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hiralpatel121 Group Title

What is the connection between critical numbers and relative extrema?

  • 2 years ago
  • 2 years ago

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  1. matheducatorMcG Group Title
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    Are you in calculus 1?

    • 2 years ago
  2. hiralpatel121 Group Title
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    Yes

    • 2 years ago
  3. matheducatorMcG Group Title
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    critical numbers can tell you where relative extrema occur on the graph. we use derivatives to find these critical numbers which in turn allows us to better sketch curves.

    • 2 years ago
  4. hiralpatel121 Group Title
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    Thank you so much! Is it possible for you to help me understand a few more problems?

    • 2 years ago
  5. matheducatorMcG Group Title
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    keep on asking.

    • 2 years ago
  6. hiralpatel121 Group Title
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    How do you find inflection points?

    • 2 years ago
  7. matheducatorMcG Group Title
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    well, this occurs on curves that change concavity. this occurs when the second derivative, f"(x) is equal to zero.

    • 2 years ago
  8. hiralpatel121 Group Title
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    Can you provide an example?

    • 2 years ago
  9. matheducatorMcG Group Title
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    just a sec.

    • 2 years ago
  10. hiralpatel121 Group Title
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    Okay

    • 2 years ago
  11. matheducatorMcG Group Title
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    determine points of inflection for \[f(x)=x^4-4x^3\] find second derivative \[f''(x)=12x^2-24x=12x(x-2)\] set equal to zero and solve. possible points are at \[x=0 \]and \[x=2\] Now test values around 0 and 2, like -1 and 1 and 3

    • 2 years ago
  12. matheducatorMcG Group Title
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    that's: possible inflection points are at

    • 2 years ago
  13. matheducatorMcG Group Title
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    if f''(x) >0 in the interval, concave up. if f''(x) <0 in the interval, concave down. if the values go concave up then concave down, you've got an inflection point.

    • 2 years ago
  14. hiralpatel121 Group Title
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    What would it be in this case?

    • 2 years ago
  15. matheducatorMcG Group Title
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    f''(-1)>0(concave up) and f''(1)<0(concave down) and f''(3)>0(concave up). Therefore, we can conclude that both are inflection points because we go concave up->concave down->concave up

    • 2 years ago
  16. hiralpatel121 Group Title
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    To list the inflection points, would I list them as: (-1,1) and (1,3)?

    • 2 years ago
  17. matheducatorMcG Group Title
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    not quite. you must substitute the values of 0 and 2 into the original function to find the y value.

    • 2 years ago
  18. matheducatorMcG Group Title
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    your points would be (0, ) and (2, )?

    • 2 years ago
  19. hiralpatel121 Group Title
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    (0,0) and (2,0)

    • 2 years ago
  20. hiralpatel121 Group Title
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    Correct?

    • 2 years ago
  21. matheducatorMcG Group Title
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    ummm... not quite again. the original was\[f(x)=x^4-4x^3\] find f(0) and f(2).

    • 2 years ago
  22. matheducatorMcG Group Title
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    (0,0) was right

    • 2 years ago
  23. hiralpatel121 Group Title
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    (2, -16)

    • 2 years ago
  24. hiralpatel121 Group Title
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    (2, -16)

    • 2 years ago
  25. matheducatorMcG Group Title
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    ooohhh... so close. I get (2,-18).

    • 2 years ago
  26. hiralpatel121 Group Title
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    I keep getting -16.

    • 2 years ago
  27. matheducatorMcG Group Title
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    wait... I'm wrong... you're right

    • 2 years ago
  28. matheducatorMcG Group Title
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    \[f(2)=(2)^4-4(2)^3\] Good Job!!

    • 2 years ago
  29. hiralpatel121 Group Title
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    Okay, thank you! The next thing I need help with is: Is the following true or false? If false, explain why and correct the statement. “If f ‘ (c)<0, then f is concave down at x=c.”

    • 2 years ago
  30. matheducatorMcG Group Title
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    sorry hold on. I miss read the question real quick. f'(c)<0 just means the slope of the curve at that point is negative. a parabola has some c where f'(c)<0 but the entire graph is always concave up. false.

    • 2 years ago
  31. hiralpatel121 Group Title
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    So how would I correct the statement?

    • 2 years ago
  32. matheducatorMcG Group Title
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    do we need to correct it to have the 'if ' statement or keep the 'then' statement?

    • 2 years ago
  33. hiralpatel121 Group Title
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    It should be like... "So, the correct statement would be:...if...then..."

    • 2 years ago
  34. matheducatorMcG Group Title
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    i'll give both "If f '(c)<0, then f is decreasing at x=c." "If f ''(c)<0, then f is concave down." i think those suffice.

