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Math Help - Continuous

  1. #1
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    Continuous

    Let f be a function such that f(x)=x^2 for all numbers x. Show that f is continuous at the point (2,4).
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  2. #2
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    Quote Originally Posted by spikedpunch View Post
    Let f be a function such that f(x)=x^2 for all numbers x. Show that f is continuous at the point (2,4).
    Notice that |x^2-4|=|x-2||x+2|.
    If |x-2|<1 then |x+2|<5.
    \varepsilon  > 0\, \Rightarrow \,\delta  = \min \left\{ {1,\varepsilon /5} \right\}
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  3. #3
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    could you explain why delta is min( 1, epsilon/5 )?
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    Quote Originally Posted by spikedpunch View Post
    could you explain why delta is min( 1, epsilon/5 )?
    Well you want it to work.
    \left| {x - 2} \right| < \delta  = \min \left\{ {1,\varepsilon /5} \right\}\, \Rightarrow \,\left| {x + 2} \right| < 5

    \left| {x^2  - 4} \right| = \left| {x + 2} \right|\left| {x - 2} \right| < 5\left( {\varepsilon /5} \right) = \varepsilon
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  5. #5
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    So I can see this works if if the min is epsilon/5, but how would it work if the min is 1? That is, how can I show that the f(x)-f(t)<epsilon?
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  6. #6
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    Quote Originally Posted by spikedpunch View Post
    [COLOR=#0000AA][FONT=monospace] <br />
\left| {x^2  - 4} \right| = \left| {x + 2} \right|\left| {x - 2} \right| < 5\left( {\varepsilon /5} \right) = \varepsilon
    So I can see this works if if the min is epsilon/5, but how would it work if the min is 1?
    Just think about it?
    If t=\min\{1,2\} then t\le1~\&~t\le 2, right?
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  7. #7
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    I guess I am confused as to why you showed -1<x-2<1 in the first place. If 1 is the min, then wouldn't it be

    absv(x^2-4)=absv(x+2)absv(x-2)=(1)(5)=5...which is not epsilon?
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  8. #8
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    Quote Originally Posted by spikedpunch View Post
    I guess I am confused as to why you showed -1<x-2<1 in the first place. If 1 is the min, then wouldn't it be absv(x^2-4)=absv(x+2)absv(x-2)=(1)(5)=5...which is not epsilon?
    I think that you need to do some background work on this.
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