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anonymous

  • one year ago

Let f(x)=3x^3−36x+3. -On which interval is f increasing? Decreasing? -Find the points at which f achieves a local min. and local max. -Find intervals on which f is concave up and concave down. -Find all points of inflection.

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  1. Jack1
    • one year ago
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    f(x)=3x^3−36x+3 -Find the points at which f achieves a local min. and local max. so solve for f'(x) = 0

  2. anonymous
    • one year ago
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    \[f'(x) = 9x^2 -36\] Setting that equal to 0 I get x=2 and x=-2

  3. anonymous
    • one year ago
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    How do I know which is a max. and which is a min. ? Do I plug it into the original function?

  4. UsukiDoll
    • one year ago
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    to find out if f is increasing or decreasing you plug in x=2 and x =-2 into the first derivative of the equation

  5. UsukiDoll
    • one year ago
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    increasing and decreasing comes from the first derivative concave up and concave down comes from the second derivative.

  6. UsukiDoll
    • one year ago
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    for increasing and decreasing if your result after plugging in x = 2 or -2 is negative, we are decreasing. If your result after plugging in x =2 or -2 is positive, we are increasing.

  7. UsukiDoll
    • one year ago
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    local min and local max... well we need to grab the absolute max (highest point in the graph) and the absolute min (lowest point the graph) before dealing with that. local min is decreasing over a certain area in the graph and local max is increasing over a certain area in the graph

  8. UsukiDoll
    • one year ago
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    inflection point (omg hope I still remember this) is the point where the graph is switching from negative to positive or positive to negative.

  9. UsukiDoll
    • one year ago
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    |dw:1434785484976:dw| this is just an example...

  10. UsukiDoll
    • one year ago
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    x_X! ok ok... wait ... x = 2 and x =-2 are critical points... we have to create a chart.... |dw:1434785632760:dw| pick a point between those intervals.

  11. UsukiDoll
    • one year ago
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    if it's negative it's decreasing and if it's positive it's increasing. pick a point between those intervals and plug it into the first derivative.

  12. UsukiDoll
    • one year ago
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    do the interval where f is increasing or decreasing(first derivative) and the interval for concavity first (2nd derivative) first... then once you obtain those results, you can do the local min, local max, and points of inflection

  13. UsukiDoll
    • one year ago
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    you can only answer the questions for local min/max and inflection after you do the first question and the third question ...otherwise, you'll be stuck

  14. princeharryyy
    • one year ago
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    still need help ?

  15. anonymous
    • one year ago
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    If I plug in x=2 or x=-2 into the first derivative I will only get 0 again, because I set f'(x)=0 in order to get the critical numbers 2 and -2.

  16. anonymous
    • one year ago
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    @UsukiDoll

  17. Jack1
    • one year ago
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  18. Jack1
    • one year ago
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  19. Jack1
    • one year ago
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  20. UsukiDoll
    • one year ago
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    you have to pick a point besides -2 and 2 in each interval

  21. UsukiDoll
    • one year ago
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    |dw:1434792689289:dw| your critical points are x = -2, and 2 remember this chart? for each interval (-oo,-2), (-2,2), and (2,oo) you have to pick a point.. like let x = -3 and plug it into the derivative equation similarly x = 1 and x =3

  22. anonymous
    • one year ago
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    Right, so then if it becomes positive then it is increasing on that interval, and if negative, it is decreasing?

  23. UsukiDoll
    • one year ago
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    \[f'(x) = 9x^2 -36 \] first derivative plug in x = -3 for the interval (-oo,-2) x = 1 for (-2,2) x=3 for (2,oo) yes @Tracy96

  24. anonymous
    • one year ago
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    Ok so f'(-3) = 45 (increasing) f'(1) = -27 (decreasing) f'(3) = 45 (increasing) Is that right?

  25. anonymous
    • one year ago
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    The calculator's giving me -27, is that wrong?

  26. Jack1
    • one year ago
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    Let \(f(x)=3x^3−36x+3\) -Find the points at which f achieves a local min. and local max. find the derivative \( \large f′(x)=9x^2−36\) derivative = gradient... so when gradient = 0, its a turning poing \( \large f′(0)=9x^2−36\) \( \large f′(0)=2~ or~ -2\) so what are the y values at these points? \(f(x)=3x^3−36x+3\) \(f(2)=3x^3−36x+3\) \(f(2)=-45\) = local minima \(f(-2)=51\) = local maxima

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  27. UsukiDoll
    • one year ago
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    sorry.. yes that's right.. -27 for f(1) I hit the wrong number by accident

  28. UsukiDoll
    • one year ago
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    |dw:1434793225103:dw|

  29. UsukiDoll
    • one year ago
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    now we need the second derivative for concavity

  30. UsukiDoll
    • one year ago
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    \[\large f′(x)=9x^2−36\] take the derivative of this equation again

  31. anonymous
    • one year ago
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    @Jack1 Ahh I see! That makes sense!

