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lilly21 Group Title

does anyone own a TI-84 calculator and know how to do Optimization?

  • 2 years ago
  • 2 years ago

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  1. jim_thompson5910 Group Title
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    Whats your question

    • 2 years ago
  2. lilly21 Group Title
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    okay i have this problem where i have sinx=2/5 and i have to find critical points........and there is no sinx never ==2/5 so my question is how do i find the critical points using the TI-84????

    • 2 years ago
  3. jim_thompson5910 Group Title
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    so you need to find the critical points of sin(x) you mean?

    • 2 years ago
  4. lilly21 Group Title
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    yes!

    • 2 years ago
  5. jim_thompson5910 Group Title
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    first plot y = sin(x) do you know how to do that?

    • 2 years ago
  6. lilly21 Group Title
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    |dw:1352867080741:dw| somethin like that right?

    • 2 years ago
  7. jim_thompson5910 Group Title
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    yes good

    • 2 years ago
  8. lilly21 Group Title
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    :)

    • 2 years ago
  9. jim_thompson5910 Group Title
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    in this case there are infinitely many critical points (because there are infinitely many max/min points)

    • 2 years ago
  10. lilly21 Group Title
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    yes

    • 2 years ago
  11. jim_thompson5910 Group Title
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    what you can do is name a few of them (algebraically, you can list the rule to generate them all, but we won't worry about that right now) So what you do is hit 2nd, then trace. Then you hit either "minimum" or "maximum" depending on whether you want the min or max point. The first point after x = 0 is a max, so pick 4: maximum Then set up the left and right bounds to the left and right of the max point. Then make the best guess you can and hit enter

    • 2 years ago
  12. jim_thompson5910 Group Title
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    The calculator will then search through this interval you gave it (starting with the guess you gave it) to find the max point

    • 2 years ago
  13. satellite73 Group Title
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    not to butt in, but i am wondering what the actual question is the equation \(\sin(x)=\frac{2}{5}\) doesn't have critical points, it is just an equation

    • 2 years ago
  14. jim_thompson5910 Group Title
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    if that's the case (as satellite73 is saying), then you would just type in arcsin(2/5) or sin^-1(2/5) to get roughly 23.578 degrees Keep in mind that 180 - 23.578 = 156.422 degrees is also a solution (assuming you're restricted from 0 to 360 degrees)

    • 2 years ago
  15. lilly21 Group Title
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    orig equation.....find the absolute extrema of the fn on the closed interval g(x)=2x+5cosx [0, 2pi]

    • 2 years ago
  16. jim_thompson5910 Group Title
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    I'm assuming you're in calculus right?

    • 2 years ago
  17. lilly21 Group Title
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    yes sir

    • 2 years ago
  18. satellite73 Group Title
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    oh that explains \(\sin(x)=\frac{2}{5}\)!!

    • 2 years ago
  19. lilly21 Group Title
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    yeah :}

    • 2 years ago
  20. jim_thompson5910 Group Title
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    So first derive g(x) to get g'(x) g(x)=2x+5cos(x) g'(x)=2-5sin(x) Then plug in g'(x) = 0 and solve for x g'(x)=2-5sin(x) 0=2-5sin(x) sin(x) = 2/5 <--- I see how you got this now x = arcsin(2/5) x = 0.4115 or x = pi - 0.4115 x = 0.4115 or x = 2.7301 So the critical points are x = 0.4115 or x = 2.7301 The endpoints are x = 0 and x = 2pi Evaluate the function g(x) at each point g(x)=2x+5cos(x) g(0)=2(0)+5cos(0) g(0) = 5 g(x)=2x+5cos(x) g(0.4115)=2(0.4115)+5cos(0.4115) g(0.4115)=5.4056 g(x)=2x+5cos(x) g(2.7301)=2(2.7301)+5cos(2.7301) g(2.7301)=0.8776 g(x)=2x+5cos(x) g(2pi)=2(2pi)+5cos(2pi) g(2pi)=17.5664 So the min is g(x) = 0.8776 and that occurs when x = 2.7301 The max is g(x) = 17.5664 and that happens when x = 2pi

    • 2 years ago
  21. lilly21 Group Title
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    okay this is EXACTLY the right answer so ok for this part sin(x) = 2/5 <--- I see how you got this now x = arcsin(2/5) x = 0.4115 or x = pi - 0.4115 x = 0.4115 or x = 2.7301 what did u do on the calculator to get these 2 critical points???

