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hiralpatel121

What is the connection between critical numbers and relative extrema?

  • one year ago
  • one year ago

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  1. matheducatorMcG
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    Are you in calculus 1?

    • one year ago
  2. hiralpatel121
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    Yes

    • one year ago
  3. matheducatorMcG
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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.

    • one year ago
  4. hiralpatel121
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    Thank you so much! Is it possible for you to help me understand a few more problems?

    • one year ago
  5. matheducatorMcG
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    keep on asking.

    • one year ago
  6. hiralpatel121
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    How do you find inflection points?

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

    • one year ago
  8. hiralpatel121
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    Can you provide an example?

    • one year ago
  9. matheducatorMcG
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    just a sec.

    • one year ago
  10. hiralpatel121
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    Okay

    • one year ago
  11. matheducatorMcG
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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

    • one year ago
  12. matheducatorMcG
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    that's: possible inflection points are at

    • one year ago
  13. matheducatorMcG
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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.

    • one year ago
  14. hiralpatel121
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    What would it be in this case?

    • one year ago
  15. matheducatorMcG
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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

    • one year ago
  16. hiralpatel121
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    To list the inflection points, would I list them as: (-1,1) and (1,3)?

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

    • one year ago
  18. matheducatorMcG
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    your points would be (0, ) and (2, )?

    • one year ago
  19. hiralpatel121
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    (0,0) and (2,0)

    • one year ago
  20. hiralpatel121
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    Correct?

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

    • one year ago
  22. matheducatorMcG
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    (0,0) was right

    • one year ago
  23. hiralpatel121
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    (2, -16)

    • one year ago
  24. hiralpatel121
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    (2, -16)

    • one year ago
  25. matheducatorMcG
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    ooohhh... so close. I get (2,-18).

    • one year ago
  26. hiralpatel121
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    I keep getting -16.

    • one year ago
  27. matheducatorMcG
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    wait... I'm wrong... you're right

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

    • one year ago
  29. hiralpatel121
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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.”

    • one year ago
  30. matheducatorMcG
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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.

    • one year ago
  31. hiralpatel121
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    So how would I correct the statement?

    • one year ago
  32. matheducatorMcG
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    do we need to correct it to have the 'if ' statement or keep the 'then' statement?

    • one year ago
  33. hiralpatel121
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    It should be like... "So, the correct statement would be:...if...then..."

    • one year ago
  34. matheducatorMcG
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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.

    • one year ago
  35. matheducatorMcG
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    the first I changed the 'then' statement. the second one i changed the 'if' statement.

    • one year ago
  36. hiralpatel121
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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.”

    • one year ago
  37. hiralpatel121
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    This is for the second derivative.

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

    • one year ago
  39. hiralpatel121
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    And why?

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

    • one year ago
  41. hiralpatel121
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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)

    • one year ago
  42. matheducatorMcG
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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?

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

    • one year ago
  44. hiralpatel121
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    Yes

    • one year ago
  45. matheducatorMcG
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    do I have it right?

    • one year ago
  46. hiralpatel121
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    Yes

    • one year ago
  47. matheducatorMcG
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    okay...just a sec

    • one year ago
  48. matheducatorMcG
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    lost internet. typing ferociously now.

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

    • one year ago
  50. matheducatorMcG
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    \[0=1-\frac{2}{\sqrt{t+1}}\]

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

    • one year ago
  52. hiralpatel121
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    How would I cross multiply this?

    • one year ago
  53. matheducatorMcG
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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.

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

    • one year ago
  55. matheducatorMcG
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    \[\sqrt{t+1}=2\]

    • one year ago
  56. matheducatorMcG
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    they become equal.

    • one year ago
  57. hiralpatel121
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    What's the next step?

    • one year ago
  58. matheducatorMcG
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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.

    • one year ago
  59. hiralpatel121
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    t=1

    • one year ago
  60. matheducatorMcG
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    \[t+1=4\]

    • one year ago
  61. hiralpatel121
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    Oh yes! I forgot to square 2.

    • one year ago
  62. matheducatorMcG
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    cool. good job fixing it.

    • one year ago
  63. matheducatorMcG
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    so your extrema is at (3, ?)

    • one year ago
  64. hiralpatel121
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    I get: y = -2

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

    • one year ago
  66. hiralpatel121
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    -1 to the root of 4 = -1 (2) = -2

    • one year ago
  67. matheducatorMcG
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    order of ops. you have to evaluate the root first, then multiplication with the 4 then subtract from the 3

    • one year ago
  68. matheducatorMcG
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    BEDMAS-brackets, exponents, division or multiplication, addition or subtraction.

