Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

gabie1121 Group Title

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

  • one year ago
  • one year ago

  • This Question is Closed
  1. gabie1121 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    \[f(x) \left\{ \frac{ \sin(cx) }{ x } if x<0 \right\}\]

    • one year ago
  2. gabie1121 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    \[f(x)=(e ^{x}-2c) if x \ge0\]

    • one year ago
  3. gabie1121 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    those go together. I just couldn't figure out how to get them all in at once

    • one year ago
  4. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • one year ago
  5. gabie1121 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Yep, that i get. So do i set both functions equal to 0 and solve for c or something?

    • one year ago
  6. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

    • one year ago
  7. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\large \lim_{x \rightarrow 0^-}f(x) \quad = \quad \lim_{x \rightarrow 0^-} \frac{\sin(cx)}{x}\]

    • one year ago
  8. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Understand why our function is sin(cx)/x when we're approaching from the left?

    • one year ago
  9. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    The tiny negative (that looks like an exponent) is letting us know we're approaching from the left.

    • one year ago
  10. gabie1121 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yep cuz it says less than 0

    • one year ago
  11. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Ok cool :) so we'll set that equal to the other piece.

    • one year ago
  12. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    And solve for c.

    • one year ago
  13. gabie1121 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    do we make the other side a limit as well?

    • one year ago
  14. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    \[\large \lim_{x \rightarrow 0^-}f(x) \qquad \qquad = \qquad \qquad \lim_{x \rightarrow 0^+}f(x)\]\[\large \lim_{x \rightarrow 0^-} \frac{\sin(cx)}{x} \qquad = \qquad \lim_{x \rightarrow 0^+}e^x-2c\]

    • one year ago
  15. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes :) the limits need to agree in order for this function to be continuous.

    • one year ago
  16. gabie1121 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    alrighty, well the left side is equal to just c, right?

    • one year ago
  17. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Yes very good ^^

    • one year ago
  18. gabie1121 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    because if i multiply by C , i can use the rule that says sinx/x=1

    • one year ago
  19. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    you remembered your identity i take it hehe

    • one year ago
  20. gabie1121 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yep! the right side is giving me fits though lol

    • one year ago
  21. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  22. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Nah it's still 1, my bad.

    • one year ago
  23. gabie1121 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    well would the right just be e^x-2? Or can i not do that cuz i'm thinking of derivative rules?

    • one year ago
  24. gabie1121 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    well the derivative is the limit as x approaches 0, so shouldn't e^x, stay e^x?

    • one year ago
  25. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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!

    • one year ago
  26. gabie1121 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    oh! well then we're left with \[c=2c\] ?

    • one year ago
  27. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Woops! Recall that if we have a 0 in the exponent, what will that change our base to?

    • one year ago
  28. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Not 0 silly! :O

    • one year ago
  29. gabie1121 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    oh well i just plugged in e^0 in my calculator and it gave me 1

    • one year ago
  30. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    hah XD that's a way to do it i guess! :D

    • one year ago
  31. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    yah 1 :3

    • one year ago
  32. gabie1121 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    lol so its actuallly c=1-2c?

    • one year ago
  33. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    Yah looks good c:

    • one year ago
  34. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • one year ago
  35. gabie1121 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  36. gabie1121 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    It looks logical enough to me. Thanks, AGAIN lol

    • one year ago
  37. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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:

    • one year ago
  38. gabie1121 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • one year ago
  39. zepdrix Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • one year ago
  40. gabie1121 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    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

    • one year ago
  41. gabie1121 Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i might be hunting you down again! lol

    • one year ago
    • Attachments:

See more questions >>>

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.