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anonymous

  • 5 years ago

Consider the function f(x) = 8 √{ x} + 9 on the interval [ 2 , 8 ] (A) Find the average or mean slope of the function on this interval (B) By the Mean Value Theorem, we know there exists at least one c in the open interval ( 2 , 8 ) such that f′( c) is equal to this mean slope. Find all values of c that work and list them (separated by commas) in the box below.

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  1. amistre64
    • 5 years ago
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    this one here?

  2. anonymous
    • 5 years ago
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    yea this is one of the ones i am having trouble on...

  3. amistre64
    • 5 years ago
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    mean slope means what? the slope of the line from end point to end point?

  4. anonymous
    • 5 years ago
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    i don't know i check all my notes and no where do i see mean slope so i don't know why the teacher assigned this problem

  5. amistre64
    • 5 years ago
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    I believe that it means the average slope between the intervals and is found by taking the slope of the line between the end points.

  6. amistre64
    • 5 years ago
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    then you find a point in the interval that touches the graph at one point with that slope

  7. anonymous
    • 5 years ago
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    so what would be the equation to find the slope between the intervals

  8. amistre64
    • 5 years ago
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    f(8) - f(2) change in y -------- over 8-2 change in x

  9. amistre64
    • 5 years ago
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    this is how you find any slope of a line :) the change in y over the change in x

  10. anonymous
    • 5 years ago
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    ok then what next?

  11. amistre64
    • 5 years ago
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    whats our mean slope? I get: 4sqrt(2) ------ 3

  12. amistre64
    • 5 years ago
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    when f'(c) = 4sqrt(2)/3 then we have found the values for "x" that satisfy the problem

  13. anonymous
    • 5 years ago
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    thats what i got...it was correct

  14. amistre64
    • 5 years ago
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    f(x) = 8x^(1/2) + 9 whats our derivative?

  15. amistre64
    • 5 years ago
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    4/sqrt(x) maybe?

  16. anonymous
    • 5 years ago
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    yes it is

  17. amistre64
    • 5 years ago
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    yay!!... 4sqrt(x) 4sqrt(2) ------ = -------- we need to solve this then x 3

  18. amistre64
    • 5 years ago
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    12 sqrt(x) = 4x sqrt(2) ; divide both sides by 4 3 sqrt(x) = x sqrt(2)

  19. anonymous
    • 5 years ago
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    the answer is 4.5....

  20. amistre64
    • 5 years ago
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    did you solve it, or is that what we are going for?

  21. anonymous
    • 5 years ago
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    i solved it...well after you help of course...i started to see the problem working out...i got another...

  22. amistre64
    • 5 years ago
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    I was gonna square both sides and do some magic :)

  23. anonymous
    • 5 years ago
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    \[f(x)=(x-3)(x-7)^3+8\] we are looking for absolute max and min for intervals at [1,4] and [1,8] and [4,9]

  24. amistre64
    • 5 years ago
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    .... i got 9/2...... so nyahh!!

  25. amistre64
    • 5 years ago
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    same function?

  26. amistre64
    • 5 years ago
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    .......ok, i can read..... sometimes :)

  27. amistre64
    • 5 years ago
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    its a quartic equation which will resemble a "W" perhaps

  28. amistre64
    • 5 years ago
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    just expand it and make it easier for ya...

  29. anonymous
    • 5 years ago
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    ok..lol... i got \[f'(x)=(x-7)^3+(x-3)(3)(x-7)^2\]

  30. amistre64
    • 5 years ago
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    f(x) = x^4 -24x^3 +210x^2 -784x +1037 f'(x) = 4x^3 -72x^2 +420x -784 :)

  31. amistre64
    • 5 years ago
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    your way might be easier to read :)

  32. anonymous
    • 5 years ago
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    ok so i would make the equation equal to zero right???

  33. amistre64
    • 5 years ago
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    yep..

  34. amistre64
    • 5 years ago
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    x = 7 is a zero, thats easy to see

  35. anonymous
    • 5 years ago
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    and would 3 be a zero

  36. amistre64
    • 5 years ago
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    3 would be a zero + (-7)^3

  37. amistre64
    • 5 years ago
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    take the second derivative to see how the graph is behaving

  38. amistre64
    • 5 years ago
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    f''(x) = 12x^2 -144x +420

  39. amistre64
    • 5 years ago
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    factor out a 12.. 12(x^2 - 12x + 35)

  40. amistre64
    • 5 years ago
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    12(x-7)(x-5) is the factored form

  41. anonymous
    • 5 years ago
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    ok and how would i use that to find the max and min between all three intervals

  42. amistre64
    • 5 years ago
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    draw a number line straight across the paper......

  43. amistre64
    • 5 years ago
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    <..................5......................7....................>

  44. amistre64
    • 5 years ago
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    at 5 and 7 we have inflection points...

  45. anonymous
    • 5 years ago
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    ok but our first interval were looking at is [1,4]

  46. amistre64
    • 5 years ago
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    <..................5......................7....................> --- +++ +++ --- --- +++ ---------------------------------- + - + when we multiply these "zones" together, we get these results

  47. amistre64
    • 5 years ago
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    do you undertand this diagram?

  48. anonymous
    • 5 years ago
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    yea...how do we apply the intervals to this...

  49. amistre64
    • 5 years ago
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    [1,4] is in an area that is concave up, so if anything it would have a MIN in it someplace

  50. amistre64
    • 5 years ago
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    Now, are we looking for the "highest and or lowest point in the intervals?

  51. anonymous
    • 5 years ago
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    i am going to break up the question like it is.... f(x)=(x−3)(x−7)^3+8 a. interval [1,4] absolute max: absolute min: this is how all of the other ones are

  52. amistre64
    • 5 years ago
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    our graph is acting like this:

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  53. amistre64
    • 5 years ago
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    solve for f(1) and f(4) to see what our values are.

  54. amistre64
    • 5 years ago
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    f(1) = 440 if I did it right

  55. amistre64
    • 5 years ago
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    f(4) = -19

  56. amistre64
    • 5 years ago
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    the max is at (1,440)

  57. amistre64
    • 5 years ago
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    here a thought, plug in 4 into the derivative and see if we get a zero :)

  58. amistre64
    • 5 years ago
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    yep, 4 is a zero for f'(x)...im a genuis :)

  59. amistre64
    • 5 years ago
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    [1,4] (1 ,440) = max and (4,-19) = min you agree?

  60. amistre64
    • 5 years ago
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    and the graph doesnt have an more "bumps in it, just some transitional bends.....

  61. anonymous
    • 5 years ago
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    yea the next interval is the same and the last one the max is (9,56) and min is (4,-19)

  62. anonymous
    • 5 years ago
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    \[f(x)=(3+9\ln (x))/x\] i need to find the x coordinate of the absolute max...

  63. amistre64
    • 5 years ago
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    start a new question, I think this ones lagging alot becasue of all the "replies" its trying to keep track of :)

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