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Math Help - showing a function is continuous and differentiable

  1. #1
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    showing a function is continuous and differentiable

    Hello,

    I have a problem in hand. I'll pose the question and include my attempt. Please correct me wherever I am wrong.

    Let f and g are functions that satisfy |f(x)| =< |g(x)| for all x and such that g is not the zero function. Give examples of what g could be to make sure that 1) f is continuous at 0 2) f is differentiable at 0.

    1) I couldn't solve it. any help?
    2) If I chose function |f(x)| =< |g(x)| = x^2, for f to be differentiable at 0,
    f'(0) = lim x->0 [f(x) - f(0)]/[x-0] = lim x->0 f(x)/x = limx->0 x^2/x = 0.
    But on 2, am I even allowed to assume f(0) = 0?

    I can rephrase things if there's a confusion. Thanks.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dgmath View Post
    Hello,

    I have a problem in hand. I'll pose the question and include my attempt. Please correct me wherever I am wrong.

    Let f and g are functions that satisfy |f(x)| =< |g(x)| for all x and such that g is not the zero function. Give examples of what g could be to make sure that 1) f is continuous at 0 2) f is differentiable at 0.

    1) I couldn't solve it. any help?
    2) If I chose function |f(x)| =< |g(x)| = x^2, for f to be differentiable at 0,
    f'(0) = lim x->0 [f(x) - f(0)]/[x-0] = lim x->0 f(x)/x = limx->0 x^2/x = 0.
    But on 2, am I even allowed to assume f(0) = 0?

    I can rephrase things if there's a confusion. Thanks.
    1) g(x)=x
    2) You know that f(0)=0. Why?
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    1) g(x)=x
    2) You know that f(0)=0. Why?
    Is it because |g(x)| is non-zero, and since |f(x)|=<|g(x)|, it is valid to say that |f(0)| can be 0?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dgmath View Post
    Is it because |g(x)| is non-zero, and since |f(x)|=<|g(x)|, it is valid to say that |f(0)| can be 0?
    In your specific examples. If |f(x)|\leqslant g(x) and if g(0)=0 then |f(0)|\leqslant g(0)=0\implies |f(0)|=0\implies f(0)=0
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    In your specific examples. If |f(x)|\leqslant g(x) and if g(0)=0 then |f(0)|\leqslant g(0)=0\implies |f(0)|=0\implies f(0)=0
    That is logical. However, what has confused me in this question is the part when it says "g is not the zero function". I kind of thought value of g(x) can never be 0? Did I completely misunderstand it?
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dgmath View Post
    That is logical. However, what has confused me in this question is the part when it says "g is not the zero function". I kind of thought value of g(x) can never be 0? Did I completely misunderstand it?
    g(x)\ne 0(x)
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  7. #7
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    Quote Originally Posted by Drexel28 View Post
    g(x)\ne 0(x)
    Great, thanks!

    Just before I move on, for part 1), how do I say that for g(x) = x, f is continuous at 0? Because I figured my answer i.e. x^2 works too, but I dont know how its shown.
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by dgmath View Post
    Great, thanks!

    Just before I move on, for part 1), how do I say that for g(x) = x, f is continuous at 0? Because I figured my answer i.e. x^2 works too, but I dont know how its shown.
    f(0)=0 by the same idea and the limit as x approaches zero of the function is zero by the squeeze theorem.
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