1. ## Question 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).

2. Dear Solid8Snake,

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

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

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

Therefore f is differentiable throught the real number system.

Can you do it from here?

3. Originally Posted by Sudharaka
Dear Solid8Snake,

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

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

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

Therefore f is differentiable throught the real number system.
This is not correct. At x= 1, the function is not differentiable. The derivative there, if it existed would have to be $\lim_{h\to 0}\frac{f(1+h)- f(1)}{h}$.

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

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

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 $tan(\theta)= 2$ and from the right at angle $tan(\theta)= 1$.

Can you do it from here?

4. Originally Posted by Sudharaka
Dear Solid8Snake,

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

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

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

Therefore f is differentiable throught the real number system.
This is not correct. At x= 1, the function is not differentiable. The derivative there, if it existed would have to be $\lim_{h\to 0}\frac{f(1+h)- f(1)}{h}$.

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

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

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 $\lim_{x\to -1} -2= -2$ and $\lim_{x\to -1} 2x= -2$ 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 $tan(\theta)= 2$ and from the right at angle $tan(\theta)= 1$.

Can you do it from here?

5. Originally Posted by HallsofIvy
This is not correct. At x= 1, the function is not differentiable. The derivative there, if it existed would have to be $\lim_{h\to 0}\frac{f(1+h)- f(1)}{h}$.

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

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

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 $\lim_{x\to -1} -2= -2$ and $\lim_{x\to -1} 2x= -2$ 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 $tan(\theta)= 2$ and from the right at angle $tan(\theta)= 1$.
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.
The content structure consists of a series of posts in a forum or discussion thread, where a user asks a question about derivatives and continuity, and other users provide answers and corrections.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
1. ## Question 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).

2. Dear Solid8Snake,

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

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

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

Therefore f is differentiable throught the real number system.

Can you do it from here?

3. Originally Posted by Sudharaka
Dear Solid8Snake,

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

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

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

Therefore f is differentiable throught the real number system.
This is not correct. At x= 1, the function is not differentiable. The derivative there, if it existed would have to be $\lim_{h\to 0}\frac{f(1+h)- f(1)}{h}$.

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

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

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 $tan(\theta)= 2$ and from the right at angle $tan(\theta)= 1$.

Can you do it from here?

4. Originally Posted by Sudharaka
Dear Solid8Snake,

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

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

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

Therefore f is differentiable throught the real number system.
This is not correct. At x= 1, the function is not differentiable. The derivative there, if it existed would have to be $\lim_{h\to 0}\frac{f(1+h)- f(1)}{h}$.

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

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

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 $\lim_{x\to -1} -2= -2$ and $\lim_{x\to -1} 2x= -2$ 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 $tan(\theta)= 2$ and from the right at angle $tan(\theta)= 1$.

Can you do it from here?

5. Originally Posted by HallsofIvy
This is not correct. At x= 1, the function is not differentiable. The derivative there, if it existed would have to be $\lim_{h\to 0}\frac{f(1+h)- f(1)}{h}$.

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

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

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 $\lim_{x\to -1} -2= -2$ and $\lim_{x\to -1} 2x= -2$ 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 $tan(\theta)= 2$ and from the right at angle $tan(\theta)= 1$.
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.
user
The content structure consists of a series of posts in a forum or discussion thread, where a user asks a question about derivatives and continuity, and other users provide answers and corrections.

TYPE: SINGLE_QUESTION_WITH_ANSWER