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

- anonymous

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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

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

- anonymous

1 -sqrt2

- anonymous

that makes x=1

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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

- anonymous

Oh okay, then I think I must be right.

- anonymous

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

- anonymous

but i dont get that

- Diyadiya

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

- anonymous

well would help if i gave you guys the right equation

- anonymous

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

- Diyadiya

Yes still the answer is 1- sqqrt2

- anonymous

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

- anonymous

thats what i need solved

- anonymous

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

- anonymous

oops squared

- anonymous

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

- anonymous

which comes out to 0 for me

- anonymous

so i assume i did something wrong

- 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

its only the x squared

- anonymous

not 2x

- Diyadiya

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

- anonymous

yes

- Diyadiya

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

- anonymous

thats what i was thinking

- anonymous

but the relative minimum given by the book is

- anonymous

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

- anonymous

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

- Diyadiya

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

- anonymous

was 1/sqrt2 but rationalized is sqrt2/2

- Diyadiya

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

- anonymous

ok ill go through the entire problem with you

- Diyadiya

ok

- Diyadiya

wait

- anonymous

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

- Diyadiya

Ok then its different

- anonymous

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

- anonymous

then i find the derivative

- Diyadiya

Wait

- Diyadiya

i dont know much about this ..relative minimum

- anonymous

oh im sure i got this part right

- anonymous

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

- anonymous

first derivative

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

thats the first derivative

- anonymous

so 2x-1=0

- anonymous

2x^2-1=0 oops

- anonymous

x^2=1/2

- anonymous

x=+/-1/sqrt2

- anonymous

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

- anonymous

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

- anonymous

but by the first derivative test

- anonymous

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

- Diyadiya

i don't understand this ..sorry
But derivative of 2x+1/x=(2x^2-1)/x^2 right?

- anonymous

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

- anonymous

yes it is

- anonymous

but since crit numbers = 0 or undefined

- anonymous

you set the numerator to 0 to find crit numbers

- anonymous

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

- anonymous

hince the 3 critical numbers

- anonymous

so this far i am correct

- anonymous

now i need the y values for my max and min

- anonymous

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

- anonymous

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

- anonymous

f(x)= 2x+1/x

- 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

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

- anonymous

well common denominator would be x

- anonymous

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

- anonymous

(2x^2+1)/x

- anonymous

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

- anonymous

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

- 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

no those two are equal

- anonymous

that just combining via common denominator

- anonymous

oh wait that is the derivative

- anonymous

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

- anonymous

not the first derivative

- 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

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

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

- anonymous

or 2x+1/x

- anonymous

both of those are the same

- Diyadiya

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

- anonymous

ok

- anonymous

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

- anonymous

2/2sqrt +sqrt2

- anonymous

2sqrt2/2+sqrt2

- anonymous

=2sqrt2

- anonymous

thats right thanks

- Diyadiya

welcome :)

- anonymous

and now the negative

- anonymous

must equal -2sqrt2

- 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

(Nah.. seems problem's solved)

- anonymous

so

- Diyadiya

yup -2sqrt2

- anonymous

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

- Diyadiya

@Callisto has given you the entire solution :)

- anonymous

-2/sqrt2 -sqrt2

- anonymous

2sqrt2/2 - sqrt2

- anonymous

-2sqrt2

- anonymous

yea just making sure i understand

- Diyadiya

yeah :)

- anonymous

i have a test in a couple of days

- anonymous

i have a test in a couple of days

- anonymous

not the kind of person to have people do my homework for me :)

- anonymous

you guys were a great help thanks

- Diyadiya

glad to help !

- 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

Yeah Goodluck!

- anonymous

no problem, i am sure what i got now is correct. thanks guys

Looking for something else?

Not the answer you are looking for? Search for more explanations.