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  1. #1
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    calc problem

    When is the function not differentiable? (see graph attached)

    The answer key says b, d, g ,h.
    I don't understand why the function is not differentiable at g. I thought that functions can be differentiated at removable discontinuities.
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  2. #2
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    Quote Originally Posted by harmonicadudee View Post
    When is the function not differentiable? (see graph attached)

    The answer key says b, d, g ,h.
    I don't understand why the function is not differentiable at g. I thought that functions can be differentiated at removable discontinuities.
    you thought wrong. if a function is differentiable at a point, then it is also continuous at that point.
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    Quote Originally Posted by harmonicadudee View Post
    When is the function not differentiable? (see graph attached)

    The answer key says b, d, g ,h.
    I don't understand why the function is not differentiable at g. I thought that functions can be differentiated at removable discontinuities.

    It's aweird question: at b it is not diff. because, apparently, there's a vertical line there...but then that's not a function!
    At d it is clear: the function has a "shpitz" or saw-tooth point there, so fine. At g and h the function isn't defined so clearly it can't be continuous there, no matter g is a removable discontinuity...

    Tonio
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    Quote Originally Posted by harmonicadudee View Post
    When is the function not differentiable? (see graph attached)

    The answer key says b, d, g ,h.
    I don't understand why the function is not differentiable at g. I thought that functions can be differentiated at removable discontinuities.
    Derivatives can't exist where a function isn't defined. How would you draw a tangent line to a point that doesn't exist?

    More formally, think about how you calculate the derivative at the point g.

    \lim_{h\to0}\frac{f(g+h)-f(g)}{h}

    But if f(g) doesn't exist, you can't calculate that limit.

    However, if you wanted to calculate \lim_{x\to g}f'(x), that would exist (in this case).
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    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by tonio View Post
    It's aweird question: at b it is not diff. because, apparently, there's a vertical line there...but then that's not a function
    That's not necessarily true. What about \sqrt[3]{x} at the origin?
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    Well I guess my calculus teacher was wrong....

    Thanks guys.
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    Quote Originally Posted by redsoxfan325 View Post
    That's not necessarily true. What about \sqrt[3]{x} at the origin?

    What EXACTLY "isn't necessarily true"? \sqrt[3]{x} is NOT a line segment at the origin. Since this is a visual exercise I am basing my opinion on what I see and that drawing shows like a vertical line segment at b. Of course, one could bring examples even more basic, as \sqrt{x} , which also isn't differentiable at the origin.

    Tonio
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  8. #8
    Super Member redsoxfan325's Avatar
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    Ok. We saw different things.
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