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anonymous

  • one year ago

NEED HELP! What are the intervals of increase and decrease of f(x) = x ( sqrt of x+3) * This is a calc problem

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  1. freckles
    • one year ago
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    Have you considered finding the derivative?

  2. UsukiDoll
    • one year ago
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    we need the first derivative test... wait we need to find critical points first!

  3. freckles
    • one year ago
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    f'>0 tells us f is increasing f'<0 tells us f is decreasing Find the critical numbers and consider the domain.

  4. UsukiDoll
    • one year ago
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    what @freckles said is correct... use product rule

  5. anonymous
    • one year ago
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    I already did all that, and I got (sqrt x+3) + x/(2(sqrt x +3)). But I just dont know how to go about solving for the critical values

  6. freckles
    • one year ago
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    I don't think we need to find the x-intercepts to find where the function is increasing/decreasing.

  7. anonymous
    • one year ago
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    is it just x= -3?

  8. UsukiDoll
    • one year ago
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    noooooo....................... freckles we need the critical points first or we can't go further T_T

  9. UsukiDoll
    • one year ago
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    sorry mary that comment was at freckles

  10. freckles
    • one year ago
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    \[f'(x)=\sqrt{x+3}+\frac{x}{2 \sqrt{x+3}} \\ \text{ solve } f'(x)=0 \text{ \to find the critical numbers }\]

  11. Zale101
    • one year ago
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    To get the critical points, we first need to set the derivative to zero

  12. anonymous
    • one year ago
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    Yea I did that already guys!!!

  13. Zale101
    • one year ago
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    and solve for x. Those x's are the crtical points.

  14. freckles
    • one year ago
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    good then what did you for the x's?

  15. anonymous
    • one year ago
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    I just need help in solving for x

  16. freckles
    • one year ago
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    \[\sqrt{x+3} \frac{2\sqrt{x+3}}{2\sqrt{x+3}}+\frac{x}{2 \sqrt{x+3}}\] find common denominator

  17. anonymous
    • one year ago
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    No, we find the first derivative and then find the critical points

  18. freckles
    • one year ago
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    \[\frac{2 (x+3)+x}{2 \sqrt{x+3}}=0\]

  19. anonymous
    • one year ago
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    hey freckles, how did you get +x in the numerator?

  20. freckles
    • one year ago
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    (sqrt x+3) + x/(2(sqrt x +3) isn't this what you got for f'?

  21. anonymous
    • one year ago
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    OH, K i see it

  22. freckles
    • one year ago
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    \[f'(x)=\sqrt{x+3}+\frac{x}{ 2 \sqrt{x+3}} \\ f'(x) =\frac{ \sqrt{x+3} }{1} \cdot \frac{2 \sqrt{x+3}}{2 \sqrt{x+3}} +\frac{x}{2 \sqrt{x+3}} \\ f'(x)=\frac{2(\sqrt{x+3})^2+x}{2 \sqrt{x+3}}\]

  23. anonymous
    • one year ago
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    So is it x=-2 for when f'(x) = 0

  24. freckles
    • one year ago
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    yes

  25. anonymous
    • one year ago
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    and then the denominator would give me when derivative is undefined?

  26. anonymous
    • one year ago
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    so the critical points are x= -2, and -3?

  27. freckles
    • one year ago
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    in by the way the domain of f is x>=-3 so we have this so far: |dw:1437451331355:dw|

  28. freckles
    • one year ago
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    x=-3 is an endpoint

  29. freckles
    • one year ago
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    since the function doesn't exist to the left of x=-3

  30. freckles
    • one year ago
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    so you only have two intervals to consider the ones in the picture above just choose a number from each and plug into f'(x)

  31. UsukiDoll
    • one year ago
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    what about the interval (-oo, -3) . Don't we have to pick a number in that interval too?

  32. freckles
    • one year ago
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    the function doesn't exist to the left of x=-3

  33. freckles
    • one year ago
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    the domain of x *sqrt(x+3) is x+3>=0 because of that square root function

  34. UsukiDoll
    • one year ago
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    yeah and we can't have x = -3. I can see that part, but how did you guys get x = -2 ?

  35. freckles
    • one year ago
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    well we can have x=-3 but we can't have anything less than x=-3 and he and she and I got x=-2 from finding the critical numbers

  36. freckles
    • one year ago
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    The critical numbers are the numbers that satisfy either f'(x)=0 or f'(x) does not exist as long as those x values are in the domain of the original function f(x).

