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Rizags

  • one year ago

Help with differentiation Please! Problem Below

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  1. Rizags
    • one year ago
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    If\[f(x) = x^2, x \le1\] and \[f(x)=2x, x>1\] is f differentiable at 1?

  2. Rizags
    • one year ago
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    wait though, it isn't continuous at 1

  3. marigirl
    • one year ago
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    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

  4. Rizags
    • one year ago
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    that isnt continuous

  5. Rizags
    • one year ago
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    also why is it not differentiable

  6. marigirl
    • one year ago
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    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)

  7. marigirl
    • one year ago
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    |dw:1444167201542:dw|

  8. marigirl
    • one year ago
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    So when we approach the discontinuity form above and below point x=1 the graph no longer approaches the same value

  9. marigirl
    • one year ago
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    same y value*

  10. Rizags
    • one year ago
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    is this valid:\[f'(1)=\lim_{h \rightarrow 0^{-}}\frac{ (1+h)^2-(1)^2 }{ h }=2\] and\[f'(1)=\lim_{h \rightarrow 0^+}\frac{2(1+h)+2(1)}{h}=2\]

  11. Rizags
    • one year ago
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    because the limits are the same, its just that the function isn't continuous

  12. marigirl
    • one year ago
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    @surjithayer

  13. anonymous
    • one year ago
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    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.

  14. Rizags
    • one year ago
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    that isn't the definition of continuity

  15. Rizags
    • one year ago
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    limits from both sides have to be the same

  16. anonymous
    • one year ago
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    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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