## mathmath333 one year ago Find the minimum and maximum value of

1. mathmath333

\large \color{black}{\begin{align} \dfrac{x^2-x+1}{x^2+x+1},\ x\in \mathbb{R}\hspace{.33em}\\~\\ \end{align}}

2. anonymous

It's hard to explain...but I will see if I can show the work.

3. anonymous

are you talking local or absolute maxes and mins?

4. mathmath333

i would also prefer a way without calculus ( idk calculus)

5. anonymous

Do you need to show your work for this problem?

6. mathmath333

not necessary i should understand the idea

7. anonymous

Based on what I know algebraically , I got a minimum of (1, 1/3) and a maximum of (-1,3).

8. mathmath333

u calculated the co-ordinates where as it should be the numerical value

9. anonymous

Ok. Let me try a different way.

10. anonymous

Ok. I got the minimum is 1/3, and the maximum is -3.

11. anonymous

oops +3. not neg

12. mathmath333

how did u calculate

13. anonymous

$\frac{x^2 - x + 1 }{ x^2 + x + 1 } = 3, when (x) = -1$ I got this after simplifying to figure out what x = for the numerator

14. anonymous

Same for the denominator, except it is when x=1

15. anonymous

It's not the way I was taught, but my aunt is a math teacher, so she taught me her way of solving these pesky sorts

16. xapproachesinfinity

eh i like your questions lol

17. xapproachesinfinity

make me struggle :)

18. mathmath333

i have tons of question but no time

19. xapproachesinfinity

i cheated haha min=1/3<y<3 let's how we get to this

20. mathmath333

one way seems to put random $$x=\pm 0, \pm 1,\pm 2, \pm 3$$ but doesnt work always http://www.wolframalpha.com/input/?i=table%5Bf%28x%29%3D%5Cdfrac%7Bx%5E2-x%2B1%7D%7Bx%5E2%2Bx%2B1%7D%2C%7Bx%2C-10%2C10%7D%5D

21. Empty

Honestly just graphing this is the best way to do this lol.

22. Empty

That's why Descartes invented graphing in the first place, to make stuff like this easier. And then we got computers, lol.

23. xapproachesinfinity

i suggest that we study $y=1+\frac{2x}{x^2+x+1} ~~x\in \mathbb{R}$

24. xapproachesinfinity

minus not plus mistake haha

25. xapproachesinfinity

$y=1-\frac{2x}{x^2+x+1}$

26. xapproachesinfinity

hmm need to find upper and lower bound m<y<M some how i think that finding the range of x for 2x/x^2+x+1 might be our way!!!

27. xapproachesinfinity

I'm just throwing stuff as i go lol

28. xapproachesinfinity

x^2+x+1=0 has no real solution yes? so this parabola is >0

29. Empty

We can factor it and for just simplicity of writing I'll make this: $\omega = e^{i 2 \pi / 3}$ $\frac{x^2-x+1}{x^2+x+1}=\frac{(x+\omega)(x+\omega^2)}{(x-\omega)(x-\omega^2)}$

30. xapproachesinfinity

y1=2x/x^2+x+1 can we find the range for this going by what i just did?

31. anonymous

let $$x=\tan z$$ with $$z \in (-\frac{\pi}{2}, \frac{\pi}{2})$$ you will get$f(\tan z)=\frac{\tan^2 z -\tan z+1}{\tan^2 z +\tan z+1}=\frac{4}{2+\sin (2z)}-1$

32. xapproachesinfinity

hmm interesting @mukushla

33. xapproachesinfinity

i feel that my way can be a lot easier haha though don't you think

34. xapproachesinfinity

just need to find the range for 2x/(x^2+x+1)?

35. anonymous

yeah, making a perfect square in denom, that's a lot better than mine

36. xapproachesinfinity

need pen and paper!

