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Math Help - differentiable, f' not continuous

  1. #1
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    differentiable, f' not continuous

    Let f(x)=x^2 \sin(\frac{1}{x}) for x \not = 0 and f(0)=0.

    a) Prove that f(x) is differentiable on \mathbb{R}.
    b) Prove that f'(x) is not continuous on \mathbb{R}.

    I need help on BOTH parts. I've tried this problem, but I think I am not being very explicit in my proofs. I am also very sure my proof in b) is not nice. Is there another way to do b) [possibly without using the fundamental theorem of calculus]?

    Attempt:
    a) Clearly f(x) is differentiable at x \not = 0. [How do I show this?] In fact, f'(x)=2x \sin(x) - \cos(\frac{1}{x}) [valid for x \not =0]. At x=0, f'(0)=\lim_{x \rightarrow 0} \frac{f(h)-f(0)}{h}=\lim_{h \rightarrow 0} h \sin(\frac{1}{h})=0. Hence, f(x) is differentiable on \mathbb{R}.

    b) f'(x)=2x \sin(x) - \cos(\frac{1}{x}). Let p(x)=2x \sin(x). Then p(0)=0 because p(0)=2 \cdot \lim_{x \rightarrow 0} x \sin(\frac{1}{x})=0. So p(0)=0. So by FTC \exists P(x) s.t. P'(x)=p(x). So f'(x)=P'(x)-cos(\frac{1}{x}). But then \cos(\frac{1}{x}) is not a continuous function. But by FTC, P(x) is differentiable. Then P(x), f(x) are differentiable functions whose derivatives are not continuous.

    Thanks.
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  2. #2
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    f'(x)\to\infty as x\to0, thus f'(x) is not continuous at x=0. f is of course, differentiable but not of class 1.
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  3. #3
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    Quote Originally Posted by chloe561 View Post
    Let f(x)=x^2 \sin(\frac{1}{x}) for x \not = 0 and f(0)=0.

    a) Prove that f(x) is differentiable on \mathbb{R}.
    b) Prove that f'(x) is not continuous on \mathbb{R}.

    I need help on BOTH parts. I've tried this problem, but I think I am not being very explicit in my proofs. I am also very sure my proof in b) is not nice. Is there another way to do b) [possibly without using the fundamental theorem of calculus]?
    For part a), to justify that f is differentiable on \mathbb{R}\setminus\{0\}, you can say that it is obtained by product and composition of the following differentiable functions: x\mapsto x^2 on \mathbb{R}, \sin on \mathbb{R} and x\mapsto\frac{1}{x} on \mathbb{R}^*. The computation at 0 is fine (well, it would be better not to write f'(0) when you don't know that it exists yet)

    For part b), you don't need to refer to FTC : just say that x\sin\frac{1}{x} tends to 0, while \cos\frac{1}{x} has no limit as x\to0^+, so that f'(x) has no limit as x\to0^+. Therefore, it can't be continuous (whatever f'(0) is).

    Quote Originally Posted by Krizalid View Post
    f'(x)\to\infty as x\to0, thus f'(x) is not continuous at x=0. f is of course, differentiable but not of class 1.
    No, f'(x) is bounded near 0 (and it has no limit when x\to0^+).
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