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anonymous

  • one year ago

Locate the absolute extrema of the function (if any exist) over the indicated variables.

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    This is how I solved a...

  3. anonymous
    • one year ago
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    I found the derivative of the function: x(4-x^2) I set it to zero... x=0 4-x^2=0 or x=2 I then plugged the following back into the original function: 2, 0, -2 And I received: 0, 2, 0 So I concluded that my max = 2 and my min = 0. Assuming that what I did was correct, how could I figure out the exact coordinates of my min and max without looking at a graph?

  4. freckles
    • one year ago
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    your derivative is just a little off

  5. freckles
    • one year ago
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    also 4-x^2=0 gives you x=2 or x=-2

  6. anonymous
    • one year ago
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    "your derivative is just a little off" What should it be? "also 4-x^2=0 gives you x=2 or x=-2" Yeah, that makes sense. I forgot about that rule.

  7. freckles
    • one year ago
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    you should have used chain rule

  8. freckles
    • one year ago
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    \[((4-x^2)^\frac{1}{2})'=\frac{1}{2}(4-x^2)^{\frac{-1}{2}} \cdot (4-x^2)'\]

  9. anonymous
    • one year ago
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    I used the chain rule. I didn't add the exponent though...and I didn't make 2x negative. Oops! O_o

  10. anonymous
    • one year ago
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    Now that I know that. How could I find the critical points for that derivative?

  11. freckles
    • one year ago
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    \[f'(x)=\frac{-x}{\sqrt{4-x^2}}\] is this what you have for f'?

  12. anonymous
    • one year ago
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    Not even close unfortunately

  13. freckles
    • one year ago
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    you can find the critical numbers by finding when it is 0 or when it does not exist (these numbers should be in the domain of the original function though) so you still have to solve the equations x=0 and 4-x^2=0 so you are right your critical number are -2,0,2

  14. freckles
    • one year ago
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    wait what?

  15. freckles
    • one year ago
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    did you not understand how I got... \[((4-x^2)^\frac{1}{2})'=\frac{1}{2}(4-x^2)^{\frac{-1}{2}} \cdot (4-x^2)'\]

  16. anonymous
    • one year ago
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    Yes, I actually solved it wrongly, and some how managed to get the right values. This is how I originally solved it...

  17. anonymous
    • one year ago
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    \[\sqrt{4-x^2}\]\[(4-x^2)^\frac{1}{2}\]\[\frac{1}{2}(4-x^2)(2x)\]\[x(4-x^2)\] I didn't make the 2x negative nor did I remember to subtract 1 from my exponent (instead I forgot about it completely.

  18. anonymous
    • one year ago
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    Then I got... x=0 4-x^2=0 or x=2 , x=-2

  19. freckles
    • one year ago
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    Ok yeah but are you understand why this is \[f'(x)=\frac{-x}{\sqrt{4-x^2}}\] ?

  20. anonymous
    • one year ago
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    I know how you get to \[-x(4-x^2)^\frac{1}{2}\] is what you have just a simplified version of that?

  21. freckles
    • one year ago
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    well 1/2-1 is -1/2 not 1/2

  22. anonymous
    • one year ago
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    Sorry, typo

  23. freckles
    • one year ago
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    \[-x(4-x^2)^\frac{-1}{2}=\frac{-x}{(4-x^2)^\frac{1}{2}} \text{ by law of exponents }\]

  24. freckles
    • one year ago
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    \[x^{-n}=\frac{1}{x^n}\]

  25. anonymous
    • one year ago
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    There are so many laws I need to re-memorize.

  26. anonymous
    • one year ago
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    But yes I do recognize what you did.

  27. freckles
    • one year ago
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    anyways you have the right critical numbers x=-2,0,2 so we need to evaluate f(-2),f(0),f(1),f(2) I put f(1) because we need to evaluate it later anyways

  28. anonymous
    • one year ago
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    1?

  29. freckles
    • one year ago
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    \[f(x)=\sqrt{4-x^2} \\ f(-2)=\sqrt{4-(-2)^2}=\sqrt{4-4}=0 \\ f(0)=? \\ f(1)=? \\ f(2)=?\]

  30. freckles
    • one year ago
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    question d has 1 as an endpoint...

  31. anonymous
    • one year ago
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    Oh okay, I was just looking at a

  32. freckles
    • one year ago
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    I just wanted to get all the evaluation done with

  33. freckles
    • one year ago
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    we can look at a first ... I want to get through evaluating the function at the critical numbers and any endpoints

  34. freckles
    • one year ago
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    anyways can you evaluate those 3 above...