    • 2 years ago
  35. matheducatorMcG Group Title
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    the first I changed the 'then' statement. the second one i changed the 'if' statement.

    • 2 years ago
  36. hiralpatel121 Group Title
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    Okay thank you. What about.. Is the following true or false? If false, explain why and correct the statement. “If f ‘’(c)<0, for all real numbers x, then f decreases without a bound.”

    • 2 years ago
  37. hiralpatel121 Group Title
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    This is for the second derivative.

    • 2 years ago
  38. matheducatorMcG Group Title
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    false "If f''(c)<0 for all real numbers x, then f is concave down."

    • 2 years ago
  39. hiralpatel121 Group Title
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    And why?

    • 2 years ago
  40. matheducatorMcG Group Title
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    your 2nd derivative being negative for all values of x implies the whole graph is concave down

    • 2 years ago
  41. hiralpatel121 Group Title
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    Okay. I don’t know how to find the relative extrema for the following function. f(t) = t – 4 to the root of ( t+1)

    • 2 years ago
  42. matheducatorMcG Group Title
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    f(t)=t-4 to the root of (t+1) to the root? does that mean multiplied by the root or to the power of the root?

    • 2 years ago
  43. matheducatorMcG Group Title
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    \[f(t)=t-4\sqrt{t+1}\] do you mean this?

    • 2 years ago
  44. hiralpatel121 Group Title
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    Yes

    • 2 years ago
  45. matheducatorMcG Group Title
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    do I have it right?

    • 2 years ago
  46. hiralpatel121 Group Title
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    Yes

    • 2 years ago
  47. matheducatorMcG Group Title
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    okay...just a sec

    • 2 years ago
  48. matheducatorMcG Group Title
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    lost internet. typing ferociously now.

    • 2 years ago
  49. matheducatorMcG Group Title
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    lets find the 1st derivative and set it equal to zero\[f(t)=t-4\sqrt{t+1}\]\[f '(t)=1-\frac{2}{\sqrt{t+1}}\]

    • 2 years ago
  50. matheducatorMcG Group Title
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    \[0=1-\frac{2}{\sqrt{t+1}}\]

    • 2 years ago
  51. matheducatorMcG Group Title
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    now add 1 to both sides, cross multiply and solve. let me know if you need the next step.

    • 2 years ago
  52. hiralpatel121 Group Title
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    How would I cross multiply this?

    • 2 years ago
  53. matheducatorMcG Group Title
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    \[1=\frac{2}{\sqrt{t+1}}\] put the 1 over 1 and multiply, remembering that the products you get are equal to each other.

    • 2 years ago
  54. hiralpatel121 Group Title
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    So I get: 2 / 1 times the root of (t+1)

    • 2 years ago
  55. matheducatorMcG Group Title
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    \[\sqrt{t+1}=2\]

    • 2 years ago
  56. matheducatorMcG Group Title
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    they become equal.

    • 2 years ago
  57. hiralpatel121 Group Title
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    What's the next step?

    • 2 years ago
  58. matheducatorMcG Group Title
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    now square both sides and solve for x. this tells you the x value where there is extrema. now substitute x into f(t) to find your y.

    • 2 years ago
  59. hiralpatel121 Group Title
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    t=1

    • 2 years ago
  60. matheducatorMcG Group Title
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    \[t+1=4\]

    • 2 years ago
  61. hiralpatel121 Group Title
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    Oh yes! I forgot to square 2.

    • 2 years ago
  62. matheducatorMcG Group Title
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    cool. good job fixing it.

    • 2 years ago
  63. matheducatorMcG Group Title
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    so your extrema is at (3, ?)

    • 2 years ago
  64. hiralpatel121 Group Title
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    I get: y = -2

    • 2 years ago
  65. matheducatorMcG Group Title
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    hmmm... I get something else. \[f(3)=(3)-4\sqrt{(3)+1}\]

    • 2 years ago
  66. hiralpatel121 Group Title
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    -1 to the root of 4 = -1 (2) = -2

    • 2 years ago
  67. matheducatorMcG Group Title
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    order of ops. you have to evaluate the root first, then multiplication with the 4 then subtract from the 3

    • 2 years ago
  68. matheducatorMcG Group Title
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    BEDMAS-brackets, exponents, division or multiplication, addition or subtraction.

    • 2 years ago
  69. hiralpatel121 Group Title
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    (3, -5)

    • 2 years ago
  70. matheducatorMcG Group Title
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    YAY! now, is this a min or a max? we can check by substituting values on either side of 3 into f '(t) to see if it goes increasing to decreasing OR decreasing to increasing. that will tell us if it is a min or a max.

