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anonymous

  • 5 years ago

Continuous Functions 1

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  1. anonymous
    • 5 years ago
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    Check if the function is continuous: \[f(x) =\] { \[{x^2 - 1, x < 1}\] \[4-x, x \ge 1\] }

  2. anonymous
    • 5 years ago
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    Okay - you'll see that the part of f that you use to evaluate the function will depend on whether x is less than 1, or greater than or equal to 1. To show that the function is continuous, you're being asked to show that the value you get for f(x) and you approach 1 from either side, is the same. I'll show you.

  3. anonymous
    • 5 years ago
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    brb

  4. anonymous
    • 5 years ago
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    ok, tks!

  5. anonymous
    • 5 years ago
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    So this is about taking things called, "left" and "right" limits. When we approach a function from the left, it's a left-hand limit. When it's from the right, it's a right-hand limit. When x<1, you to get to 1, you need to approach f(x) from the left. So we write\[\lim_{x^- \rightarrow 1}f(x)=\lim_{x^- \rightarrow 1}(x^2-1)=1-1=0\]

  6. anonymous
    • 5 years ago
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    This is because f(x)=x^2-1 is only defined for x<1.

  7. anonymous
    • 5 years ago
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    We couldn't use f(x) = 4-x since this is only true for x>=1.

  8. anonymous
    • 5 years ago
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    *Now* we look at the right-hand limit.

  9. anonymous
    • 5 years ago
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    I begin to understand.

  10. anonymous
    • 5 years ago
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    \[\lim_{x^+ \rightarrow 1}f(x)=\lim_{x^+ \rightarrow 1}(4-x)=4-1=3\]

  11. anonymous
    • 5 years ago
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    Now, the function can only be continuous if it has the same value when x approaches 1 from either side (otherwise, if you look at a plot, the graph will be cut in two). Since your function does not reach the same value as we approach from the left and the right, it cannot be continuous. If you were to plot f(x), you could see it visually.

  12. anonymous
    • 5 years ago
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    UNDERSTAND! 0 <> 3. I have another problem of that kind, I'll try to do it. Thank you!

  13. anonymous
    • 5 years ago
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    Excellent, I'm glad! Become a fan :p

  14. anonymous
    • 5 years ago
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    I am already! :)

  15. anonymous
    • 5 years ago
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    :D

  16. anonymous
    • 5 years ago
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    Ah... What kind is this discontinuity? Removable, infinite, ...?

  17. anonymous
    • 5 years ago
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    First kind, irremovable discontinuity.

  18. anonymous
    • 5 years ago
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    Otherwise known as a 'jump' discontinuity.

  19. anonymous
    • 5 years ago
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    Thanks again!

  20. anonymous
    • 5 years ago
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    np

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