A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • one year ago

Need help! Got through finding the first derivative, but not too sure about my answer. Here is the question: When f(x) = (x-2)e^x A) Find the intervals of increase of decrease B) Find the local max and mins C) Find the intervals of concavity and inflection points D) Use the information from parts A-C to sketch the graph.

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

    For the first derivative I got: f'(x) = e^x (x-1)

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

    yes

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

    D

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

    So for the critical values, for when f'(x) = 0 I got x=1. But for the other critical value when f'(X)= undefined... I tried to put the first derivative in a fraction form but Im not sure Its right.

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

    When the first derivative in fraction form, I got: e^x - (x-2)* (e^x) all over (e^-x)^2

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

    so when f'(x) is undefined, does x= 0? Im not sure...

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

    sorry my mouse died

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

    yes x=1 has a horizontal tangent point, test either side of it to see what the function is doing

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

    f'(x)=0 gives x=1

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

    is f '(x) positive(increasing) or negative(decreasing) function on etiher side of x=1, test any point

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

    Just for practice, here is the actual graph of f, to compare your answers to..

    1 Attachment
  12. DanJS
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    since x=1 is the only critical point here, you have two intervals to test for inc/dec (-inf, 1) and (1, +inf)

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

    Really? what about when f'(x) = undefined, does that give us any critical values?

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

    where does that happen?

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

    so you would have to do the first derivative in a fraction form and then whatever is at the bottom of that fraction, you set that to zero and solve...and whatever you get from that is another critical value. But i'm not even sure that I put the first derivative in its right fraction form

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

    f ' (x) = (x-1)e^x what did you do next

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

    So I actually got what you got, but then I did it all over again trying to put it into fraction form and this is what I got: e^x - (x-2)* (e^x) all over (e^-x)^2

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

    Oh, do you mean the original function f(x)

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

    Yea from the original function, I tried to find the first derivative in a fraction form..and well I got e^x - (x-2)* (e^x) all over (e^-x)^2

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

    f(x) = (x-2)e^x = xe^x - 2e^x f(x) = xe^x - 2e^x

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

    i dont understand what you did

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

    using the product rule here , no quotient

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

    Okay, so I basically did: f(x)= (x-2)/ e^-x and then I did the quotient rule on this to find the first derivative in fraction form But I assumed to use the quotient rule because I took the e^x down

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

    If I took e^x down to become e^-x, wouldnt I be able to use the quotient rule? That seems logical to me

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

    yes, that gets messy with ln functions

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

    but it will come out to x=1 the same

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

    when you set it to zero

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

    I didnt knowthat when (e^-x)^2 =0 it solves out to be x=1

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

    e^(-2x) never is zero

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

    So we get out nothing when f'(x)= undefined? Either way I got inc from: (1,infinity) and decreasing from: (negative infinity, 1).

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

    there is no undefined value for the derivative

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

    Okay good. And for local max, i got none. But for local min I got x= 1

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

    Then to find the intervals of concavity and inflection points I know I have to find the second derivative and I got f''(x) = e^x (x) which basically gives me x=0

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

    And if I am right with the critical value... do I do a number line that has 1 and 0 and test out points between them

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

    |dw:1437606668108:dw|

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

    yes so far so good

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

    so far you have from first derivative... |dw:1437606825963:dw|

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

    you can also calculate the x and y intercepts from original function

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

    and test the limits as x goes to + and - infinity for horizontal asmytotes

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

    So I got for CU: (0,1) U (1,infinity) and CD: (negative infinity, 0)

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

    looks righ

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

    t

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

    And for the inflection points, do I plug in zero into the second derivative equation or original equation>

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

    original, you want to add it to the graph of F(X)

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

    right

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

    So I got for inflection points: (0,-2)

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

    |dw:1437607095039:dw|

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

    right, so that is what your graph looks kinda like with the info so far

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

    maybe calculate an X intercept when f(x)=0

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

    and test for any limits when x goes off to infinities

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

    I put the original equation into my graphing calc, and it gave me something diff

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

    |dw:1437607248105:dw|

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

    yes, what is the limit as x goes off to neg infinity for the original function...

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

    e^(x) becomes very very small when x is larger negative numbers

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

    close to zero?

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

    yes \[\lim_{x \rightarrow - \infty}=0\]

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

    so the function approaches zero as x goes off to negative infinity .. like you drew

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

    plug in smaller and smaller values for x in the original function and you can see how e^(x) behaves for smaller and smaller values of x, and it is multiplied by (x-2) to make it negative

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

    Yea, totes see it! thanks

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

    that's about it.. less you want to calculate the x intercept and add that point, and show that the limit at positive infinity is infinity

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

    Nah, dont need that. Do you mind helping me go through another similar problem like this?

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

    ill try. hehe

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

    open new thread and close it , ill move over

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

    kk

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

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.