  35. anonymous
    • one year ago
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    x=2 would end up being 0 x=-1 would also be 0 x=0 would be 2 x=1 would be sqrt(3)

  36. anonymous
    • one year ago
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    That was supposed to say -2 not -1

  37. freckles
    • one year ago
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    f(-2)=0 f(0)=2 f(2)=0 f(1)=sqrt(3) great work

  38. freckles
    • one year ago
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    so look at the outputs above

  39. freckles
    • one year ago
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    0,2,0

  40. freckles
    • one year ago
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    what is biggest number there

  41. anonymous
    • one year ago
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    2

  42. freckles
    • one year ago
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    so that is your absolute max

  43. freckles
    • one year ago
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    0,2,0 and of these numbers 0 is the absolute min

  44. anonymous
    • one year ago
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    That makes sense, but how would I represent that max as coordinates?

  45. freckles
    • one year ago
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    which this makes sense... because the graph is the top part of a circle... |dw:1444500172873:dw|

  46. freckles
    • one year ago
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    well you already found the coordinates

  47. freckles
    • one year ago
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    f(x)=y means (x,y) is a coordinate on f

  48. freckles
    • one year ago
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    for example we said f(2)=0 so that means (2,0) is a coordinate on f

  49. anonymous
    • one year ago
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    Oh so max would be 0,2

  50. freckles
    • one year ago
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    (0,2) would be max ordered pairs have ( ) around them

  51. freckles
    • one year ago
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    |dw:1444500273762:dw|

  52. anonymous
    • one year ago
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    so could (-2,0) and (2,0) both be min values?

  53. freckles
    • one year ago
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    yep

  54. anonymous
    • one year ago
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    Because b has a parentheses instead of a bracket, would it be solved for differently?

  55. freckles
    • one year ago
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    now we can look at b we have [-2,0) we only care about f(-2) and f(0) here but we cannot actually include what happens at x=0 f(-2)=0 and f(0)=2 is what we found earlier of 0 and 2 the smallest is 0 so your absolute min is at x=0 and we will not include an absolute max for this question because we cannot include what happens at x=0

  56. anonymous
    • one year ago
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    Why can't we include 0?

  57. freckles
    • one year ago
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    [-2,0) means we have everything between -2 and 0 including -2 but not including 0

  58. freckles
    • one year ago
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    |dw:1444500569732:dw| this function on [-2,0) looks like: |dw:1444500587078:dw| my quarter of a circle was horribly drawn I blame it on the cat licking me

  59. anonymous
    • one year ago
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    [ means included ( means to that point but the point isn't included itself

  60. anonymous
    • one year ago
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    Correct?

  61. freckles
    • one year ago
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    see the absolute max doesn't actually exist there is a hole at x=0 because the domain didn't allow us to include what happens at x=0

  62. anonymous
    • one year ago
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    Ah okay

  63. anonymous
    • one year ago
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    and for C it would just be solving for 0?

  64. freckles
    • one year ago
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    brackets means we include the endpoint parenthesis means we don't include the endpoint

  65. freckles
    • one year ago
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    we already looked at a and said the absolute mins were (-2,0) and (2,0) and the absolute max was (0,2) c would be the same answer except the absolute mins would not exist since we have (-2,2) and not [-2,2]

  66. anonymous
    • one year ago
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    So only the max would exist in that particular problem.

  67. freckles
    • one year ago
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    yep

  68. anonymous
    • one year ago
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    Should i just state that it cannot be determined or is there a specific term I should use?

  69. freckles
    • one year ago
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    I think the word choice I used was good "does not exist" |dw:1444500908510:dw| this graph on [1,2) looks like: |dw:1444500919987:dw|

  70. anonymous
    • one year ago
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    Thank you so much for your help! :D

  71. freckles
    • one year ago
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    what did you get for b again

  72. anonymous
    • one year ago
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    max: does not exist min: (-2, 0)

  73. freckles
    • one year ago
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    great

  74. freckles
    • one year ago
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    and what about d

  75. anonymous
    • one year ago
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    just a second

  76. anonymous
    • one year ago
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    max: sqrt(3) min: does not exist That may or may not be right.

  77. freckles
    • one year ago
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    did you still want the coordinates or will the y value just be okay ?

  78. freckles
    • one year ago
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    like you can say the max is sqrt(3) but if your teacher prefers coordinates then you should say (1,sqrt(3))

  79. anonymous
    • one year ago
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    Yeah that's what I'll put

  80. freckles
    • one year ago
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    ok ok that sounds great to me

  81. anonymous
    • one year ago
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    excellent! It's surprising how simple this stuff really is compared to how difficult it seems at times

  82. freckles
    • one year ago
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    It is really not that bad... You do have to train yourself back on your algebra though.. The calculus I think is easier than the algebra :p

  83. anonymous
    • one year ago
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    I do the worst on trig identities personally

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