## lilly21 Group Title does anyone own a TI-84 calculator and know how to do Optimization? 2 years ago 2 years ago

1. jim_thompson5910

2. lilly21

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????

3. jim_thompson5910

so you need to find the critical points of sin(x) you mean?

4. lilly21

yes!

5. jim_thompson5910

first plot y = sin(x) do you know how to do that?

6. lilly21

|dw:1352867080741:dw| somethin like that right?

7. jim_thompson5910

yes good

8. lilly21

:)

9. jim_thompson5910

in this case there are infinitely many critical points (because there are infinitely many max/min points)

10. lilly21

yes

11. jim_thompson5910

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

12. jim_thompson5910

The calculator will then search through this interval you gave it (starting with the guess you gave it) to find the max point

13. satellite73

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

14. jim_thompson5910

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)

15. lilly21

orig equation.....find the absolute extrema of the fn on the closed interval g(x)=2x+5cosx [0, 2pi]

16. jim_thompson5910

I'm assuming you're in calculus right?

17. lilly21

yes sir

18. satellite73

oh that explains $$\sin(x)=\frac{2}{5}$$!!

19. lilly21

yeah :}

20. jim_thompson5910

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

21. lilly21

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???

22. Dido525

You guys always write such through explanations :P . I wish I could get to your level one day : ) .

23. jim_thompson5910

I hit 2nd ---> sin then typed 2/5, closed the parenthesis and hit enter This types in sin^-1(2/5)

24. Dido525

@lilly21 : Look on a unit circle. Sin ALWAYS has solution ffrom 0 to pi.

25. jim_thompson5910

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.

26. Dido525

0 to 2pi sorry.

27. lilly21

it s ok :]

28. jim_thompson5910

in this case, 0 < x < pi or 0 < x < 2pi gives the same number of solutions

29. Dido525

Indeed. I just wanted to be Concise @jim_thompson5910 .

30. jim_thompson5910

i gotcha

31. lilly21

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?

32. jim_thompson5910

if you look at the unit circle |dw:1352868694165:dw|

33. jim_thompson5910

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|

34. jim_thompson5910

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

35. jim_thompson5910

So that's why there are 2 solutions to sin(x) = 2/5 where 0 < x < 2pi

36. lilly21

oooooooooooo okay i understand now ::}}}]]

37. jim_thompson5910

ok great

38. lilly21

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???

39. jim_thompson5910

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

40. jim_thompson5910

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)

41. lilly21

okay gotchA!!!!!! thanks a million! :]

42. jim_thompson5910

yw