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j814wong

  • 3 years ago

Find a number δ > 0 such that for all x, 0 < absolute value (x-x0) < δ → absolute value (f(x) - L) < ϵ f(x)=4/x, x0=2, ϵ=0.4 Please explain in detail. Unfortunately, my school took a long time to fix my schedule so I was not put into AP BC Calculus until 2 or so weeks in today so much of the foundation is missing but I'm studying on my own to catch up.

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  1. j814wong
    • 3 years ago
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    Also, please explain the symbols.

  2. anonymous
    • 3 years ago
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    they are just variable they use greek letters to annoy you

  3. mahmit2012
    • 3 years ago
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    |dw:1348198061023:dw|

  4. j814wong
    • 3 years ago
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    What do those variables represent?

  5. mahmit2012
    • 3 years ago
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    |dw:1348198204753:dw|

  6. j814wong
    • 3 years ago
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    δ Is delta right? But does it mean the same as the triangle delta?

  7. mahmit2012
    • 3 years ago
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    what is the triangle delta ?

  8. anonymous
    • 3 years ago
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    you want to make \(f\) close to \(L\) how close? within \(\epsilon\) for arbitrary \(\epsilon\) in your case (\epsilon\) is not arbitrary, it is 0.4

  9. mahmit2012
    • 3 years ago
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    delta can be 02 or less to satisfy .

  10. j814wong
    • 3 years ago
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    Δ I mean. Seems it's upper case delta.

  11. anonymous
    • 3 years ago
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    it is just a variable, you could use anything \(a\), \(b\), \(\alpha\) \(\beta\) etc

  12. j814wong
    • 3 years ago
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    Ok. I'm frantically flipping through my textbook trying to comprehend all this.

  13. j814wong
    • 3 years ago
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    In this case, what would epsilon be representing specifically?

  14. j814wong
    • 3 years ago
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    What is the name of these parts of calculus?

  15. j814wong
    • 3 years ago
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    Also, my textbook says δ = 1/3 is the answer but how they got it I have no idea.

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