and check your assertion that the limits on each side exist. — Mark Bennet, Jun 15 '19 at 19:46
Guessing where things went wrong. "Since the gradient is the same I think it is trivial": What's trivial is that $\lim_{x\to a}f'(x)=m$. That does not imply that $f'(a)$ exists - that's not the limit in the definition of $f'(a)$. — David C. Ullrich, Jun 15 '19 at 20:08

score 0 · Accepted Answer · answered Jun 15 '19 at 20:17

0

The right hand limit does not exist. Remember, $f(a) = ma$, not $ma + c$, even when you're finding the right hand limit.

answered Jun 15 '19 at 20:17

JonathanZ

10,615

Thank you so much – Lasan Manujitha Ukwaththa Liya Jun 15 '19 at 20:24

$f(X)$ is differentiable at $x=a$ but it is discontinuous

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