anonymous
  • anonymous
HELPPPPPPP!!!! locate all critical points (both types) of h(x)= radical 8-6*x^2 The critical point(s):
Calculus1
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I FOUND THE X WHICH IS 0. THEN I TRIED TO FIND Y BY PLUGGING IN ZERO BUT IT IS INCORRECT... :(
anonymous
  • anonymous
i know the steps to this problem. which is find the derivative and i did...
anonymous
  • 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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zepdrix
  • 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
  • anonymous
yes
zepdrix
  • 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
  • anonymous
okay.
anonymous
  • anonymous
yeah I determined one critical point, but what about y?
zepdrix
  • 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
  • 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
  • anonymous
yes i found 2 radical 2
anonymous
  • anonymous
and i wrote the answer like this... (0,2 radical 2)
zepdrix
  • zepdrix
Hmm I think you end up with \(\large x=\pm \dfrac{2}{\sqrt3}\)
anonymous
  • anonymous
(0,[2\sqrt{2}\])
anonymous
  • anonymous
how?
anonymous
  • anonymous
so x is not 0?
zepdrix
  • 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
  • anonymous
to find the critical points.. dont we find the derivative first?
zepdrix
  • 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
  • anonymous
okay... then
anonymous
  • anonymous
we solve the denominator = to 0 but what about the radical we ignore it?
anonymous
  • anonymous
have you used webwork?
anonymous
  • anonymous
im not sure how to plug in the answer
anonymous
  • anonymous
(0,
zepdrix
  • zepdrix
The square root? I squared both sides. 0^2 = 0
anonymous
  • anonymous
no the answer; how do we write it if there a positive and a negative 2 radical 3.
zepdrix
  • zepdrix
I've used webassign, not webworks. Hmm Is there a panel on the left somewhere where you can use tools like sqrt?
anonymous
  • anonymous
the critical points are:
anonymous
  • anonymous
ive tried every way but it's "Wrong"
anonymous
  • anonymous
yes sqrt( )
zepdrix
  • 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
  • anonymous
Operands of '*' are not of compatible types
anonymous
  • anonymous
because its between of (0, sqrt(8)) (2/sqrt(3),0) (-2/sqrt(3),0)
anonymous
  • anonymous
hold on im so confused.... lol how did you get (0,8)
anonymous
  • anonymous
(0, sqrt8)?
zepdrix
  • 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
  • zepdrix
"Operands of '*' are not of compatible types" I don't understand what you're saying :o
anonymous
  • anonymous
not me, thats what webworks says..
zepdrix
  • zepdrix
Does it tell you to list them as ordered pairs? Usually you just list critical points as x=
anonymous
  • 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
  • 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
  • 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
  • anonymous
correctttt :)
anonymous
  • 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
  • anonymous
i understand the 0 and sqrt 8 order pair...
anonymous
  • anonymous
why couldnt i write 0, 2 radical 2 as a critical point?
zepdrix
  • 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
  • anonymous
nvm but now... how about the +&- 2 radical 3.
zepdrix
  • 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
  • anonymous
i understand how you made the cal. but why did you ignore the radical and wrote it as 8-6x^2
anonymous
  • 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
  • zepdrix
I didn't ignore the radical.\[\large 0=\sqrt{8-6x^2}\]Step 1, square both sides.
anonymous
  • anonymous
thank you so much! :)
anonymous
  • anonymous
thank you thank you! ^_^
anonymous
  • anonymous
i'll keep that the radicals need a + and - in mind.

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