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About LinkBacksGreeting 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!
This is not correct. At x= 1, the function is not differentiable. The derivative there, if it existed would have to beOriginally Posted by Sudharaka
Dear Solid8Snake,
If x<-1,
If -1<x<1,
If x>1,
Therefore f is differentiable throught the real number system..
From below that is-f(1)}{h}= \lim_{h\to 0}\frac{(1+h)^2- 1}{h})
.
From above that is- f(1)}{h}= \lim_{h\to 0}\frac{1+h- 1}{h})
.
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 angleand from the right at angle
.
Can you do it from here?
This is not correct. At x= 1, the function is not differentiable. The derivative there, if it existed would have to beDear Solid8Snake,
If x<-1,
If -1<x<1,
If x>1,
Therefore f is differentiable throught the real number system.
From below that is
From above that is
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 areand
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
Can you do it from here?
Dear HallsofIvy,This is not correct. At x= 1, the function is not differentiable. The derivative there, if it existed would have to be
From below that is
From above that is
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
There will be a "corner" in the graph at (1, 1) with the curve coming in from the left at angle
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.
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