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anonymous
 4 years ago
Solve. f(x)=2x^21/x ; x=sqrt2/2
anonymous
 4 years ago
Solve. f(x)=2x^21/x ; x=sqrt2/2

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[2x^21/\sqrt{x} ; x=\sqrt{2/2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh okay, then I think I must be right.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well the book says the answer is \[2\sqrt{2}\]

Diyadiya
 4 years ago
Best ResponseYou've already chosen the best response.4\[2( \frac{1}{\sqrt{2}})^2 \frac{1}{1/\sqrt{2}}=1\sqrt{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well would help if i gave you guys the right equation

Diyadiya
 4 years ago
Best ResponseYou've already chosen the best response.4Yes still the answer is 1 sqqrt2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(2(\sqrt{2}/2)^21)/\sqrt{2}/2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thats what i need solved

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(4(\sqrt{2}/2)2)/\sqrt{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[4(2/4)2/\sqrt{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0which comes out to 0 for me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so i assume i did something wrong

Diyadiya
 4 years ago
Best ResponseYou've already chosen the best response.4oh ok \[\frac{(2x)^21}{x}\]\[\frac{(2 \times \frac{1}{\sqrt{2}})^21}{ \frac{1}{\sqrt{2}}}\]\[(4 \times 1/2 \ 1)\sqrt{2}\]\[=\sqrt{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0its only the x squared

Diyadiya
 4 years ago
Best ResponseYou've already chosen the best response.4\[\frac{2(x)^21}{x}\]is this the equation?

Diyadiya
 4 years ago
Best ResponseYou've already chosen the best response.4It can't be because it gives zero unless zero is the answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thats what i was thinking

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but the relative minimum given by the book is

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[(\sqrt{2}/2, 2\sqrt{2})\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0which means the f(sqrt2/2) somehow is not 0

Diyadiya
 4 years ago
Best ResponseYou've already chosen the best response.4Wait \[x=\frac{\sqrt{2}}{2} \ or \ \frac{2}{\sqrt{2}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0was 1/sqrt2 but rationalized is sqrt2/2

Diyadiya
 4 years ago
Best ResponseYou've already chosen the best response.4i'm getting the same answer sorry! once again check your question

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok ill go through the entire problem with you

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i have to find the critical numbers of f(x)=2x+1/x using the first derivative test

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so first critical number is 0 because f(x) is undefinedat 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then i find the derivative

Diyadiya
 4 years ago
Best ResponseYou've already chosen the best response.4i dont know much about this ..relative minimum

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh im sure i got this part right

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thats the first derivative

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so critical numbers = 0,sqrt2/2, and sqrt2/2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.00 is a vertical asymtope so its not a min or max

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but by the first derivative test

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i can see the interval ( infinity, sqrt2/2) is increasing which tells me sqrt2/2 is the max

Diyadiya
 4 years ago
Best ResponseYou've already chosen the best response.4i don't understand this ..sorry But derivative of 2x+1/x=(2x^21)/x^2 right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the interval (sqrt2/2, infinty) is decreasing which means minimum

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but since crit numbers = 0 or undefined

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you set the numerator to 0 to find crit numbers

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and since x=0 would make the denominator 0 that would be undefined

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hince the 3 critical numbers

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so this far i am correct

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now i need the y values for my max and min

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0which means i plug my crit numbers into f(x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so i have to solve for f(sqrt2/2) and f(sqrt2/2)

Diyadiya
 4 years ago
Best ResponseYou've already chosen the best response.4Ok so\[2x+ \frac{1}{x}= 2 \frac{1}{\sqrt{2}}+\frac{1}{1/\sqrt{2}}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0which equals (2x^21)/x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well common denominator would be x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now plug in \[x=\sqrt{2}/2; x=\sqrt{2}/2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and the answers in the book i verified via cramster and calcchat.com

Diyadiya
 4 years ago
Best ResponseYou've already chosen the best response.4But in your question f(x) was \[2x+ \frac{1}{x}\]thats how you found derivative to be \[\frac{2x^21}{x^2}\] i'm sorry if i'm wrong i have no idea about this i'm just trying

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no those two are equal

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that just combining via common denominator

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh wait that is the derivative

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but we plug the crit numbers back into f(x) to find y

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0not the first derivative

Diyadiya
 4 years ago
Best ResponseYou've already chosen the best response.4Yes when we plug into f(x)\[2x+ \frac{1}{x}= 2 \frac{1}{\sqrt{2}}+\frac{1}{1/\sqrt{2}}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\]

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.1I'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)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0and f(x) = (2x^2+1)/x

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0both of those are the same

Diyadiya
 4 years ago
Best ResponseYou've already chosen the best response.4ok so when we plug in 1/sqrt2 to f(x) we get 2sqrt2

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.1It's so messy here ... I dunno if this is sth you want 'cause i don't even know the question

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.1(Nah.. seems problem's solved)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.02(1/sqrt2) +1/1/sqrt2

Diyadiya
 4 years ago
Best ResponseYou've already chosen the best response.4@Callisto has given you the entire solution :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yea just making sure i understand

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i have a test in a couple of days

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i have a test in a couple of days

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0not the kind of person to have people do my homework for me :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you guys were a great help thanks

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.1lol that 'solution' serves as a reference only.. I don' t even know if it is correct, BTW, Good luck with your test :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no problem, i am sure what i got now is correct. thanks guys
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