    • one year ago
  69. hiralpatel121
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    (3, -5)

    • one year ago
  70. matheducatorMcG
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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.

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

    • one year ago
  72. hiralpatel121
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    Plug 2 and 4 into the first derivative?

    • one year ago
  73. matheducatorMcG
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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]

    • one year ago
  74. hiralpatel121
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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.

    • one year ago
  75. hiralpatel121
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    2 gives a negative answer and 4 gives a negative answer.

    • one year ago
  76. matheducatorMcG
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    |dw:1341599783811:dw|

    • one year ago
  77. hiralpatel121
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    2 gives a negative answer and 4 gives a negative answer.

    • one year ago
  78. matheducatorMcG
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    does 4 give us a negative in the 1st derivative? check again.

    • one year ago
  79. hiralpatel121
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    Wait nvm, it's positive. So it's negative to positive.

    • one year ago
  80. hiralpatel121
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    (3,-5) is a min.

    • one year ago
  81. matheducatorMcG
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    yay! good job.

    • one year ago
  82. hiralpatel121
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    What about the maximum? There isn't one?

    • one year ago
  83. matheducatorMcG
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    not unless you got more than one critical point. check the graph.

    • one year ago
  84. hiralpatel121
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    I did, I didn't think there was a max. I wanted to make sure I was right.

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

    • one year ago
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  86. hiralpatel121
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    okay

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

    • one year ago
  88. matheducatorMcG
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    do you all use L'Hopital's rule yet?

    • one year ago
  89. hiralpatel121
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    No

    • one year ago
  90. hiralpatel121
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    I have to find the limit analytically, without the L Hospital Rule.

    • one year ago
  91. matheducatorMcG
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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, \[\lim_{x \rightarrow \infty} \frac{3x}{\sqrt{x^2+4}}=3\]

    • one year ago
  92. hiralpatel121
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    Why wouldn't it be + and - 2?

    • one year ago
  93. hiralpatel121
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    OH! Wait, I got it.

    • one year ago
  94. hiralpatel121
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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

    • one year ago
  95. matheducatorMcG
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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..

    • one year ago
  96. hiralpatel121
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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?

    • one year ago
  97. matheducatorMcG
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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.

    • one year ago
  98. hiralpatel121
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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)

    • one year ago
  99. matheducatorMcG
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    \[g(x)=x \sqrt{x+5}\] this is it?

    • one year ago
  100. hiralpatel121
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    yes

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

    • one year ago
  102. matheducatorMcG
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    but we need the 2nd derivative, so...

    • one year ago
  103. matheducatorMcG
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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.

    • one year ago
  104. hiralpatel121
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    Okayy

    • one year ago
  105. matheducatorMcG
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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. \[0=\frac{3x+20}{4(x+5)^{\frac{3}{2}}}\]

    • one year ago
  106. matheducatorMcG
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    the only way this is zero is if 3x+20 is equal to 0.

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

    • one year ago
  108. hiralpatel121
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    3x+20 = 0 gives us -20/3

    • one year ago
  109. matheducatorMcG
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    kew

    • one year ago
  110. hiralpatel121
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    So how would we find the inflection here?

    • one year ago
  111. matheducatorMcG
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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.

    • one year ago
  112. matheducatorMcG
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    if positive, concave up. if negative, concave down.

    • one year ago
  113. hiralpatel121
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    I plugged it into my calculator and it keeps saying "domain error."

    • one year ago
  114. matheducatorMcG
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    into the original function, right?

    • one year ago
  115. matheducatorMcG
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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

    • one year ago
  116. matheducatorMcG
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    this means, there are no inflection points.

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

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

    • one year ago
  119. hiralpatel121
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    And for the concavity?

    • one year ago
  120. matheducatorMcG
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    it is a global min, meaning there are no other min's. there are no global or relative max's.

    • one year ago
  121. matheducatorMcG
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    if there is no inflection point and the graph has a min, then it is concave up.

    • one year ago
  122. matheducatorMcG
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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

    • one year ago
  123. hiralpatel121
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    Okay that's all?

    • one year ago
  124. matheducatorMcG
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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.

    • one year ago
  125. hiralpatel121
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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.

    • one year ago
  126. matheducatorMcG
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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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    \[h(t)=t-4\sqrt{t+1}\] \[h'(t)=1-\frac{2}{\sqrt{t+1}}\] \[0=1-\frac{2}{\sqrt{t+1}}\] \[1=\frac{2}{\sqrt{t+1}}\] \[2=\sqrt{t+1}\] \[4=t+1\] \[t=3\] \[h''(t)=\frac{1}{(t+1)^{\frac{3}{2}}}\] \[h''(3)=\frac{1}{8}\] 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)

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