  32. anonymous
    • one year ago
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    @UsukiDoll Great! So the 2nd derivative is 18x right?

  33. UsukiDoll
    • one year ago
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    yes... so wee need to find the critical point f''(x) = 18x set the equation to 0 and solve for x

  34. anonymous
    • one year ago
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    x=0 ?

  35. UsukiDoll
    • one year ago
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    yeah... so there is only one critical point this time x =0 so we only find intervals for (-oo,0) and (0,oo) pick a point.. now the only problem is I tend to have the concave up or concave down vocabulary mixed up :/

  36. anonymous
    • one year ago
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    Haha no worries... so can we choose x=-1 and x=1 ?

  37. UsukiDoll
    • one year ago
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    but again depends on the number oh here it is if f''(x) > 0 concave up f''(x) <0 concave down.. yes we can choose those values as long as it's in the interval (-oo,0) for x = -1 and (0,oo) for x =1

  38. anonymous
    • one year ago
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    And I have to plug them in the 2nd derivative am I right?

  39. UsukiDoll
    • one year ago
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    yeah

  40. anonymous
    • one year ago
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    f''(-1) = -18 and f''(1) = 18

  41. UsukiDoll
    • one year ago
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    so for the interval (-oo,0) we have concave down because f''(x) <0 concave down.. and for the interval (0,oo) we have concave up because f''(x) > 0 is concave up

  42. anonymous
    • one year ago
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    Ohhh okie dokie! That makes a lot of sense now! :D ahaha

  43. UsukiDoll
    • one year ago
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    now I sort of forgot how the drawing is.. well I know it .. but I keep switching them unintentionally

  44. anonymous
    • one year ago
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    By this point have we found the points of inflection, or is there more to calculate?

  45. Jack1
    • one year ago
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    -Find intervals on which f is concave up and concave down. find the points either side of the local minima/maxima to work out if they're higher or lower of either side is lower, if the turning point is the highest point (highest y value), therefore it's concave up if the turning point is the lowest point (lowest y value), its concave down f(x)=3x^3−36x+3. 1st turning point = (-2, 51) so a point either side of -2 is -1.9 and -2.1 y value at x = -1.9 \(\large f(x)=3x^3−36x+3\) \(\large f(-1.9)=3x^3−36x+3\) \(\large f(-1.9)=50.823\) y value at x = -2.1 \(\large f(x)=3x^3−36x+3\) \(\large f(-2.1)=3x^3−36x+3\) \(\large f(-2.1)=50.817\) each point either side is slightly lower than 51, so it's concave down do the same for each side of the other turning point (2,-45) to work out what it is

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  46. UsukiDoll
    • one year ago
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    I don't think so ... but I think |dw:1434793917549:dw|

  47. UsukiDoll
    • one year ago
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    what the heck? I think he's doing it a different way. The way I learned it is to create a chart and plug in values in the intervals after solving for 0 and finding the critical points.

  48. anonymous
    • one year ago
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    Hmmm I understand :) @Jack1 That makes sense!

  49. UsukiDoll
    • one year ago
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    a. find first and second derivatives b. find critical points c. do sign chart --- plug in whatever value you pick in the first and second derivative.

  50. anonymous
    • one year ago
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    My prof is teaching me the way with the chart which confuses me at times, but the method @Jack1 is using was the 1st method I learned back in high school

  51. UsukiDoll
    • one year ago
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    why is the chart confusing? I'm curious... that's how my professor taught me and it was straight forward.

  52. UsukiDoll
    • one year ago
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    actually I went into a debate with my Mathematical Biology professor last semester..because I used the sign chart and first derivative to find increasing and decreasing but he was all like derrr grab the second derivative and look at the sign. what is that? >:/

  53. UsukiDoll
    • one year ago
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    I know there is more than one method to solve math problems as long as it doesn't break math rules... but creating math shortcuts like what my professor did (last semester) threw me off. But, I did used it in the midterms.

