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gabie1121

  • 3 years ago

find the values for c such that the function is continuous at all x. FUNCTION BELOW

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  1. gabie1121
    • 3 years ago
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    \[f(x) \left\{ \frac{ \sin(cx) }{ x } if x<0 \right\}\]

  2. gabie1121
    • 3 years ago
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    \[f(x)=(e ^{x}-2c) if x \ge0\]

  3. gabie1121
    • 3 years ago
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    those go together. I just couldn't figure out how to get them all in at once

  4. zepdrix
    • 3 years ago
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    In order for this function to be continuous, we need to pick a \(c\) value that makes the two pieces connect together nicely at \(x=0\). Imagine railroad tracks, the track needs to be continuous, and can't have any sharp corners or the train will fly off the tracks.

  5. gabie1121
    • 3 years ago
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    Yep, that i get. So do i set both functions equal to 0 and solve for c or something?

  6. zepdrix
    • 3 years ago
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    We want to look at them in the limit. We want to see what is happening when we get closer and closer to 0 from the left side. And also what is happening when we get closer and closer to 0 from the right side. For this function to be continuous, they must be approaching the same value. So we'll set these limits equal to one another.

  7. zepdrix
    • 3 years ago
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    \[\large \lim_{x \rightarrow 0^-}f(x) \quad = \quad \lim_{x \rightarrow 0^-} \frac{\sin(cx)}{x}\]

  8. zepdrix
    • 3 years ago
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    Understand why our function is sin(cx)/x when we're approaching from the left?

  9. zepdrix
    • 3 years ago
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    The tiny negative (that looks like an exponent) is letting us know we're approaching from the left.

  10. gabie1121
    • 3 years ago
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    yep cuz it says less than 0

  11. zepdrix
    • 3 years ago
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    Ok cool :) so we'll set that equal to the other piece.

  12. zepdrix
    • 3 years ago
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    And solve for c.

  13. gabie1121
    • 3 years ago
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    do we make the other side a limit as well?

  14. zepdrix
    • 3 years ago
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    \[\large \lim_{x \rightarrow 0^-}f(x) \qquad \qquad = \qquad \qquad \lim_{x \rightarrow 0^+}f(x)\]\[\large \lim_{x \rightarrow 0^-} \frac{\sin(cx)}{x} \qquad = \qquad \lim_{x \rightarrow 0^+}e^x-2c\]

  15. zepdrix
    • 3 years ago
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    Yes :) the limits need to agree in order for this function to be continuous.

  16. gabie1121
    • 3 years ago
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    alrighty, well the left side is equal to just c, right?

  17. zepdrix
    • 3 years ago
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    Yes very good ^^

  18. gabie1121
    • 3 years ago
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    because if i multiply by C , i can use the rule that says sinx/x=1

  19. zepdrix
    • 3 years ago
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    you remembered your identity i take it hehe

  20. gabie1121
    • 3 years ago
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    yep! the right side is giving me fits though lol

  21. zepdrix
    • 3 years ago
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    Or would it be -1 since we're coming from the left? Hmm I didn't think about that. lemme check real quick.

  22. zepdrix
    • 3 years ago
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    Nah it's still 1, my bad.

  23. gabie1121
    • 3 years ago
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    well would the right just be e^x-2? Or can i not do that cuz i'm thinking of derivative rules?

  24. gabie1121
    • 3 years ago
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    well the derivative is the limit as x approaches 0, so shouldn't e^x, stay e^x?

  25. zepdrix
    • 3 years ago
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    no we're not thinking of this as a derivative :) We're looking at the limit and saying to ourselves, "If I plug x=0 directly into this function, does it cause a problem?" If the answer is no, then we can do just that!

  26. gabie1121
    • 3 years ago
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    oh! well then we're left with \[c=2c\] ?

  27. zepdrix
    • 3 years ago
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    Woops! Recall that if we have a 0 in the exponent, what will that change our base to?

  28. zepdrix
    • 3 years ago
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    Not 0 silly! :O

  29. gabie1121
    • 3 years ago
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    oh well i just plugged in e^0 in my calculator and it gave me 1

  30. zepdrix
    • 3 years ago
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    hah XD that's a way to do it i guess! :D

  31. zepdrix
    • 3 years ago
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    yah 1 :3

  32. gabie1121
    • 3 years ago
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    lol so its actuallly c=1-2c?

  33. zepdrix
    • 3 years ago
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    Yah looks good c:

  34. zepdrix
    • 3 years ago
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    Do you by chance have an answer key that we can check this against? This is one of those annoying problems that it's easy to make a mistake on c: lol

  35. gabie1121
    • 3 years ago
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    c=1/3! , and I don't have one yet. I will monday so its not a huge deal

  36. gabie1121
    • 3 years ago
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    It looks logical enough to me. Thanks, AGAIN lol

  37. zepdrix
    • 3 years ago
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    This type of problem becomes a little bit harder when they throw `2 unknown constants` at you. Because then you have to also look at the limits of their derivatives. But this was a good problem to get a feel for the concept c:

  38. gabie1121
    • 3 years ago
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    well i just started calc1 this semester, so i haven't learned much yet. Just dipping my toes in the water. I have to take all the way through calculus 3 though, bleh

  39. zepdrix
    • 3 years ago
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    Hmm you'll do quite well, I can tell. You seem quite smart. You're very quick on remember how to do little steps. Calc 2 is a doozy!! Power Series made me want to rip my hair out! :O

  40. gabie1121
    • 3 years ago
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    Yeah, i'm told i'll want to murder myself with Calc 2. Not looking forward to it, but thanks! That makes me feel a little better

  41. gabie1121
    • 3 years ago
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    i might be hunting you down again! lol

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