  37. anonymous
    • one year ago
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    why cant we say the interval is from (-3, Infinity)

  38. freckles
    • one year ago
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    remember @mary_am you found x=-2 as a critical number so we have to consider the intervals (-3,-2) and (-2,infty) separately

  39. anonymous
    • one year ago
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    gotcha, I think im overthinking it

  40. freckles
    • one year ago
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    choose a number from each of the intervals (not an endpoint)

  41. freckles
    • one year ago
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    plug into f'(x)

  42. anonymous
    • one year ago
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    so its increasing not decreasing on those two diff intervals, right?

  43. UsukiDoll
    • one year ago
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    \[\frac{2 (x+3)+x}{2 \sqrt{x+3}}=0 \] \[\frac{ 3x+6}{2 \sqrt{x+3}}=0\] \[\frac{ 3(x+2)}{2 \sqrt{x+3}}=0\] x=-3, -2 our intervals are based on the critical points (-3,-2) and (-2,oo) then pick a number between those intervals and plug it into f'(x)

  44. anonymous
    • one year ago
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    ya plugged the numbers in and got increasing for both those intervals, but just wanted to make sure I was correct

  45. UsukiDoll
    • one year ago
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    if it's negative <- decreasing if it's positive <- increasing

  46. anonymous
    • one year ago
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    and also would there be any local max or min values for this problem?

  47. anonymous
    • one year ago
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    because if its just increasing....I would believe that x=-3 would just be the local min...? but no so sure actually

  48. UsukiDoll
    • one year ago
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    we'll you were asked to find just increasing and decreasing so finding local max and local max isn't required.

  49. UsukiDoll
    • one year ago
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    which is the first derivative test

  50. anonymous
    • one year ago
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    lol the second part of the questions asks for the local max and min

  51. UsukiDoll
    • one year ago
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    I don't even see that you've typed that.

  52. Zale101
    • one year ago
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    You can find local min and max after you find what intervals are decreasing or increasing.

  53. anonymous
    • one year ago
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    yea I just did that and I got increasing for both invtervals and now im like super confused

  54. Zale101
    • one year ago
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    |dw:1437451995489:dw| If the graph goes from increasing to decreasing, then that point can be a local max, if it goes from decreasing to increasing then it is local min.

  55. freckles
    • one year ago
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    \[f'(x)=\frac{3x+6}{2 \sqrt{x+3}} \\ \text{ a number from } (-2,-3) \text{ is } -2.5 \\ f'(-2.5)=\frac{3(-2.5)+6}{2 \sqrt{-2.5+3}}=\frac{-6.5+6}{2 \sqrt{.5}}\] is this positive or negative @mary_am ?

  56. anonymous
    • one year ago
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    should be negative

  57. freckles
    • one year ago
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    oops I meant (-3,-2)

  58. freckles
    • one year ago
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    but yes f'<0 on (-3,-2)

  59. freckles
    • one year ago
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    f'<0 means f' is negative

  60. freckles
    • one year ago
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    if f' is negative on (-3,-2) then f is increasing/decreasing on (-3,-2)?

  61. anonymous
    • one year ago
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    Ifound my mistake, I plugged in 0...

  62. freckles
    • one year ago
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    yeah 0 cannot be chosen from (-3,-2)

  63. anonymous
    • one year ago
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    lol yea....

  64. UsukiDoll
    • one year ago
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    for (-3,-2) we can only choose a very small selection

  65. UsukiDoll
    • one year ago
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    x = 0 is more appropriate in (-2,oo) though

  66. freckles
    • one year ago
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    you could choose some of the following: -2.99 -2.88 -2.8 -2.7 -2.6 -2.4 -2.1 any many more...

  67. anonymous
    • one year ago
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    okay, so its increasing from -2, infitnity and decreasing from (-3,-2), right?

  68. anonymous
    • one year ago
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    Yea I know I just dont know what happened... brain isnt functioning to properly

  69. UsukiDoll
    • one year ago
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    mine isn't either . It's in force to be awake mode.

  70. freckles
    • one year ago
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    ok sounds great to me except I would say on not from increasing on (-2,infty) decreasing on (-3,-2)

  71. freckles
    • one year ago
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    |dw:1437452317669:dw| so the function may look sorta like this I just used the information from f' and the domain of the original function to help me draw a rough graph of f

  72. freckles
    • one year ago
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    |dw:1437452372080:dw|

  73. freckles
    • one year ago
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    reading the graph from left to right of course

  74. anonymous
    • one year ago
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    But I am right when I say that the fucntion inreases from -2 to infinity and decreases from -3 to -2, right? and that there is a local minumum at x=-2

  75. freckles
    • one year ago
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    yes but that local min can also be a ?