  54. anonymous
    • one year ago
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    Hahaha idk, I guess the 1st way you learn things always sits in the back of your mind better :D lol

  55. UsukiDoll
    • one year ago
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    hahahaa yeah... I still don't agree with his method XD

  56. anonymous
    • one year ago
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    LOL XD

  57. anonymous
    • one year ago
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    I agree, it's quite hard to agree with profs sometimes :P haha

  58. UsukiDoll
    • one year ago
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    I'm glad I don't have to deal with him anymore!!!!!!!!!!! :D not too many people are in his classes for the upcoming Fall semester. . . I can't wait until it gets canceled which most likely is! :D He was such a jerk. I don't care if I passed with an above average grade

  59. Jack1
    • one year ago
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    -Find all points of inflection. Let \(\large f(x)=3x^3−36x+3\) find 2nd derivative \(\large f(x)=3x^3−36x+3\) \(\large f'(x)=9x^2−36\) \(\large f''(x)=18x\) when f''(x) = 0, it's a point of inflection \(\large f''(x) = 18x\) \(\large f''(0) = 18x\) \(\large f''(0) = 0\) ==> so x = 0 is a point of inflection what is the y value here? \(\large f(x)=3x^3−36x+3\) \(\large f(0)=3x^3−36x+3\) \(\large f(0)=3\) so point of inflection is at ( 0, 3)

  60. anonymous
    • one year ago
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    Hahahaha right on!! XD above average grade!! I hope the same happens with me! :D

  61. UsukiDoll
    • one year ago
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    whoa.. nice one @Jack1

  62. Jack1
    • one year ago
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    -On which interval is f increasing? Decreasing? psshh... graph it or freehand draw it after u have worked out the other main points... if y value is increasing from left to right )if it looks uphill) then f is increasing

  63. UsukiDoll
    • one year ago
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    have you seen the questions from this one guy in the Qualified Helpers section? It's crazy! :/

  64. anonymous
    • one year ago
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    @Jack1 Ok so whatever x value you find when you set 2nd derivative = 0 is an inflection point?

  65. UsukiDoll
    • one year ago
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    we have the interval increasing or decreasing already.. via the chart.

  66. anonymous
    • one year ago
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    @Jack1 haha yeah don't worry about the graphing... if I'm good at one thing , its graphing xD lol

  67. Jack1
    • one year ago
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    yep @Tracy96 ... 2nd derivative shows the rate of change of the gradient... so if it's 0, then at that point it's transitioning from a decreasing gradient to an increasing gradient, or vice versa

  68. anonymous
    • one year ago
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    Ahhh alrighty!! @Jack1 very helpful tips thanks so much!!

  69. anonymous
    • one year ago
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    And @UsukiDoll thank to you too!! I got everything you said! Including the 'chart' method! :D haha

  70. anonymous
    • one year ago
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    And thanks for the medal! :D @UsukiDoll

  71. Jack1
    • one year ago
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    @UsukiDoll 's way of doing charts and finding the critical points will probably be the quicker way to do it overall... I just find it more logical in my mind to do the other way

  72. UsukiDoll
    • one year ago
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    xD thanks... everyone got a medal on here

  73. anonymous
    • one year ago
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    Yeah, logic is always a saviour when it comes to exams >.< lol thanks again!!

  74. Jack1
    • one year ago
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    np =)

  75. anonymous
    • one year ago
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    :D

  76. anonymous
    • one year ago
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    @UsukiDoll mind me asking which school you go to?

  77. UsukiDoll
    • one year ago
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    university

  78. anonymous
    • one year ago
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    yeah lol same thing here in Canada... Which year are you? I'm a 1st year xD

  79. UsukiDoll
    • one year ago
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    I'm known as a super senior.. still going to school despite reaching over 120 credits

  80. anonymous
    • one year ago
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    Well its gonna be my 2nd year this fall :D

  81. anonymous
    • one year ago
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    Oh wow!!! 120 credits!!! Thats a lot!!!

  82. UsukiDoll
    • one year ago
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    I actually have 128...

  83. anonymous
    • one year ago
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    Lucky you!!! :O

  84. UsukiDoll
    • one year ago
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    but I went to community college for 6 years because I was horrendous at Math, my community college screwed me over for having a requirement that's impossible to achieve...plus I was still iffy about what I wanted to do after college... but now I know and that's to teach... hopefully if I CAN GET OVER THE STUPID ANALYTICAL LOGIC MATH STUFF I can try Graduate school so I can teach adults.

  85. UsukiDoll
    • one year ago
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    I need a MA in Math to teach in a college campus ...otherwise I would be stuck with high schoolers.

  86. anonymous
    • one year ago
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    Hahahaha teaching high schoolers is a serious challenge XD lol

  87. anonymous
    • one year ago
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    Specially when it comes to Math lol

  88. UsukiDoll
    • one year ago
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    oh yeah my state is the worst.

  89. anonymous
    • one year ago
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    And wow! 6 years of college! That should take a lot hard work!!

  90. anonymous
    • one year ago
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    lot of*

  91. UsukiDoll
    • one year ago
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    I think part of it was do to me being in a different Major which messed up my gpa so I stopped enrolling in Business courses and focused on Math

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