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This is from a textbook for high school pupils. Derivatives have just been explained. And it is said that if the derivative exists for an argument, then the function is continuous at this point.

The task is: determine whether the function is continuous or not at 1. x=0, and 2. x = -1.

The function and my attempt is in the picture.

Well, for me this function is continuous in for both points. Well, the derivative exists from what I can see to the best of my ability.

But from the picture we can see that as if the function is not continuous where x = -1.

Anyway, I seem to have come to my wits end. Could you give me a kick here?

enter image description here

Jeremy Dover
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Michael
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    Your picture seems to show a jump discontinuity at $x=-1$. – Angina Seng Sep 17 '17 at 18:16
  • I myself drew that picture for better understanding. This is not required in the task. What is required is to check derivatives, as far as I can understand. – Michael Sep 17 '17 at 18:20
  • If you assume that polynomials are continuous (which they are), then you don't even need derivatives. Just a two sided limit at $x=-1$. – Dando18 Sep 17 '17 at 18:21
  • ....ion is continuous or not if x1=0, x2 = -1 please explain what do you mean by x1&x2 – neonpokharkar Sep 17 '17 at 18:22
  • That is very peculiar: the question (as you state it) makes no mention of derivatives. Also you draw a function as if it has a discontinuity, yet state it is continuous. – Angina Seng Sep 17 '17 at 18:23
  • Go back to the definition of derivative. Notice that you cannot compute the derivative just looking at the value of the function at $x=-1$, you need all the points in a neighbourhood. So your answer that $f'(x)=1$ is wrong because you are not taking into account the points on the right of $-1$. – Miguel Sep 17 '17 at 18:27
  • How do you conclude that the derivative exists ??? –  Sep 17 '17 at 18:30

1 Answers1

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Aat $x=-1$, this function has a left-hand side derivative: $\;f'_s(-1)=1$, and no right-hand side derivative since the limit of the rate of variation for $h>0$: $$ \dfrac{f(-1+h)-f(-1)}h=\dfrac{h^2-3h}h=h-\frac3h=-\infty.$$ So the function has no derivative at this point.

Bernard
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  • At $x=-1$, the function has no right-hand derivative. – Angina Seng Sep 17 '17 at 18:24
  • Oh! Yes. I stupidly used the limit of the derivative of the second formula. Thanks for correcting me – Bernard Sep 17 '17 at 18:26
  • If I'm not mistaken, |x| also has no right-hand derivative at 0. But it is continuous at 0. So, I still can't understand this task. – Michael Sep 17 '17 at 18:43
  • @Michael: No, it has a derivative at every point, except at $-1$, since is it locally a polynomial function. – Bernard Sep 17 '17 at 18:47
  • Well, I accept your logic. But then I can't catch the contrast with the function y=|x|. As I wrote, it is continuous but has no derivative at 0. – Michael Sep 17 '17 at 18:54
  • This one is not continuous at $-1$. It is continuous on the left-hand side, not on the right-hand side. – Bernard Sep 17 '17 at 19:01