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clara1223

  • one year ago

At x=3, is the function given by f(x)={x^2 when x<3 and 6x-9 when x>=3} continuous and/or differentiable? I know that the graph looks like a parabola but just continues at x=3, so I know it is continuous but would it be considered a sharp turn (would it be differentiable)?

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  1. clara1223
    • one year ago
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    |dw:1442957985635:dw| That's what it looks like.

  2. zepdrix
    • one year ago
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    Oh these are kinda neat :o So ummmm

  3. zepdrix
    • one year ago
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    Ya so you have a line segment, and a curve. For `continuous` you need, what I like to call, `connectedness`. The pieces have to connect together. For `differentiable` you need, uh I guess we'll call it, `smoothness`. The slope from the left side needs to be equal to the slope from the right side.

  4. zepdrix
    • one year ago
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    So for continuity, you need:\[\large\rm \lim_{x\to3^-}f(x)=\lim_{x\to3^+}f(x)\]

  5. zepdrix
    • one year ago
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    The 3^- shows that we're approaching from the `left side`. When we're on the left side of 3, are we dealing with the line segment, or the parabola?

  6. zepdrix
    • one year ago
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    Oh you already figured out continuous :p Woops I didn't read the whole post lol

  7. zepdrix
    • one year ago
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    For differentiable, you need the slopes to be the same from the left and right,\[\large\rm \lim_{x\to3^-}f'(x)=\lim_{x\to3^+}f'(x)\]

  8. zepdrix
    • one year ago
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    Take some derivatives :o

  9. clara1223
    • one year ago
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    @zepdrix what do you mean? Like take the derivative and plug in x=2 and 4? or what?

  10. zepdrix
    • one year ago
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    The only value you'll end up plugging in is x=3. That's the only one that matters. When we're on the left side of 3, the function is \(\large\rm f(x)=x^2\) So on the left side of 3, \(\large\rm f'(x)=2x\). \[\large\rm \lim_{x\to3^-}f'(x)=\lim_{x\to3^+}f'(x)\]\[\large\rm \lim_{x\to3^-}2x=\lim_{x\to3^+}f'(x)\] How bout from the right side of 3? the function \(\large\rm f(x)=?\)

  11. clara1223
    • one year ago
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    well if f'(x)=2x, wouldn't it just be 2x?

  12. clara1223
    • one year ago
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    Wait I don't think that was what you were asking, I'm confused.

  13. zepdrix
    • one year ago
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    We have a `piece-wise function`. When we're on the RIGHT of 3, our function is defined by a different `piece`.

  14. clara1223
    • one year ago
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    yes, I get that

  15. clara1223
    • one year ago
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    So would we take f'(x) of 6x-9?

  16. clara1223
    • one year ago
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    which would be 6

  17. zepdrix
    • one year ago
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    |dw:1442961574141:dw|Yes, good. We're using that other piece to define our function on the right side o f3.

  18. zepdrix
    • one year ago
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    Ugh, I got a new mouse. It messed up my tablet settings :c Grr I'll have to fix that later..

  19. clara1223
    • one year ago
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    Haha still completely legible, continue

  20. clara1223
    • one year ago
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    if f'(x)=2(3)=6 then they're equal, does that make it differentiable?

  21. zepdrix
    • one year ago
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    Maybe I should have written it like this. We are trying to see if these are equal, not just say that they are.\[\large\rm \lim_{x\to3^-}2x\stackrel{?}{=}\lim_{x\to3^+}f'(x)\]\[\large\rm \lim_{x\to3^-}2x\stackrel{?}{=}\lim_{x\to3^+}6\]

  22. zepdrix
    • one year ago
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    So you plug in your 3, and you determined that the slope values are the same from the left and right. So yayyy it's differentiable! \c:/

  23. clara1223
    • one year ago
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    Thanks!

  24. zepdrix
    • one year ago
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    Looks kinda cool when you graph it :o https://www.desmos.com/calculator/klqptvzagg

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