    • 2 years ago
  22. Dido525 Group Title
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    You guys always write such through explanations :P . I wish I could get to your level one day : ) .

    • 2 years ago
  23. jim_thompson5910 Group Title
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    I hit 2nd ---> sin then typed 2/5, closed the parenthesis and hit enter This types in sin^-1(2/5)

    • 2 years ago
  24. Dido525 Group Title
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    @lilly21 : Look on a unit circle. Sin ALWAYS has solution ffrom 0 to pi.

    • 2 years ago
  25. jim_thompson5910 Group Title
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    Keep in mind that sin^-1 or arcsine is a function, so it's only going to spit out one answer. However, there are 2 solutions to sin(x) = 2/5 where 0 < x < pi as Dido525 is saying So that's why I'm subtracting that first result from pi.

    • 2 years ago
  26. Dido525 Group Title
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    0 to 2pi sorry.

    • 2 years ago
  27. lilly21 Group Title
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    it s ok :]

    • 2 years ago
  28. jim_thompson5910 Group Title
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    in this case, 0 < x < pi or 0 < x < 2pi gives the same number of solutions

    • 2 years ago
  29. Dido525 Group Title
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    Indeed. I just wanted to be Concise @jim_thompson5910 .

    • 2 years ago
  30. jim_thompson5910 Group Title
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    i gotcha

    • 2 years ago
  31. lilly21 Group Title
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    okay so let me get this straight..... .411 is subtracted from pi to get the 2nd critical point right.....:( and y is there a 2nd critical pnt?

    • 2 years ago
  32. jim_thompson5910 Group Title
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    if you look at the unit circle |dw:1352868694165:dw|

    • 2 years ago
  33. jim_thompson5910 Group Title
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    If the sine of some angle is positive, then we're only focusing on the upper half (since the positive y coordinates point to a positive sine value) There are 2 ways to generate a triangle in which the sine of the reference angle is 2/5 and they look like this |dw:1352868786503:dw|

    • 2 years ago
  34. jim_thompson5910 Group Title
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    In quadrant I, the sine of that angle theta is 2/5, but you can easily mirror it over (to quadrant II) and find another angle that gives you the same sine value

    • 2 years ago
  35. jim_thompson5910 Group Title
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    So that's why there are 2 solutions to sin(x) = 2/5 where 0 < x < 2pi

    • 2 years ago
  36. lilly21 Group Title
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    oooooooooooo okay i understand now ::}}}]]

    • 2 years ago
  37. jim_thompson5910 Group Title
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    ok great

    • 2 years ago
  38. lilly21 Group Title
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    okay so then to find abs extrema or min and max all i would do is plug back in the points i was able to get and find my min and max points right???

    • 2 years ago
  39. jim_thompson5910 Group Title
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    yes, once you find the critical points and the endpoints, you evaluate them all to see which ones give you the smallest and the largest outputs smallest output ---> min largest output ---> max

    • 2 years ago
  40. jim_thompson5910 Group Title
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    I should clarify smallest output ---> absolute min largest output ---> absolute max This is only true if you're working with a finite interval (ie the curve of the graph has a specific length)

    • 2 years ago
  41. lilly21 Group Title
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    okay gotchA!!!!!! thanks a million! :]

    • 2 years ago
  42. jim_thompson5910 Group Title
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    yw

    • 2 years ago
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