Question

Solve. f(x)=2x^2-1/x ; x=sqrt2/2

Answer 1

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

Answer 2

1 -sqrt2

Answer 3

that makes x=1

Answer 4

oops\[x=\sqrt{2}/2\]

Answer 5

Oh okay, then I think I must be right.

Answer 6

well the book says the answer is \[2\sqrt{2}\]

Answer 7

but i dont get that

Answer 8

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

Answer 9

well would help if i gave you guys the right equation

Answer 10

\[2x^2-1/x\]

Answer 11

Yes still the answer is 1- sqqrt2

Answer 12

\[(2(\sqrt{2}/2)^2-1)/\sqrt{2}/2\]

Answer 13

thats what i need solved

Answer 14

\[(4(\sqrt{2}/2)-2)/\sqrt{2}\]

Answer 15

oops squared

Answer 16

\[4(2/4)-2/\sqrt{2}\]

Answer 17

which comes out to 0 for me

Answer 18

so i assume i did something wrong

Answer 19

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

Answer 20

its only the x squared

Answer 21

not 2x

Answer 22

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

Answer 23

yes

Answer 24

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

Answer 25

thats what i was thinking

Answer 26

but the relative minimum given by the book is

Answer 27

\[(\sqrt{2}/2, 2\sqrt{2})\]

Answer 28

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

Answer 29

Wait \[x=\frac{\sqrt{2}}{2} \ or \ \frac{2}{\sqrt{2}}\]

Answer 30

was 1/sqrt2 but rationalized is sqrt2/2

Answer 31

i'm getting the same answer
sorry!
once again check your question

Answer 32

ok ill go through the entire problem with you

Answer 33

ok

Answer 34

wait

Answer 35

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

Answer 36

Ok then its different

Answer 37

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

Answer 38

then i find the derivative

Answer 39

Wait

Answer 40

i dont know much about this ..relative minimum

Answer 41

oh im sure i got this part right

Answer 42

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

Answer 43

first derivative

Answer 44

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

Answer 45

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

Answer 46

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

Answer 47

thats the first derivative

Answer 48

so 2x-1=0

Answer 49

2x^2-1=0 oops

Answer 50

x^2=1/2

Answer 51

x=+/-1/sqrt2

Answer 52

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

Answer 53

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

Answer 54

but by the first derivative test

Answer 55

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

Answer 56

i don't understand this ..sorry

But derivative of 2x+1/x=(2x^2-1)/x^2 right?

Answer 57

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

Answer 58

yes it is

Answer 59

but since crit numbers = 0 or undefined

Answer 60

you set the numerator to 0 to find crit numbers

Answer 61

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

Answer 62

hince the 3 critical numbers

Answer 63

so this far i am correct

Answer 64

now i need the y values for my max and min

Answer 65

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

Answer 66

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

Answer 67

f(x)= 2x+1/x

Answer 68

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

Answer 69

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

Answer 70

well common denominator would be x

Answer 71

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

Answer 72

(2x^2+1)/x

Answer 73

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

Answer 74

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

Answer 75

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

Answer 76

no those two are equal

Answer 77

that just combining via common denominator

Answer 78

oh wait that is the derivative

Answer 79

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

Answer 80

not the first derivative

Answer 81

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

Answer 82

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)?

Answer 83

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

Answer 84

or 2x+1/x

Answer 85

both of those are the same

Answer 86

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

Answer 87

ok

Answer 88

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

Answer 89

2/2sqrt +sqrt2

Answer 90

2sqrt2/2+sqrt2

Answer 91

=2sqrt2

Answer 92

thats right thanks

Answer 93

welcome :)

Answer 94

and now the negative

Answer 95

must equal -2sqrt2

Answer 96

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

Answer 97

(Nah.. seems problem's solved)