Rizags
  • Rizags
Help with differentiation Please! Problem Below
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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Rizags
  • Rizags
If\[f(x) = x^2, x \le1\] and \[f(x)=2x, x>1\] is f differentiable at 1?
Rizags
  • Rizags
wait though, it isn't continuous at 1
marigirl
  • marigirl
your write cuz it looks like this |dw:1444166725638:dw| IT is continuous but not differentable at x=1 sorry if i confused u before

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Rizags
  • Rizags
that isnt continuous
Rizags
  • Rizags
also why is it not differentiable
marigirl
  • marigirl
Limits: usually we look at a specific value when the graph approaches a number that we are after. A limit won't exist when we get different solutions from above and below graph (See my picture)
marigirl
  • marigirl
|dw:1444167201542:dw|
marigirl
  • marigirl
So when we approach the discontinuity form above and below point x=1 the graph no longer approaches the same value
marigirl
  • marigirl
same y value*
Rizags
  • Rizags
is this valid:\[f'(1)=\lim_{h \rightarrow 0^{-}}\frac{ (1+h)^2-(1)^2 }{ h }=2\] and\[f'(1)=\lim_{h \rightarrow 0^+}\frac{2(1+h)+2(1)}{h}=2\]
Rizags
  • Rizags
because the limits are the same, its just that the function isn't continuous
marigirl
  • marigirl
@surjithayer
anonymous
  • anonymous
It is Continuous because the x value exists at all points. x=1 is subsidized for the leap. just because there is a leap doesn't make it not continuous.
Rizags
  • Rizags
that isn't the definition of continuity
Rizags
  • Rizags
limits from both sides have to be the same
anonymous
  • anonymous
a function f is said to be continuous at x=a 1. Limit exists Means left hand limit =Right hand limit. 2. f(x)=f(a) Here limit does not exist at x=1 so it is not continuous at x=1 so it is not differentiable at x=1

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