# Thread: how to show a function is not derivable at it's critical point?

1. ## how to show a function is not derivable at it's critical point?

I'm looking at the derivative of |-5x+8| WRT x

The critical point is 8/5. The derivative is = -5 if x<8/5 and 5 if x>8/5

this shows that the lim f'(x) as x-> 8/5 from above 8/5 is 5

and lim f'(x) as x-> 8/5 from below 8/5 is -5

This, in turn, shows that the derivative is discontinuous at 8/5 and therefore the function is not derivable at that point.

Does the above constitute a proof? Have I shows that the function is not derivable at its critical point? Have I successfully proven the limit does not exist?

thanks

2. ## Re: how to show a function is not derivable at it's critical point?

You were good until line 5. Just show the limit of the difference quotient doesn't exist at 8/5 and that shows the derivative doesn't exist at that point.

3. ## Re: how to show a function is not derivable at it's critical point?

Showing that a derivative is not continuous at a point does NOT show that the derivative does not exist there! A derivative does not necessarily have to be continuous.

For example, if $\displaystyle y= x^2sin(1/x)$ if $\displaystyle x\ne 0$, 0 if x= 0, then one can show that the derivative, for x not 0, is $\displaystyle 2x sin(1/x)- cos(1/x)$ so that the limit, as x goes to 0, does not exist. But the derivative at x= 0 exists. It is, from the basic definition, $\displaystyle \lim_{h\to 0}\frac{h^2sin(1/h)}{h}= \lim_{h\to 0} h sin(1/h)= 0$.

You can show, using the mean value theorem, that a derivative always has the "intermediate value property" so if a derivative is NOT continuous at a point, then it must be very complicated. The two one-sided limits must not exist. So if you can show the limit exists, the derivative must be the same as the limit.