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Math Help - f(x) = x^3

  1. #1
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    f(x) = x^3

    Let f(x) = x^3.

    1.) Prove (using the \epsilon-\delta definition) that at x = 2, f is continuous on \mathbb{R}.

    2.) Prove (using the def. of differentiability) that at x = 4, f is differentiable.

    Well, I don't know if this is helpful, but I know f is not uniformly continuous on \mathbb{R}. There is a similar example in my book proving that x^2 is continuous on \mathbb{R}. They do:

    Given c \in \mathbb{R}:

    |f(x) - f(c)| = |x^2 - c^2| = |x-c||x+c|

    Now they find an upperbound on |x+c| by makings ure \delta does not exceed 1. Apparently this insures all values of x will fall in the interval (c-1,c+1).

    I don't entirely get that ste.

    Then, they get:

    |x + c| \leq |x| + |c| \leq (|c|+1) + |c| = 2|c| + 1

    They let \epsilon > 0 be given, then choose \delta = min\left(1,\frac{\epsilon}{2|c|+1}\right), then |x - c| < \delta

    \implies |f(x) - f(c)| = |x-c||x+c|< \frac{\epsilon}{2|c|+1}(2|c|+1) = \epsilon.

    I'm assuming it's similar for x^3?
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  2. #2
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    Idea:

    Can I use the Algebraic Continuity Theorem and show f(x) = x is continuous, and therefore f(x) = x^3 will be continuous too? Since we can take f(x)*f(x)*f(x) = g(x), and by ACT, g(x) should be cont too.

    If so, how does this proof look:

    Let \epsilon > 0 be given. Further, choose \delta = \epsilon. Then, for x,y \in \mathbb{R}, |x - y| < \delta \implies |f(x) - f(y)| =  |x - y| < \epsilon
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    see here
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  4. #4
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    Oh nice! So, I can't make \delta = \epsilon like my proof above?
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by seerTneerGevoLI View Post
    Oh nice! So, I can't make \delta = \epsilon like my proof above?
    correct. use the structure of the proof given there
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by seerTneerGevoLI View Post
    2.) Prove (using the def. of differentiability) that at x = 4, f is differentiable.
    How rigorous was your definition of differentiability

    if not very rigorous, you can just show that the derivative limit has a numerical value using the limit definition of the derivative. that is, find

    \lim_{h \to 0} \frac {f(4 + h) - f(4)}{h}
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  7. #7
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    Let f : A \rightarrow \mathbb{R} be a func. that's defined on an interval A. Given c \in A, the derivative of f at c is defined by:

    f'(c) = \lim_{x\rightarrow c}{\frac{f(x) - f(c)}{x-c}}

    provided that the limit exists.

    f is diff'able on A if f' exists for all points c \in A.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by seerTneerGevoLI View Post
    Let f : A \rightarrow \mathbb{R} be a func. that's defined on an interval A. Given c \in A, the derivative of f at c is defined by:

    f'(c) = \lim_{x\rightarrow c}{\frac{f(x) - f(c)}{x-c}}

    provided that the limit exists.

    f is diff'able on A if f' exists for all points c \in A.
    fine, so all you need to do is show that f'(4) = \lim_{x \to 4} \frac {f(x) - f(4)}{x - 4} exists, where f(x) = x^3 of course.

    Tip: You can use "\to" instead of "\rightarrow" for the \to symbol in your limit
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  9. #9
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    For the second one:

    f'(4) = \lim_{x \to 4} \frac {f(x) - f(4)}{x - 4}

    So...

    f'(4) = \lim_{x \to 4} \frac {x^3 - 64}{x - 4}

    = \lim_{x \to 4} \frac{(x-4)(x^2 + 4x + 16)}{x-4}

     = \lim_{x \to 4} x^2 + 4x + 16 ONLY IF x \neq 4.

    So I don't see how to prove this is dif. at 4.

    Suppose I could do this. The limit then is 48. Is this really a proof showing it's dif'able?
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