  76. UsukiDoll
    • one year ago
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    it's backwards I would write, our function is decreasing from (-3,-2) and increasing from (-2,oo)

  77. anonymous
    • one year ago
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    OH it can also be an inflection point....?

  78. freckles
    • one year ago
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    Well I was going for global max inflection point you might want to consider when f''=0

  79. anonymous
    • one year ago
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    I also have to find the intervals for cancavity and inflection points

  80. freckles
    • one year ago
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    global min*

  81. anonymous
    • one year ago
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    so I know that Im supposed to the second derivative

  82. anonymous
    • one year ago
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    I've never heard of global min...

  83. freckles
    • one year ago
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    |dw:1437452577062:dw|

  84. freckles
    • one year ago
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    it is just the lowest point on the graph

  85. UsukiDoll
    • one year ago
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    for concavity take the derivative again... set that second derivative to 0 and find the critical points

  86. anonymous
    • one year ago
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    So is the second derivative = (x+3^-1/2) (2) - (2x+3)(-1/2x+3^-3/2)

  87. anonymous
    • one year ago
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    or am I like waaaay off....? :/

  88. freckles
    • one year ago
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    \[f'(x)=\frac{3x+6}{2 \sqrt{x+3}} \\ f'(x)=\frac{(3x+6)' \cdot 2 \sqrt{x+3}-(2 \sqrt{x+3})' \cdot (3x+6)}{(2 \sqrt{x+3})^2}\]

  89. freckles
    • one year ago
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    can you find (3x+6)' and (2 sqrt(x+3))'

  90. anonymous
    • one year ago
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    yea, hollup

  91. anonymous
    • one year ago
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    its 3 and the other one is x+3 ^-1/2

  92. freckles
    • one year ago
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    that is right \[(3x+6)'=(3x)'+(6)'=3+0=3 \\ (2 \sqrt{x+3})'=2 (\sqrt{x+3})'=2 \frac{1}{2 \sqrt{x+3}} = \frac{1}{\sqrt{x+3}}\]

  93. freckles
    • one year ago
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    \[f''(x)=\frac{3 \cdot 2 \sqrt{x+3}-\frac{1 }{\sqrt{x+3}} \cdot (3x+6)}{4(x+3)}\]

  94. freckles
    • one year ago
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    you multiply bottom and top by sqrt(x+3) to clear the compound fraction

  95. anonymous
    • one year ago
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    \[(2\sqrt{X+3) (3) - }\]

  96. anonymous
    • one year ago
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    what the heck...why didnt my answer show up

  97. anonymous
    • one year ago
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    okay ill do what you just asked

  98. anonymous
    • one year ago
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    so I got 6(sqrt x+3) * (sqrt x+3) - (3x-6)(sqrt x+3) all over 4(x+3)*(sqrt x+3)

  99. anonymous
    • one year ago
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    should I simplify more?

  100. freckles
    • one year ago
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    \[f''(x)=\frac{6(x+3)-(3x+6)}{4 (x+3) \sqrt{x+3}}\]

  101. anonymous
    • one year ago
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    yea, basicaly the same thing!

  102. UsukiDoll
    • one year ago
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    now we need critical points.

  103. freckles
    • one year ago
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    except you have a sqrt(x+3) next to the (3x+6)

  104. anonymous
    • one year ago
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    yea thanks for pointing that out!

  105. freckles
    • one year ago
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    anyways do you know where to go from here?

  106. anonymous
    • one year ago
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    so x= -4 is one of the critical points

  107. freckles
    • one year ago
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    find the POSSIBLE inflection points by finding where f''=0 f'' does not exist at x=-3 we already know at x=-3 we do not have an inflection point since it is an endpoint

  108. freckles
    • one year ago
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    \[6(x+3)-3x-6=0 \\ 3x+12=0 \\ x=-4\] yeah sounds great

  109. freckles
    • one year ago
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    now to find if (-4,f(-4)) is an inflection points test the intervals to see if the concavity switches

  110. anonymous
    • one year ago
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    and for the undefined crtical point its going to x= -3?

  111. anonymous
    • one year ago
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    Do I also put x= -2 on the number line when testing out the points?

