anonymous
  • anonymous
Solve. f(x)=2x^2-1/x ; x=sqrt2/2
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
\[2x^2-1/\sqrt{x} ; x=\sqrt{2/2}\]
anonymous
  • anonymous
1 -sqrt2
anonymous
  • anonymous
that makes x=1

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

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