anonymous
  • anonymous
Does there exist a function which is continuous everywhere but not differentiable at exactly two points ?
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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merchandize
  • merchandize
Yes it can exist. If you have a piecewise function with three parts, the endpoints can be continuous but not differentiable. For instance, say that you have a linear function that goes from negative infinity to negative 2 and stops. Then for the second part of the piecewise function, from negative 2 to positive 2 you would have a quadratic function. And from positive 2 to infinity, you could have another linear function. At x-values of negative and positive 2, the graphs intersect; therefore, they are continuous. However, at these points, cusps could exist, thereby making the function/curve not differentiable at those two points.
anonymous
  • anonymous
So, if there is a function like \[f(x)=\sqrt{x^{2}-4}\]then \[x \epsilon [-2,2]\]and at -2 and 2 the function is not differentiable.
anonymous
  • anonymous
@merchandize ?

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anonymous
  • anonymous
Yes but it is not continuous everywhere.
anonymous
  • anonymous
Think of setting up a piecewise function as merchandize suggested using the absolute value function.
merchandize
  • merchandize
yes.
anonymous
  • anonymous
what is a piecewise function actually ?
anonymous
  • anonymous
Example: \(f(x) = \left\{\begin{array}{ccc}0 & ; & x \leq 0 \\ 1 & ; & x > 0\end{array}\right.\)
merchandize
  • merchandize
You can create functions that behave differently depending on the input (x) value
merchandize
  • merchandize
it is same in the form as Alchemista mentioned in the example.
anonymous
  • anonymous
|dw:1370069696347:dw| then the graph would be this way right ? @Alchemista
anonymous
  • anonymous
Correct can you think of a way to modify my example using the absolute value function?
anonymous
  • anonymous
We want to use the absolute value function to create two points where the function is not differentiable.
anonymous
  • anonymous
I will give you half of what you need. $$f(x) = \left\{\begin{array}{ccc}|x + 1| & ; & x \leq 0 \\ 1 & ; & x > 0\end{array}\right.$$
anonymous
  • anonymous
Actually that's all you need.
anonymous
  • anonymous
Graph this function carefully and try to understand why its not differentiable at -1 and 0 $$f(x) = \left\{\begin{array}{ccc}|x + 1| & ; & x \leq 0 \\ 1 & ; & x > 0\end{array}\right.$$
anonymous
  • anonymous
|dw:1370070439326:dw|
anonymous
  • anonymous
these are the points where the graph turns.
anonymous
  • anonymous
And they are cusps. In other words they are not smooth so the limit of the difference function will not exist.
anonymous
  • anonymous
oh yes the left and the right limits won't coincide.
anonymous
  • anonymous
In less rigorous terms you will end up with a limit where the slope is 0 in one direction and nonzero in the other direction.
anonymous
  • anonymous
At zero that is.
anonymous
  • anonymous
now i get it :D
anonymous
  • anonymous
thank you so much @Alchemista and @merchandize :)

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