    • 2 years ago
  71. matheducatorMcG Group Title
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    so find f ' (2) and f ' (4) to see which pattern it takes on.

    • 2 years ago
  72. hiralpatel121 Group Title
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    Plug 2 and 4 into the first derivative?

    • 2 years ago
  73. matheducatorMcG Group Title
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    yup. the 1st derivative is the slope of the curve. if it goes decreasing to increasing, its a min. if it goes increasing to decreasing, it is a max. you could also look at the 2nd derivative to see if it is positive or negative at that point, but this doesn't always work. it works here though. ;0]

    • 2 years ago
  74. hiralpatel121 Group Title
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    When I plugged 2 in, I got 1 - 2 times root of 3. When I plugged 4 in, I got 1 - 2 times root of 5.

    • 2 years ago
  75. hiralpatel121 Group Title
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    2 gives a negative answer and 4 gives a negative answer.

    • 2 years ago
  76. matheducatorMcG Group Title
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    |dw:1341599783811:dw|

    • 2 years ago
  77. hiralpatel121 Group Title
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    2 gives a negative answer and 4 gives a negative answer.

    • 2 years ago
  78. matheducatorMcG Group Title
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    does 4 give us a negative in the 1st derivative? check again.

    • 2 years ago
  79. hiralpatel121 Group Title
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    Wait nvm, it's positive. So it's negative to positive.

    • 2 years ago
  80. hiralpatel121 Group Title
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    (3,-5) is a min.

    • 2 years ago
  81. matheducatorMcG Group Title
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    yay! good job.

    • 2 years ago
  82. hiralpatel121 Group Title
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    What about the maximum? There isn't one?

    • 2 years ago
  83. matheducatorMcG Group Title
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    not unless you got more than one critical point. check the graph.

    • 2 years ago
  84. hiralpatel121 Group Title
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    I did, I didn't think there was a max. I wanted to make sure I was right.

    • 2 years ago
  85. matheducatorMcG Group Title
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    there is a vertical asymptote at x=-1. that would be the only other thing I would want to know

    • 2 years ago
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  86. hiralpatel121 Group Title
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    okay

    • 2 years ago
  87. hiralpatel121 Group Title
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    How do I find the limit for: limit as x approaches infinity 3x / (root of x^2+4)

    • 2 years ago
  88. matheducatorMcG Group Title
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    do you all use L'Hopital's rule yet?

    • 2 years ago
  89. hiralpatel121 Group Title
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    No

    • 2 years ago
  90. hiralpatel121 Group Title
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    I have to find the limit analytically, without the L Hospital Rule.

    • 2 years ago
  91. matheducatorMcG Group Title
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    okay... how about the fact that were going off to infinity so, \[\sqrt{x^2} \approx \sqrt{x^2+4} \] as x-> infinity, therefore, we can evaluate \[\lim_{x \rightarrow \infty} \frac{3x}{\sqrt{x^2+4}}\] as \[\lim_{x \rightarrow \infty} \frac{3x}{\sqrt{x^2}}\] which algebraically simplifies to be \[\lim_{x \rightarrow \infty} \frac{3x}{x}\] which again simplifies to 3. so, \[\lim_{x \rightarrow \infty} \frac{3x}{\sqrt{x^2+4}}=3\]

    • 2 years ago
  92. hiralpatel121 Group Title
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    Why wouldn't it be + and - 2?

    • 2 years ago
  93. hiralpatel121 Group Title
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    OH! Wait, I got it.

    • 2 years ago
  94. hiralpatel121 Group Title
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    Next problem... Find any vertical and horizontal asymptotes of the graph of the function. g(x) = 5x^2 / x^2+2

    • 2 years ago
  95. matheducatorMcG Group Title
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    horizontal asymptote is what value you approach as x->infinity, mainly 5. vertical asymptotes are what makes the denominator zero, so none here. asymptotes are based off limits as x approaches infinity for horizontal or as x approaches some value a for vertical asymptotes..

    • 2 years ago
  96. hiralpatel121 Group Title
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    One question.. There are times when we just set the top and bottom equal to 0 and find asymptotes. Does this occur all the time?

    • 2 years ago
  97. matheducatorMcG Group Title
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    Rational functions are the graphs that have the most asymptotes. But be careful. Read on. With rational functions the denominator will tell you where vertical asymptotes are located. setting the numerator equal to zero will tell you where it crosses the x axis. finding horizontal asymptotes are all about the limit of the graph as x approaches infinity.

    • 2 years ago
  98. hiralpatel121 Group Title
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    Okay, one last question I need help with. What would be the points of inflection and concavity of the following graph of the function? g(x) = x to the root of (x+5)

    • 2 years ago
  99. matheducatorMcG Group Title
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    \[g(x)=x \sqrt{x+5}\] this is it?

