Greeting I have a question regarding continuity and derivatives can someone help me with this question if possible. The question I am referring to is 6).

Thx in advance!

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- Mar 11th 2010, 09:54 AMSolid8SnakeQuestion Regarding derivatives
Greeting I have a question regarding continuity and derivatives can someone help me with this question if possible. The question I am referring to is 6).

Thx in advance! - Mar 12th 2010, 12:02 AMSudharaka
Dear Solid8Snake,

If x<-1, $\displaystyle \frac{dy}{dx}=-2$

If -1<x<1, $\displaystyle \frac{dy}{dx}=2x$

If x>1, $\displaystyle \frac{dy}{dx}=1$

Therefore f is differentiable throught the real number system.

Can you do it from here? - Mar 12th 2010, 03:41 AMHallsofIvy
This is not correct. At x= 1, the function is not differentiable. The derivative there, if it existed would have to be $\displaystyle \lim_{h\to 0}\frac{f(1+h)- f(1)}{h}$.

From below that is $\displaystyle \lim_{h\to 0^-}\frac{f(1+h)-f(1)}{h}= \lim_{h\to 0}\frac{(1+h)^2- 1}{h}$$\displaystyle = \lim_{h\to 0}\frac{2h+ h^2}{h}= 2$.

From above that is $\displaystyle \lim_{h\to 0^+}\frac{f(1+h)- f(1)}{h}= \lim_{h\to 0}\frac{1+h- 1}{h}$$\displaystyle = \lim_{h\to 0}\frac{h}{h}= 1$.

Since those two one sided limits are not the same the limit itself and so the derivative does not exist at x= 1.

While the derivative of a function is not necessarily continuous, it does satisfy the "intermediate value property" so**if**the limits of the derivative function, from either side, exist, they must be the same. Here, the limit of 2x, as x goes to 1 is 2 and the limit of 1, as x goes to 1, is 1. They exist but are not the same so the function is not differentiable at x= 1.

There will be a "corner" at (1, 1) with the curve coming in from the left at angle $\displaystyle tan(\theta)= 2$ and from the right at angle $\displaystyle tan(\theta)= 1$.

Quote:

Can you do it from here?

- Mar 12th 2010, 03:43 AMHallsofIvy
This is not correct. At x= 1, the function is not differentiable. The derivative there, if it existed would have to be $\displaystyle \lim_{h\to 0}\frac{f(1+h)- f(1)}{h}$.

From below that is $\displaystyle \lim_{h\to 0^-}\frac{f(1+h)-f(1)}{h}= \lim_{h\to 0}\frac{(1+h)^2- 1}{h}$$\displaystyle = \lim_{h\to 0}\frac{2h+ h^2}{h}= 2$.

From above that is $\displaystyle \lim_{h\to 0^+}\frac{f(1+h)- f(1)}{h}= \lim_{h\to 0}\frac{1+h- 1}{h}$$\displaystyle = \lim_{h\to 0}\frac{h}{h}= 1$.

Since those two one sided limits are not the same the limit itself and so the derivative does not exist at x= 1.

While the derivative of a function is not necessarily continuous, it does satisfy the "intermediate value property" so**if**the limits of the derivative function, from either side, exist, they must be the same. Here, the limit of 2x, as x goes to 1 is 2 and the limit of 1, as x goes to 1, is 1. They exist but are not the same so the function is not differentiable at x= 1.

Notice that at x= -2, the limits of the two derivative functions are $\displaystyle \lim_{x\to -1} -2= -2$ and $\displaystyle \lim_{x\to -1} 2x= -2$ so the function**is**differentiable at x= -1.

There will be a "corner" in the graph at (1, 1) with the curve coming in from the left at angle $\displaystyle tan(\theta)= 2$ and from the right at angle $\displaystyle tan(\theta)= 1$.

Quote:

Can you do it from here?

- Mar 12th 2010, 06:05 PMSudharaka
Dear HallsofIvy,

Thank you very much for showing my mistake. This is what I always expect from MHF, to know things I don't know and to further improve the knowledge of what I already know. Thanks a lot. (I suppose you are not angry with me for posting an incorrect answer!!)

Goodbye.