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

Ishaan94Best ResponseYou've already chosen the best response.0
Oh okay, then I think I must be right.
 2 years ago

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

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

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

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

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

ChrisVBest ResponseYou've already chosen the best response.0
thats what i need solved
 2 years ago

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

ChrisVBest ResponseYou've already chosen the best response.0
which comes out to 0 for me
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
so i assume i did something wrong
 2 years ago

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

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

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

ChrisVBest ResponseYou've already chosen the best response.0
thats what i was thinking
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
but the relative minimum given by the book is
 2 years ago

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

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

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

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

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

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

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

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

ChrisVBest ResponseYou've already chosen the best response.0
then i find the derivative
 2 years ago

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

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

ChrisVBest ResponseYou've already chosen the best response.0
thats the first derivative
 2 years ago

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

ChrisVBest ResponseYou've already chosen the best response.0
0 is a vertical asymtope so its not a min or max
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
but by the first derivative test
 2 years ago

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

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

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

ChrisVBest ResponseYou've already chosen the best response.0
but since crit numbers = 0 or undefined
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
you set the numerator to 0 to find crit numbers
 2 years ago

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

ChrisVBest ResponseYou've already chosen the best response.0
hince the 3 critical numbers
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
so this far i am correct
 2 years ago

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

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

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

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

ChrisVBest ResponseYou've already chosen the best response.0
well common denominator would be x
 2 years ago

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

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

DiyadiyaBest ResponseYou've already chosen the best response.4
But 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
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
that just combining via common denominator
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
oh wait that is the derivative
 2 years ago

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

ChrisVBest ResponseYou've already chosen the best response.0
not the first derivative
 2 years ago

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

CallistoBest ResponseYou've already chosen the best response.1
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)?
 2 years ago

ChrisVBest ResponseYou've already chosen the best response.0
both of those are the same
 2 years ago

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

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

CallistoBest ResponseYou've already chosen the best response.1
(Nah.. seems problem's solved)
 2 years ago

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

ChrisVBest ResponseYou've already chosen the best response.0
yea just making sure i understand
 2 years ago

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

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

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

ChrisVBest ResponseYou've already chosen the best response.0
you guys were a great help thanks
 2 years ago

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

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