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I FOUND THE X WHICH IS 0. THEN I TRIED TO FIND Y BY PLUGGING IN ZERO BUT IT IS INCORRECT... :(

i know the steps to this problem. which is find the derivative and i did...

then = to zero. and i did that too. and I goy x but then im not sure how i got the answer wrong....

yes

okay.

yeah I determined one critical point, but what about y?

yes i found 2 radical 2

and i wrote the answer like this... (0,2 radical 2)

Hmm I think you end up with \(\large x=\pm \dfrac{2}{\sqrt3}\)

(0,[2\sqrt{2}\])

how?

so x is not 0?

to find the critical points.. dont we find the derivative first?

okay... then

we solve the denominator = to 0 but what about the radical we ignore it?

have you used webwork?

im not sure how to plug in the answer

(0,

The square root? I squared both sides. 0^2 = 0

no the answer; how do we write it if there a positive and a negative 2 radical 3.

the critical points are:

ive tried every way but it's "Wrong"

yes sqrt( )

Operands of '*' are not of compatible types

because its between of (0, sqrt(8)) (2/sqrt(3),0) (-2/sqrt(3),0)

hold on im so confused.... lol how did you get (0,8)

(0, sqrt8)?

"Operands of '*' are not of compatible types"
I don't understand what you're saying :o

not me, thats what webworks says..

Does it tell you to list them as ordered pairs?
Usually you just list critical points as x=

correctttt :)

i understand the 0 and sqrt 8 order pair...

why couldnt i write 0, 2 radical 2 as a critical point?

nvm but now... how about the +&- 2 radical 3.

i understand how you made the cal. but why did you ignore the radical and wrote it as 8-6x^2

I didn't ignore the radical.\[\large 0=\sqrt{8-6x^2}\]Step 1, square both sides.

thank you so much! :)

thank you thank you! ^_^

i'll keep that the radicals need a + and - in mind.