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

Mathematics
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\[2x^2-1/\sqrt{x} ; x=\sqrt{2/2}\]
1 -sqrt2
that makes x=1

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Other answers:

oops\[x=\sqrt{2}/2\]
Oh okay, then I think I must be right.
well the book says the answer is \[2\sqrt{2}\]
but i dont get that
\[2( \frac{1}{\sqrt{2}})^2- \frac{1}{1/\sqrt{2}}=1-\sqrt{2}\]
well would help if i gave you guys the right equation
\[2x^2-1/x\]
Yes still the answer is 1- sqqrt2
\[(2(\sqrt{2}/2)^2-1)/\sqrt{2}/2\]
thats what i need solved
\[(4(\sqrt{2}/2)-2)/\sqrt{2}\]
oops squared
\[4(2/4)-2/\sqrt{2}\]
which comes out to 0 for me
so i assume i did something wrong
oh ok \[\frac{(2x)^2-1}{x}\]\[\frac{(2 \times \frac{1}{\sqrt{2}})^2-1}{ \frac{1}{\sqrt{2}}}\]\[(4 \times 1/2 \ -1)\sqrt{2}\]\[=\sqrt{2}\]
its only the x squared
not 2x
\[\frac{2(x)^2-1}{x}\]is this the equation?
yes
It can't be because it gives zero unless zero is the answer
thats what i was thinking
but the relative minimum given by the book is
\[(\sqrt{2}/2, 2\sqrt{2})\]
which means the f(sqrt2/2) somehow is not 0
Wait \[x=\frac{\sqrt{2}}{2} \ or \ \frac{2}{\sqrt{2}}\]
was 1/sqrt2 but rationalized is sqrt2/2
i'm getting the same answer sorry! once again check your question
ok ill go through the entire problem with you
ok
wait
i have to find the critical numbers of f(x)=2x+1/x using the first derivative test
Ok then its different
so first critical number is 0 because f(x) is undefinedat 0
then i find the derivative
Wait
i dont know much about this ..relative minimum
oh im sure i got this part right
2x/=1/x=(2x^2+1)/x
first derivative
(x(4x)-(2x^2+1))/X^2
(4x^2-2x*2-1)/x^2
(2x^2-1)/x^2
thats the first derivative
so 2x-1=0
2x^2-1=0 oops
x^2=1/2
x=+/-1/sqrt2
so critical numbers = 0,-sqrt2/2, and sqrt2/2
0 is a vertical asymtope so its not a min or max
but by the first derivative test
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?
the interval (sqrt2/2, infinty) is decreasing which means minimum
yes it is
but since crit numbers = 0 or undefined
you set the numerator to 0 to find crit numbers
and since x=0 would make the denominator 0 that would be undefined
hince the 3 critical numbers
so this far i am correct
now i need the y values for my max and min
which means i plug my crit numbers into f(x)
so i have to solve for f(sqrt2/2) and f(-sqrt2/2)
f(x)= 2x+1/x
Ok so\[2x+ \frac{1}{x}= 2 \frac{1}{\sqrt{2}}+\frac{1}{1/\sqrt{2}}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\]
which equals (2x^2-1)/x
well common denominator would be x
so 2x(x)/x +1/x
(2x^2+1)/x
now plug in \[x=\sqrt{2}/2; x=-\sqrt{2}/2\]
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
no those two are equal
that just combining via common denominator
oh wait that is the derivative
but we plug the crit numbers back into f(x) to find y
not the first derivative
Yes when we plug into f(x)\[2x+ \frac{1}{x}= 2 \frac{1}{\sqrt{2}}+\frac{1}{1/\sqrt{2}}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\]
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)?
and f(x) = (2x^2+1)/x
or 2x+1/x
both of those are the same
ok so when we plug in 1/sqrt2 to f(x) we get 2sqrt2
ok
2(1/sqrt2)+1/1/sqrt2
2/2sqrt +sqrt2
2sqrt2/2+sqrt2
=2sqrt2
thats right thanks
welcome :)
and now the negative
must equal -2sqrt2
It's so messy here ... I dunno if this is sth you want 'cause i don't even know the question
1 Attachment
(Nah.. seems problem's solved)
so
yup -2sqrt2
2(-1/sqrt2) +1/-1/sqrt2
@Callisto has given you the entire solution :)
-2/sqrt2 -sqrt2
2sqrt2/2 - sqrt2
-2sqrt2
yea just making sure i understand
yeah :)
i have a test in a couple of days
i have a test in a couple of days
not the kind of person to have people do my homework for me :)
you guys were a great help thanks
glad to help !
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!
no problem, i am sure what i got now is correct. thanks guys

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