  112. anonymous
    • one year ago
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    |dw:1437454017504:dw|

  113. anonymous
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    and then I test points in between those numbers with the second derivative to find where its CU and CD?

  114. freckles
    • one year ago
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    you have one possible inflection point so you only have two intervals to check

  115. anonymous
    • one year ago
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    so just stick with -3 and -2 ?

  116. anonymous
    • one year ago
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    okay

  117. freckles
    • one year ago
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    lol

  118. freckles
    • one year ago
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    wait sorry x=-4 wasn't on (-3,infty) which is the domain of f

  119. freckles
    • one year ago
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    we don't have an inflection point

  120. freckles
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    x=-4 wasn't in the domain of the original function

  121. anonymous
    • one year ago
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    thats super odd

  122. freckles
    • one year ago
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    the function is either concave up or concave down

  123. freckles
    • one year ago
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    it doesn't have a combination of the two

  124. anonymous
    • one year ago
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    so the answer is like nothing

  125. freckles
    • one year ago
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    test any number in the domain of f that is test any number from (-3,inf) to see which it is

  126. anonymous
    • one year ago
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    I plugged in -3.5 to test it out between the intervals of (-3,-4)

  127. freckles
    • one year ago
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    put that number into f''

  128. freckles
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    (-3,-4) doesn't make sense because -4 isn't greater than -3

  129. anonymous
    • one year ago
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    -4,-3

  130. UsukiDoll
    • one year ago
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    the bigger the negative number, the smaller its values is

  131. freckles
    • one year ago
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    but (-4,-3) isn't in the domain of f

  132. anonymous
    • one year ago
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    kk lemme plug in -2 then

  133. freckles
    • one year ago
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    you are forgetting the function has domain (-3,infty)

  134. UsukiDoll
    • one year ago
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    is it because at x =-3 the whole function is undefined?

  135. freckles
    • one year ago
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    you can choose a number from (-3,infty) to test the concavity of f by using f''

  136. anonymous
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    so when I used -2 I got a negative number out which was -1.2

  137. freckles
    • one year ago
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    \[f''(x)=\frac{6(x+3)-(3x+6)}{4 (x+3) \sqrt{x+3}}\] \[f''(x)=\frac{3x+12}{4(x+3)\sqrt{x+3} } \\ f''(0)=\frac{3(0)+12}{4(0+3)\sqrt{0+3}} \\\ f''(0)=\frac{12}{4(3)\sqrt{3}}>0 \\ \text{ hmm... you got a negative number \let me see } \\ f''(-2)=\frac{3(-2)+12}{4(-2+3)\sqrt{-2+3}}=\frac{6}{4(1)\sqrt{1}}\] don't see how you got a negative number :(

  138. anonymous
    • one year ago
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    wow thats odd, I plugged the number into the first equation you put above but that shouldnt have given me a diff answer

  139. freckles
    • one year ago
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    you mean the unsimplfied version I have of f'' ?

  140. anonymous
    • one year ago
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    basicalllly

  141. anonymous
    • one year ago
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    okay i did it again and i got 6/4

  142. anonymous
    • one year ago
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    so this is how my number line looks like

  143. freckles
    • one year ago
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    how does it look?

  144. anonymous
    • one year ago
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    |dw:1437454970480:dw|

  145. freckles
    • one year ago
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    right because our function only exists to the right of and including x=-3

  146. anonymous
    • one year ago
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    sorry for the ratchetness of it, still getting used to it

  147. freckles
    • one year ago
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    what is all the + signs mean are you just saying it is concave up on (-3,infty)?

  148. anonymous
    • one year ago
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    ya

  149. freckles
    • one year ago
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    ok cool

  150. anonymous
    • one year ago
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    so is there also an inflection point at x=-3?

  151. freckles
    • one year ago
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    anyways I must leave you it is passed my bed time

  152. freckles
    • one year ago
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    no there is no inflection point

  153. freckles
    • one year ago
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    the concavity does not switch ever

  154. nincompoop
    • one year ago
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    laughing out loud

  155. freckles
    • one year ago
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    the function is concave up everywhere on the function's domain

  156. anonymous
    • one year ago
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    thanks a lot, you really are a great person! Bless yo soul dude

  157. anonymous
    • one year ago
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    LOL I really wish you were my calc teacher instead

  158. freckles
    • one year ago
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    it was no problem peace for now

  159. anonymous
    • one year ago
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    Peace freckles

  160. nincompoop
    • one year ago
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    ye a teacher that solves your own homework! that would be nice

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