    • 2 years ago
  100. hiralpatel121 Group Title
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    yes

    • 2 years ago
  101. matheducatorMcG Group Title
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    \[g'(x)=x \frac{1}{2\sqrt{x+5}}+\sqrt{x+5}\]

    • 2 years ago
  102. matheducatorMcG Group Title
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    but we need the 2nd derivative, so...

    • 2 years ago
  103. matheducatorMcG Group Title
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    \[g"(x)=-\frac{x}{4(x+5)^{\frac{3}{2}}}+\frac{1}{\sqrt{x+5}} \] now set equal to zero and solve. whooo.... give me a sec here.

    • 2 years ago
  104. hiralpatel121 Group Title
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    Okayy

    • 2 years ago
  105. matheducatorMcG Group Title
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    combine the two fractions to get \[g"(x)=\frac{3x+20}{4(x+5)^{\frac{3}{2}}}\] then set equal to 0. \[0=\frac{3x+20}{4(x+5)^{\frac{3}{2}}}\]

    • 2 years ago
  106. matheducatorMcG Group Title
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    the only way this is zero is if 3x+20 is equal to 0.

    • 2 years ago
  107. matheducatorMcG Group Title
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    so inflection at x=? values on either side of the inflection value will tell you concave up or concave down.

    • 2 years ago
  108. hiralpatel121 Group Title
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    3x+20 = 0 gives us -20/3

    • 2 years ago
  109. matheducatorMcG Group Title
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    kew

    • 2 years ago
  110. hiralpatel121 Group Title
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    So how would we find the inflection here?

    • 2 years ago
  111. matheducatorMcG Group Title
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    x=-20/3 gets substituted into the original function to find the y value of the point. then test values on either side of the inflection point in the 2nd derivative to see if positive or negative.

    • 2 years ago
  112. matheducatorMcG Group Title
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    if positive, concave up. if negative, concave down.

    • 2 years ago
  113. hiralpatel121 Group Title
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    I plugged it into my calculator and it keeps saying "domain error."

    • 2 years ago
  114. matheducatorMcG Group Title
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    into the original function, right?

    • 2 years ago
  115. matheducatorMcG Group Title
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    oh wait... the calculator is right. -20/3 would make the inside of the root negative, which is no no in the real world. hmm... contemplating

    • 2 years ago
  116. matheducatorMcG Group Title
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    this means, there are no inflection points.

    • 2 years ago
  117. matheducatorMcG Group Title
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    it has a min at -10/3 but is always concave up. check graph. when in doubt, graph it out.

    • 2 years ago
  118. matheducatorMcG Group Title
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    your answer should be: there are no inflection points and there is a min at (-10/3, -4.303314829)

    • 2 years ago
  119. hiralpatel121 Group Title
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    And for the concavity?

    • 2 years ago
  120. matheducatorMcG Group Title
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    it is a global min, meaning there are no other min's. there are no global or relative max's.

    • 2 years ago
  121. matheducatorMcG Group Title
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    if there is no inflection point and the graph has a min, then it is concave up.

    • 2 years ago
  122. matheducatorMcG Group Title
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    *sorry, previous entry should read: if there is no inflection point and the graph has a global min, then it is concave up

    • 2 years ago
  123. hiralpatel121 Group Title
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    Okay that's all?

    • 2 years ago
  124. matheducatorMcG Group Title
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    http://www.wolframalpha.com/ can help you immensely if your looking to get quick answers. more explanation may be needed but its a great start.

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    Thank you so much for your help! I have a test next week and I might come across a few problems like this. Will you be online next week? I'd love if you could help me solve a few more problems so I have more to study from.

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    I should be as this website is my new internet addiction. I am teaching a small class this summer but I love answering math questions, so I'll be around. become a fan of mine and send me a message if you get on and see me.

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    Okay, thank you so much once again!

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    Have a nice day.

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    you too.

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    \[h(t)=t-4\sqrt{t+1}\] \[h'(t)=1-\frac{2}{\sqrt{t+1}}\] \[0=1-\frac{2}{\sqrt{t+1}}\] \[1=\frac{2}{\sqrt{t+1}}\] \[2=\sqrt{t+1}\] \[4=t+1\] \[t=3\] \[h''(t)=\frac{1}{(t+1)^{\frac{3}{2}}}\] \[h''(3)=\frac{1}{8}\] which is positive, so by the 2nd derivative test, this is a min. there is no relative max. the min occurs at \[h(3)=-5\] (3,-5)

    • 2 years ago
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