37. mathmath333

i like that tan substitution looks something creative

38. anonymous

let me continue with yours$f(x)=1-\frac{2x}{x^2+x+1}=1-\frac{2}{x+\frac{1}{x}+1}$actually there is no need to make perfect square

39. anonymous

we just need to note that$\left|x+\frac{1}{x} \right| \ge 2$

40. xapproachesinfinity

hmm seems you a different approach

41. xapproachesinfinity

i was trying to use parabola and eyeball the range

42. xapproachesinfinity

how is |x+1/x|>=2

43. xapproachesinfinity

first time i see that :)

44. anonymous

AM-GM inequality$x+\frac{1}{x} \ge 2\sqrt{x.\frac{1}{x}}$

45. anonymous

if $$x<0$$ it will be$x+\frac{1}{x} \le -2$

46. xapproachesinfinity

hmm i see

47. anonymous

equality occurs when x=-1 and x=1 for two cases

48. xapproachesinfinity

so for x=-1 we have x+1/x<-2==> x+1/x+1<-1 then y<3

49. xapproachesinfinity

for x=1 by the same reasoning 1/3<y

50. xapproachesinfinity

excellent @mukushla

51. Empty

That's a nice trick with that AM-GM inequality.

52. xapproachesinfinity

beautiful indeed :)

53. anonymous

@mathmath333 ty, actually it is a nice trick to use trig sub for finding range, you may try this one$f(x)=\sqrt{x-2}+\sqrt{4-x}$

54. mathmath333

thats incorrect

55. xapproachesinfinity

hmm if one realize they can use trig sub the problem becomes really easy since sin and cos are easily bounded

56. anonymous

that's right

57. mathmath333

the range of x here is $$2<x<4$$

58. anonymous

when you wanna do a trig sub, find the domain, it will give you the clue for the form of sub

59. xapproachesinfinity

well domain above is R how can that help

60. anonymous

tan covers R

61. xapproachesinfinity

but tan has pi/2 singularity?

62. xapproachesinfinity

does not matter ?

63. anonymous

no we will treat bounds in limit, so it doesn't matter

64. anonymous

now, what about the second function?

65. anonymous

note that$2 \le x \le4$is a restricted domain

66. xapproachesinfinity

67. xapproachesinfinity

so what sub is that telling us to use?

68. xapproachesinfinity

sin or cos ?

69. xapproachesinfinity

hold on don't tell me yet

70. anonymous

both of then can be

71. anonymous

you know why?

72. xapproachesinfinity

hmm no really because they are both have same bounds -1 and 1

73. anonymous

yes and because$-1\le x-3 \le 1$

74. xapproachesinfinity

oh i see

75. Empty

They're both the same function just shifted argument by 90 degrees. $$\cos x + 3$$ will work just as well as $$\sin x + 3$$

76. anonymous

quite right

77. xapproachesinfinity

so in that we put x-3=sinz yes?

78. xapproachesinfinity

damn i feel weak in trig lol :)

79. anonymous

oh yeah, $$3+\cos z$$ will be better maybe

80. xapproachesinfinity

hmm you said they all work fine No? okay lets go with x=3+cosz

81. anonymous

well derive with cos will be easier

82. xapproachesinfinity

so $y=\sqrt{\cos z+1}+\sqrt{1-\cos z}$

83. xapproachesinfinity

so $0\leq y \leq 2\sqrt{2}$

84. anonymous

oh no you can't just plug in there values, your trig function is not alone, you'll need more simplification

85. xapproachesinfinity

if we simplify we just get radical 2?

86. xapproachesinfinity

if i didn't make an error that is

87. anonymous

$f=\sqrt{2} \left( \left|\sin \frac{z}{2} \right|+\left|\cos \frac{z}{2} \right| \right)$

88. xapproachesinfinity

cosz=2cos^2x/2-1 oh i see i threw the square when i did my simplification haha

89. xapproachesinfinity

i didn't i mean*

90. anonymous

ok, good discussion, tnx guys

91. xapproachesinfinity

well isn't that the same result above? 0<y<2root2

92. xapproachesinfinity

thanks to you learned bunch of stuff here :)

93. xapproachesinfinity

this one i never remember $\frac{a+b}{2}\leq \sqrt{ab}$