At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

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

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the **expert** answer you'll need to create a **free** account at **Brainly**

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

1 -sqrt2

that makes x=1

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

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

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

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

(Nah.. seems problem's solved)

so

yup -2sqrt2

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

-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 !

Yeah Goodluck!

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