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Thread: f(x) = x^3

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

    Let $\displaystyle f(x) = x^3$.

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

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

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

    Given $\displaystyle c \in \mathbb{R}$:

    $\displaystyle |f(x) - f(c)| = |x^2 - c^2| = |x-c||x+c|$

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

    I don't entirely get that ste.

    Then, they get:

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

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

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

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

    Can I use the Algebraic Continuity Theorem and show $\displaystyle f(x) = x$ is continuous, and therefore $\displaystyle f(x) = x^3$ 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 $\displaystyle \epsilon > 0$ be given. Further, choose $\displaystyle \delta = \epsilon$. Then, for $\displaystyle x,y \in \mathbb{R}, |x - y| < \delta \implies |f(x) - f(y)| = |x - y| < \epsilon$
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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 $\displaystyle \delta = \epsilon$ 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 $\displaystyle \delta = \epsilon$ like my proof above?
    correct. use the structure of the proof given there
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  6. #6
    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 $\displaystyle x = 4, f$ 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

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

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

    provided that the limit exists.

    $\displaystyle f$ is diff'able on $\displaystyle A$ if $\displaystyle f'$ exists for all points $\displaystyle c \in A$.
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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 $\displaystyle f : A \rightarrow \mathbb{R}$ be a func. that's defined on an interval A. Given $\displaystyle c \in A$, the derivative of f at c is defined by:

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

    provided that the limit exists.

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

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

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

    So...

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

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

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

    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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