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anonymous

  • 4 years ago

Hey guys! Problem 1D8a... f(x)=ax+b, x>0 ; f(x)=sin2x, x</=o. What are the values of constants a and b so the function is not differentiable. Trouble when trying to find the derivative f'(0-)...

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  1. TuringTest
    • 4 years ago
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    f'(0)=a=2cos(0)=2 so if a is not =2 it's not differentiable at zero

  2. TuringTest
    • 4 years ago
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    plus it's not differentiable if it's not continuous, which is the case for b not=0

  3. anonymous
    • 4 years ago
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    Ok, just saw a topic below and realized and haven't got to the part when I can calculate f'(0-), so for me it didn't make sense that f'(x)=2cos(2x)... Just gonna move on and tackle it again later! I got the b part nonetheless... Thanks a lot for the help ;)

  4. TuringTest
    • 4 years ago
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    oh yeah, the chain rule is later in the series but there are questions that require it in the beginning. They should fix that :/

  5. TuringTest
    • 4 years ago
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    ...you could perhaps use a trig formula instead

  6. anonymous
    • 4 years ago
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    |dw:1328490764769:dw| |dw:1328491513309:dw|

  7. anonymous
    • 4 years ago
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    Got it! Thanks a lot :)

  8. anonymous
    • 4 years ago
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    This is the value of a and b where it is differentiable....are you looking for the values where it is not?

  9. anonymous
    • 4 years ago
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    Therefore if a is not equal to 2 or if b is not equal to zero, then the function is not differentiable.

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