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Math Help - Derivatives/Continuity

  1. #1
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    Derivatives/Continuity

    1.)Show that there is one real number x such that 2cosx = x^2. Justify your answer

    2.)Let g(x) = x^(2/5)
    a.) is g continuous at x=0? Use the definition of continuity to justify.
    b.) is g differentiable at x=0? Use the definition of differentiability to justify.
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  2. #2
    Senior Member JaneBennet's Avatar
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    (1) Let f(x)=2\cos{x}-x^2. Note that f(0)>0 and f\left(\frac{\pi}{2}\right)<0. Now invoke intermediate-value theorem.

    (2)(a) Hint: Given \epsilon>0, take \delta=\epsilon^{\frac{5}{2}}.

    (2)(b) Hint: Does \lim_{x\,\to\,0}{x^{-\frac{3}{5}}} exist?
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  3. #3
    Super Member 11rdc11's Avatar
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    You can show the first question grahically
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  4. #4
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    Quote Originally Posted by JaneBennet View Post
    (1) Let f(x)=2\cos{x}-x^2. Note that f(0)>0 and f\left(\frac{\pi}{2}\right)<0. Now invoke intermediate-value theorem.

    (2)(a) Hint: Given \epsilon>0, take \delta=\epsilon^{\frac{5}{2}}.

    (2)(b) Hint: Does \lim_{x\,\to\,0}{x^{-\frac{3}{5}}} exist?
    in 2b.) why x^3/5?
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  5. #5
    Senior Member JaneBennet's Avatar
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    Work it out and see.
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    Quote Originally Posted by JaneBennet View Post
    Work it out and see.
    yes, it approaches 0. just a little lost as to the reason -why- i am taking the limit to the 3/5 power
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  7. #7
    Moo
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    Hello,
    Quote Originally Posted by NotEinstein View Post
    yes, it approaches 0. just a little lost as to the reason -why- i am taking the limit to the 3/5 power
    Because \left(x^{\frac 25}\right)'=\frac 25 \cdot x^{-\frac 35}
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  8. #8
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by Moo View Post
    \left(x^{\frac 25}\right)'=\frac 25 \cdot x^{-\frac 35}
    No, donít do that. We are supposed to be using the definition of a derivative here.

    The derivative of x^{\frac25} at 0 is \lim_{x\,\to\,0}{\frac{(0+x)^{\frac25}-0^{\frac25}}{(0+x)-0}}.
    Last edited by JaneBennet; September 11th 2008 at 04:56 AM.
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by JaneBennet View Post
    (1) Let f(x)=2\cos{x}-x^2. Note that f(0)>0 and f\left(\frac{\pi}{2}\right)<0. Now invoke intermediate-value theorem.
    This shows that there is at least one root in (0,\pi/2), we still need to show that there is only one real root (that is if there is), which can be done by showing that f(x) is strictly decreasing for x positive. Since f(x) is symmetric there are infact two real roots.

    RonL
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  10. #10
    Senior Member JaneBennet's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    This shows that there is at least one root in (0,\pi/2), we still need to show that there is only one real root.

    RonL
    Does the question say we have to do it?
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