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yes

i mean x and y

critical points..

Oh, okay. First of all, do you know how to take the derivative of h(x)?

so far im here: -12x/ 2 radical 8-6x^2

The derivative of h(x) is: \[\frac{ -6x }{ \sqrt{8-6x ^{2}} }\]

So, you're going to set f'(x) to 0 and solve for x.

okay makes sense... then to find the critical points i equal that to zero... and just solve for x?

I got zero as well... not how do i find y

i plugged in zero into the normal question... and got 2 radical 2

answer: (0,2 radical 2) is wrong://

Gah, I was doing something more complicated. Ignore the part about test intervals, sorry. :(

its okay can we do after we find the x

the y..

Plug in the critical numbers into the f(x).

i plug in 0 in the original function? to find y right

Yes, sorry. Working it out over here, too. ^^;

\[2\sqrt{2}\]...

i keep getting wrong :(

\[f(0) = \sqrt{8-6(0)^2} = \sqrt{8}= 2\sqrt{2}\]

I was actually getting kinda confused, and I ended up graphing it to check.