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Math Help - Continuity.

  1. #1
    Junior Member pearlyc's Avatar
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    Continuity.

    Question 1.
    Prove using \epsilon - \delta arguments that

    (i) f(x) = x^4 is continuous at x=a, where a not equal to 0.
    (ii) Deduce (quoting an appropriate theorem) that g(y) = y^{1/4} is continuous on the interval [0,\infty).

    Question 2.
    Consider the function

    f(x)=\left\{\begin{array}{cc}cos(\pi x), &\mbox{  } x\leq 2\\x-1, &\mbox{  } 2<x<4\\4, &\mbox{  } x\geq 4\end{array}\right.

    (i) At which points is f(x) continuous?
    (ii) Are there points at which f is continuous on the left or on the right but not continuous?
    (iii) Give the smallest value of c for which the function f is an injection if its domain is restricted to the interval [c,3]

    Thanks a lot for your guidance guys!
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by pearlyc View Post
    Question 1.
    Prove using \epsilon - \delta arguments that

    (i) f(x) = x^4 is continuous at x=a, where a not equal to 0.
    (ii) Deduce (quoting an appropriate theorem) that g(y) = y^{1/4} is continuous on the interval [0,\infty).
    why don't you show us what you have done..

    Quote Originally Posted by pearlyc View Post
    Question 2.
    Consider the function

    f(x)=\left\{\begin{array}{cc}cos(\pi x), &\mbox{  } x\leq 2\\x-1, &\mbox{  } 2<x<4\\4, &\mbox{  } x\geq 4\end{array}\right.

    (i) At which points is f(x) continuous?
    (ii) Are there points at which f is continuous on the left or on the right but not continuous?
    (iii) Give the smallest value of c for which the function f is an injection if its domain is restricted to the interval [c,3]

    Thanks a lot for your guidance guys!
    \cos is continuous, polynomials and constants are continuous..

    so check whether f(a) = \lim_{x\rightarrow a^+} f(x) = \lim_{x\rightarrow a^-} f(x) where a are the possible values of x which can give discontinuities, in short, 2 and 4.
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  3. #3
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    Quote Originally Posted by pearlyc View Post
    (i) f(x) = x^4 is continuous at x=a, where a not equal to 0.
    Say |x-a|<\delta then |x| \leq |x-a|+|a| < \delta + |a|.
    This means,
    |f(x)-f(a)| = |x^4-a^4| = |x-a||x^3+x^2a+xa^2+a^3| < \delta[ (\delta+|a|)^3+|a|(\delta+|a|)^2+|a|^2(\delta+|a|)  +|a|^3].

    Then for \epsilon > 0 choose \delta = \min \left\{ 1 , \frac{\epsilon}{(1+|a|)^3+|a|(1+|a|)^2+|a|^2(1+|a|  )+|a|^3} \right\}
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  4. #4
    Junior Member pearlyc's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Say |x-a|<\delta then |x| \leq |x-a|+|a| < \delta + |a|.
    This means,
    |f(x)-f(a)| = |x^4-a^4| = |x-a||x^3+x^2a+xa^2+a^3| < \delta[ (\delta+|a|)^3+|a|(\delta+|a|)^2+|a|^2(\delta+|a|)  +|a|^3].

    Then for \epsilon > 0 choose \delta = \min \left\{ 1 , \frac{\epsilon}{(1+|a|)^3+|a|(1+|a|)^2+|a|^2(1+|a|  )+|a|^3} \right\}
    is that all we have to do for proving this question? they gave me a hint for this question saying that it may help to commence with the restriction that | x - a | < |a|/2. what does that mean?

    can you please explain how did \delta = \min \left\{ 1 , \frac{\epsilon}{(1+|a|)^3+|a|(1+|a|)^2+|a|^2(1+|a|  )+|a|^3} \right\} came about?
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  5. #5
    Junior Member pearlyc's Avatar
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    anyone else that is willing to give me a light in this?
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