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hiralpatel121

  • 2 years ago

What is the connection between critical numbers and relative extrema?

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  1. matheducatorMcG
    • 2 years ago
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    Are you in calculus 1?

  2. hiralpatel121
    • 2 years ago
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    Yes

  3. matheducatorMcG
    • 2 years ago
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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.

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

  5. matheducatorMcG
    • 2 years ago
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    keep on asking.

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

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

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

  9. matheducatorMcG
    • 2 years ago
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    just a sec.

  10. hiralpatel121
    • 2 years ago
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    Okay

  11. matheducatorMcG
    • 2 years ago
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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

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

  13. matheducatorMcG
    • 2 years ago
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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.

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

  15. matheducatorMcG
    • 2 years ago
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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

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

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

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

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

  20. hiralpatel121
    • 2 years ago
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    Correct?

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

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

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

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

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

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

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

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

  29. hiralpatel121
    • 2 years ago
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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.”

  30. matheducatorMcG
    • 2 years ago
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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.

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

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

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

  34. matheducatorMcG
    • 2 years ago
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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.

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

  36. hiralpatel121
    • 2 years ago
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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.”

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

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

  39. hiralpatel121
    • 2 years ago
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    And why?

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

  41. hiralpatel121
    • 2 years ago
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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)

  42. matheducatorMcG
    • 2 years ago
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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?

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

  44. hiralpatel121
    • 2 years ago
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    Yes

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

  46. hiralpatel121
    • 2 years ago
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    Yes

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

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

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

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

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

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

  53. matheducatorMcG
    • 2 years ago
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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.

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

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

  56. matheducatorMcG
    • 2 years ago
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    they become equal.

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

  58. matheducatorMcG
    • 2 years ago
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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.

  59. hiralpatel121
    • 2 years ago
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    t=1

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

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

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

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

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

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

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

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

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

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

  70. matheducatorMcG
    • 2 years ago
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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.

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

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

  73. matheducatorMcG
    • 2 years ago
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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]

  74. hiralpatel121
    • 2 years ago
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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.

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

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

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

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

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

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

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

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

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

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

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

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  86. hiralpatel121
    • 2 years ago
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    okay

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

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

  89. hiralpatel121
    • 2 years ago
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    No

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

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

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

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

  94. hiralpatel121
    • 2 years ago
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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

  95. matheducatorMcG
    • 2 years ago
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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..

  96. hiralpatel121
    • 2 years ago
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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?

  97. matheducatorMcG
    • 2 years ago
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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.

  98. hiralpatel121
    • 2 years ago
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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)

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

  100. hiralpatel121
    • 2 years ago
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    yes

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

  102. matheducatorMcG
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    but we need the 2nd derivative, so...

  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.

  104. hiralpatel121
    • 2 years ago
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    Okayy

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

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

  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.

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

  109. matheducatorMcG
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    kew

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

  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.

  112. matheducatorMcG
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    if positive, concave up. if negative, concave down.

  113. hiralpatel121
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    I plugged it into my calculator and it keeps saying "domain error."

  114. matheducatorMcG
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    into the original function, right?

  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

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

  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.

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

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

  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.

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

  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

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

  124. matheducatorMcG
    • 2 years ago
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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.

  125. hiralpatel121
    • 2 years ago
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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.

  126. matheducatorMcG
    • 2 years ago
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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.

  127. hiralpatel121
    • 2 years ago
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    Okay, thank you so much once again!

  128. hiralpatel121
    • 2 years ago
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    Have a nice day.

  129. matheducatorMcG
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    you too.

  130. matheducatorMcG
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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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