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

Bomull Group Title

prove that the function f(x)=....has a derivative at x=0

  • one year ago
  • one year ago

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

    \[f(x)=x²\cos \frac{ 1 }{ 2 }, when x \neq 0; 0, when x=0\]

    • one year ago
  2. Bomull Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    so a graphing calculator shows that x²cos(1/2) fluctuates near 0, and I believe it might not be continuous at 0. So the problem is to say that the piecewise function is defined and continuous and therefore has a derivative?

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

    Is that function typed correctly? I'm getting a very smooth looking parabola as my graph for that.

    • one year ago
  4. Bomull Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    you're right, I must have made a mistake before. but still the problem would be that if f(x) wasn't defined to be 0 at 0, it wouldn't be continuous and have a derivate at 0?

    • one year ago
  5. Bomull Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    *derivative at 0

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

    Could it have been \[x^2\cos\left(\frac{1}{2x}\right)\]

    • one year ago
  7. Bomull Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    oh, it's actually x²cos(1/x)

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

    Ah. Well, to show that it's differentiable, you basically have to show that it's continuous, and that the limit \[\lim_{h\to0} \frac{f(0+h)-f(0)}{h}\]exists.

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

    To show continuity, you need to show\[\lim_{x\to 0} x^2\cos\left(\frac{1}{x}\right)=0.\]

    • one year ago
  10. Bomull Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    so cos (1/x) at x->0 would be undefined because of (1/0), right? but then I don't know how to go further

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

    What you could do, is since \(\cos(x)\) is bounded by -1 and 1, when you multiply by \(x^2\), that term will dominate. So \[\lim_{x\to 0} x^2\cos\left(\frac{1}{x}\right)=0.\]

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

    In the first limit, \[\lim_{h\to0} \frac{f(0+h)-f(0)}{h} =\lim_{h\to0} \frac{h^2\cos\left(\frac{1}{h}\right)}{h}\]We just showed that the numerator goes to 0, so we can use L'hopital's on this.

    • one year ago
  13. Bomull Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    L'hopital's rule comes only later in the course!

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

    In that, case, try multiplying by \[\Large\frac{\frac{1}{h}}{\frac{1}{h}}\]

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

    Then we get \[\lim_{h\to0}\frac{h\cos\left(\frac{1}{h}\right)}{1}=\lim_{h\to0}h\cos\left(\frac{1}{h}\right)\]

    • one year ago
  16. Bomull Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    "What you could do, is since cos(x) is bounded by -1 and 1, when you multiply by x2, that term will dominate." but does it not matter that 1/0 would be undefined?

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

    But you aren't dividing by 0. You're just getting very close. So you have \[\cos(\text{A very large number})\] which is bounded by -1 and 1.

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

    I've got to go now. Hopefully that made enough sense. To do that very last limit, use the same fact that cosine is bounded.

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

    Refer to the attached plot of\[\left\{x^2 \cos \left(\frac{1}{x}\right),\sin \left(\frac{1}{x}\right)+2 x \cos \left(\frac{1}{x}\right)\right\} \]in blue and red respectively.

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

    Sorry, the tiff graphic above is not viewable unless your have access to Apple's Safari browser or perhaps photoshop.

    • one year ago
  21. Bomull Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    @robtobey I can see it on Ubuntu no problem. Now I just need to figure out what youre trying to say ;)

    • one year ago
  22. Bomull Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    why is \[\lim_{h \rightarrow 0}\frac{ f(0+h)-f(0) }{ h }=\lim_{h \rightarrow 0}\frac{ h²\cos(\frac{ 1 }{ h }) }{ h }\] ?

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

    Because \(f(0+h)=f(h)=h^2cos(\frac{1}{h})\) (just replace x with h) Also, \(f(0)=0\), so\[\lim_{h \rightarrow 0}\frac{ f(0+h)-f(0) }{ h }=\lim_{h \rightarrow 0}\frac{ f(h)-f(0) }{ h }=\lim_{h \rightarrow 0}\frac{ h^2\cos(\frac{1}{h})-0 }{ h }=\lim_{h \rightarrow 0}\frac{ h^2\cos(\frac{1}{h})}{ h }\]

    • 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.