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Thread: show that the function is continuous?

  1. #1
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    show that the function is continuous?

    How would i show that f(x)=$\displaystyle x^3$-3$\displaystyle x^2$+17 is continuous on [-1,1]?
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  2. #2
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    Quote Originally Posted by tn11631 View Post
    How would i show that f(x)=$\displaystyle x^3$-3$\displaystyle x^2$+17 is continuous on [-1,1]?
    there is a theorem said if "f" and "g" continuous functions then f+g is continuous

    so I'd like to split the function into three functions

    $\displaystyle g(x) = x^3 $ , $\displaystyle h(x) = -3x^2 $ , $\displaystyle j(x) = 17 $

    I will prove the first one and leave the others for you

    given $\displaystyle \epsilon >0 $ and take $\displaystyle c \in [-1,1] $

    we need to find delta depending on epsilon and x

    $\displaystyle \mid x^3 - c^3 \mid $

    $\displaystyle \mid (x-c)(x^2+cx+c^2) \mid$

    $\displaystyle \mid (x-c)(x^2+cx+c^2) \mid \leq \mid x-c \mid \mid (x^2+cx+c^2) \mid $

    $\displaystyle \mid x-c \mid \mid (x^2+cx+c^2) \mid < \epsilon$

    $\displaystyle \delta \mid (x^2+cx+c^2) \mid < \epsilon$

    choose $\displaystyle \delta = \frac{\epsilon }{(x^2+cx+c^2)} $

    this delta works for any epsilon
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