text: 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.
explanation: The document is a series of posts from a forum or discussion board, primarily focused on a math question regarding the differentiability of a function. The first post introduces the question, and subsequent posts provide an incorrect answer, a correction to that answer, and a response acknowledging the correction. The document contains a question and multiple answers.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
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  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: 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.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a series of posts from a forum or discussion board, primarily focused on a math question regarding the differentiability of a function. The first post introduces the question, and subsequent posts provide an incorrect answer, a correction to that answer, and a response acknowledging the correction. The document contains a question and multiple answers.
  
  Label: question-and-answer