Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Dodo1

  • one year ago

Calculus question ; photo attached

  • This Question is Closed
  1. Dodo1
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  2. abb0t
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Start by taking the derivative of your function. That will be your first step. Can you do that?

  3. Dodo1
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    9+(2/x^2)=0?

  4. abb0t
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Great. Now that you have that, solve for "x".

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

    @abb0t that derivative is incorrect.

  6. abb0t
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Hmm, is it now? Well, @Dodo1 you need to try again then.

  7. Dodo1
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how is it wrong?

  8. stamp
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @abb0t Good luck evaluation that term = 0 without using imaginary numbers. @Dodo1 Forgot his - sign when taking the derivative using the power rule

  9. Dodo1
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    9+(2x^(-2))=0?

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

    no

  11. stamp
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[9x+2x^{-1}\]\[9+(-1)2x^{-1-1}\]\[9-2x^{-2}\]\[9-2/x^2\]

  12. stamp
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @Dodo1 How did you plan on solving 9 + 2/x^2 = 0? It is not possible using real numbers because any real number squared is positive, this was the first clue that something was wrong with the derivative

  13. Dodo1
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh not plus ok. I see

  14. Dodo1
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    next step is mutiply x^2 both side?

  15. Dodo1
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @stamp

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

    @Dodo1 It is algebra I at this point. Solve for x.

  17. Dodo1
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    x=Sqrt(9/2) @stamp

  18. stamp
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[9-2/x^2=0\]\[9=2/x^2\]\[x^2=2/9\]\[x=+/-(\sqrt2/3)\]

  19. Dodo1
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ops sorry i did 9X^2+2=0

  20. Dodo1
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    whats the next step? max at _sart(2)/3? @stamp

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

    You do not need to tag me in every response, Dodo, I am here. These values, x = +/- c, are critical points. They are max or mins, we do not know until we consider each point with respect to the second derivative. You have been familiarized with the second derivative test in lecture, now we are using it to see which of these critical points are max, mins, etc.

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

    I have to go, but this should help you enough to get started. Somebody else should be able to help you.

  23. Dodo1
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yea, I do need to prepare derivative problems. Are you going?

  24. stamp
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    f(x) = http://www.wolframalpha.com/input/?i=9x%2B2%2Fx f'(x) = http://www.wolframalpha.com/input/?i=derivative+of+9x%2B2%2Fx f''(x) = http://www.wolframalpha.com/input/?i=second+derivative+of+9x%2B2%2Fx critical points where f'(x) is 0 = http://www.wolframalpha.com/input/?i=solve+derivative+of+9x%2B2%2Fx graph of f(x) http://www.wolframalpha.com/input/?i=plot+9x%2B2%2Fx Visually one can observe that the negative critical point is a maximum, while the positive critical point is a minimum. Analytically, you can observe the change of f''(x) to the left of the critical point and to the right of the critical point. Since f'( - c ) goes from + to - , - c is a maximum. Since f'( + c ) goes from - to +, + c is a minimum.

  25. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

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.