HELPPPPPPP!!!! locate all critical points (both types) of
h(x)= radical 8-6*x^2
The critical point(s):

- anonymous

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- anonymous

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

- anonymous

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

- anonymous

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

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## More answers

- zepdrix

It's hard to read the function when it's not formatted.
Does this look accurate?\[\large h(x)=\sqrt{8-6x^2}\]Everything under the square root like that?

- anonymous

yes

- zepdrix

Hmm so we get something like this, yes?
\[\large h'(x)=\frac{-6x}{\sqrt{8-6x^2}}\]
So you determined that \(\large x=0\) is a critical point.
There is also another.
A critical point of the function also exists where \(\large f'(x)\) is undefined. (Assuming that value is included in the domain of \(\large f\).

- anonymous

okay.

- anonymous

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

- zepdrix

I'm saying that there is another critical point,\[\large \sqrt{8-6x^2}\]When this thing equals zero, we're dividing by zero, so f'(x) is undefined, DNE.
\[\large 0=\sqrt{8-6x^2}\]
Solve for x to find your other critical point(s).

- zepdrix

As for your y values, if you want to, you can plug x=0 into your original function, to find the corresponding y value. That will allow you to write it as an ordered pair if you want.

- anonymous

yes i found 2 radical 2

- anonymous

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

- zepdrix

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

- anonymous

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

- anonymous

how?

- anonymous

so x is not 0?

- zepdrix

\[\large 0=8-6x^2\]\[\large 8=6x^2\]\[\large \frac{8}{6}=x^2 \qquad \rightarrow \qquad \frac{4}{3}=x^2\]\[\large x=\pm \sqrt{\frac{4}{3}}\qquad \rightarrow \qquad x=\pm \frac{2}{\sqrt3}\]

- anonymous

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

- zepdrix

Yes, this was our derivative,\[\large h'(x)=\frac{-6x}{\sqrt{8-6x^2}}\]We set the numerator equal to zero to find one critical point.
We need to also set the denominator equal to zero to find the other(s).
Which is what I was showing.

- anonymous

okay... then

- anonymous

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

- anonymous

have you used webwork?

- anonymous

im not sure how to plug in the answer

- anonymous

(0,

- zepdrix

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

- anonymous

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

- zepdrix

I've used webassign, not webworks. Hmm
Is there a panel on the left somewhere where you can use tools like sqrt?

- anonymous

the critical points are:

- anonymous

ive tried every way but it's "Wrong"

- anonymous

yes sqrt( )

- zepdrix

Write it as three separate critical points:
(0, sqrt(8))
(2/sqrt(3),0)
(-2/sqrt(3),0)
Does that work maybe? D:

- anonymous

Operands of '*' are not of compatible types

- anonymous

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

- anonymous

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

- anonymous

(0, sqrt8)?

- zepdrix

We found that x=0 is a critical point.
If we plug x=0 back into the original function it gives us, \(\large f(0)=\sqrt8\) right? We can write that as an ordered pair (0,sqrt(8))

- zepdrix

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

- anonymous

not me, thats what webworks says..

- zepdrix

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

- anonymous

Locate all critical points ( both types )
The critical point(s) is (are) .
(The answers are to be points. Use parentheses in your answer(s). If there are no critical points, enter none .)

- anonymous

now i tried to put the answer like this: (0,sqrt(8))U(2/sqrt(3),0)U(-2/sqrt(3),0) but it says: Left endpoint must be less than right endpoint

- zepdrix

Hmm I'm not sure how your assignment wants them formatted.
I would guess that you just separate them with commas.
(0,sqrt(8)), (2/sqrt(3),0), (-2/sqrt(3),0)

- anonymous

correctttt :)

- anonymous

okay, so after we find x, how do we get the + and - 2 radical 3? i mean we plug in 0 to the original but i got 2 radical 2

- anonymous

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

- anonymous

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

- zepdrix

Scroll up to see my steps. I don't want to write them out again.
Set the denominator equal to zero, solve for x.

- anonymous

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

- zepdrix

We have an \(\large x^2\). When we take the root of a variable, we get two solutions, the positive and negative root.

- anonymous

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

- anonymous

yes ,right. we have a postive and negative. i understand that but why did you ignore the radical and wrote it as 8-6x^2? when the original one had a radical .

- zepdrix

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

- anonymous

thank you so much! :)

- anonymous

thank you thank you! ^_^

- anonymous

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

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