## ChrisV 3 years ago Solve. f(x)=2x^2-1/x ; x=sqrt2/2

1. ChrisV

$2x^2-1/\sqrt{x} ; x=\sqrt{2/2}$

2. Ishaan94

1 -sqrt2

3. Ishaan94

that makes x=1

4. ChrisV

oops$x=\sqrt{2}/2$

5. Ishaan94

Oh okay, then I think I must be right.

6. ChrisV

well the book says the answer is $2\sqrt{2}$

7. ChrisV

but i dont get that

$2( \frac{1}{\sqrt{2}})^2- \frac{1}{1/\sqrt{2}}=1-\sqrt{2}$

9. ChrisV

well would help if i gave you guys the right equation

10. ChrisV

$2x^2-1/x$

Yes still the answer is 1- sqqrt2

12. ChrisV

$(2(\sqrt{2}/2)^2-1)/\sqrt{2}/2$

13. ChrisV

thats what i need solved

14. ChrisV

$(4(\sqrt{2}/2)-2)/\sqrt{2}$

15. ChrisV

oops squared

16. ChrisV

$4(2/4)-2/\sqrt{2}$

17. ChrisV

which comes out to 0 for me

18. ChrisV

so i assume i did something wrong

oh ok $\frac{(2x)^2-1}{x}$$\frac{(2 \times \frac{1}{\sqrt{2}})^2-1}{ \frac{1}{\sqrt{2}}}$$(4 \times 1/2 \ -1)\sqrt{2}$$=\sqrt{2}$

20. ChrisV

its only the x squared

21. ChrisV

not 2x

$\frac{2(x)^2-1}{x}$is this the equation?

23. ChrisV

yes

It can't be because it gives zero unless zero is the answer

25. ChrisV

thats what i was thinking

26. ChrisV

but the relative minimum given by the book is

27. ChrisV

$(\sqrt{2}/2, 2\sqrt{2})$

28. ChrisV

which means the f(sqrt2/2) somehow is not 0

Wait $x=\frac{\sqrt{2}}{2} \ or \ \frac{2}{\sqrt{2}}$

30. ChrisV

was 1/sqrt2 but rationalized is sqrt2/2

32. ChrisV

ok ill go through the entire problem with you

ok

wait

35. ChrisV

i have to find the critical numbers of f(x)=2x+1/x using the first derivative test

Ok then its different

37. ChrisV

so first critical number is 0 because f(x) is undefinedat 0

38. ChrisV

then i find the derivative

Wait

41. ChrisV

oh im sure i got this part right

42. ChrisV

2x/=1/x=(2x^2+1)/x

43. ChrisV

first derivative

44. ChrisV

(x(4x)-(2x^2+1))/X^2

45. ChrisV

(4x^2-2x*2-1)/x^2

46. ChrisV

(2x^2-1)/x^2

47. ChrisV

thats the first derivative

48. ChrisV

so 2x-1=0

49. ChrisV

2x^2-1=0 oops

50. ChrisV

x^2=1/2

51. ChrisV

x=+/-1/sqrt2

52. ChrisV

so critical numbers = 0,-sqrt2/2, and sqrt2/2

53. ChrisV

0 is a vertical asymtope so its not a min or max

54. ChrisV

but by the first derivative test

55. ChrisV

i can see the interval ( -infinity, -sqrt2/2) is increasing which tells me -sqrt2/2 is the max

i don't understand this ..sorry But derivative of 2x+1/x=(2x^2-1)/x^2 right?

57. ChrisV

the interval (sqrt2/2, infinty) is decreasing which means minimum

58. ChrisV

yes it is

59. ChrisV

but since crit numbers = 0 or undefined

60. ChrisV

you set the numerator to 0 to find crit numbers

61. ChrisV

and since x=0 would make the denominator 0 that would be undefined

62. ChrisV

hince the 3 critical numbers

63. ChrisV

so this far i am correct

64. ChrisV

now i need the y values for my max and min

65. ChrisV

which means i plug my crit numbers into f(x)

66. ChrisV

so i have to solve for f(sqrt2/2) and f(-sqrt2/2)

67. ChrisV

f(x)= 2x+1/x

Ok so$2x+ \frac{1}{x}= 2 \frac{1}{\sqrt{2}}+\frac{1}{1/\sqrt{2}}=\sqrt{2}+\sqrt{2}=2\sqrt{2}$

69. ChrisV

which equals (2x^2-1)/x

70. ChrisV

well common denominator would be x

71. ChrisV

so 2x(x)/x +1/x

72. ChrisV

(2x^2+1)/x

73. ChrisV

now plug in $x=\sqrt{2}/2; x=-\sqrt{2}/2$

74. ChrisV

and the answers in the book i verified via cramster and calcchat.com

But in your question f(x) was $2x+ \frac{1}{x}$thats how you found derivative to be $\frac{2x^2-1}{x^2}$ i'm sorry if i'm wrong i have no idea about this i'm just trying

76. ChrisV

no those two are equal

77. ChrisV

that just combining via common denominator

78. ChrisV

oh wait that is the derivative

79. ChrisV

but we plug the crit numbers back into f(x) to find y

80. ChrisV

not the first derivative

Yes when we plug into f(x)$2x+ \frac{1}{x}= 2 \frac{1}{\sqrt{2}}+\frac{1}{1/\sqrt{2}}=\sqrt{2}+\sqrt{2}=2\sqrt{2}$

82. Callisto

I'm so confused... Is f(x) = (2x^2 -1)/x or f(x) = (2x^2 +1)/x or f(x) = 2x^2 + (1/x) or f(x) = 2x^2 - (1/x)?

83. ChrisV

and f(x) = (2x^2+1)/x

84. ChrisV

or 2x+1/x

85. ChrisV

both of those are the same

ok so when we plug in 1/sqrt2 to f(x) we get 2sqrt2

87. ChrisV

ok

88. ChrisV

2(1/sqrt2)+1/1/sqrt2

89. ChrisV

2/2sqrt +sqrt2

90. ChrisV

2sqrt2/2+sqrt2

91. ChrisV

=2sqrt2

92. ChrisV

thats right thanks

welcome :)

94. ChrisV

and now the negative

95. ChrisV

must equal -2sqrt2

96. Callisto

It's so messy here ... I dunno if this is sth you want 'cause i don't even know the question

97. Callisto

(Nah.. seems problem's solved)

98. ChrisV

so

yup -2sqrt2

100. ChrisV

2(-1/sqrt2) +1/-1/sqrt2

@Callisto has given you the entire solution :)

102. ChrisV

-2/sqrt2 -sqrt2

103. ChrisV

2sqrt2/2 - sqrt2

104. ChrisV

-2sqrt2

105. ChrisV

yea just making sure i understand

yeah :)

107. ChrisV

i have a test in a couple of days

108. ChrisV

i have a test in a couple of days

109. ChrisV

not the kind of person to have people do my homework for me :)

110. ChrisV

you guys were a great help thanks

112. Callisto

lol that 'solution' serves as a reference only.. I don' t even know if it is correct, BTW, Good luck with your test :)

Yeah Goodluck!

114. ChrisV

no problem, i am sure what i got now is correct. thanks guys