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HELPPPPPPP!!!! locate all critical points (both types) of
h(x)= radical 86*x^2
The critical point(s):
 one year ago
 one year ago
HELPPPPPPP!!!! locate all critical points (both types) of h(x)= radical 86*x^2 The critical point(s):
 one year ago
 one year ago

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mathcalculusBest ResponseYou've already chosen the best response.0
I FOUND THE X WHICH IS 0. THEN I TRIED TO FIND Y BY PLUGGING IN ZERO BUT IT IS INCORRECT... :(
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
i know the steps to this problem. which is find the derivative and i did...
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
then = to zero. and i did that too. and I goy x but then im not sure how i got the answer wrong....
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
It's hard to read the function when it's not formatted. Does this look accurate?\[\large h(x)=\sqrt{86x^2}\]Everything under the square root like that?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Hmm so we get something like this, yes? \[\large h'(x)=\frac{6x}{\sqrt{86x^2}}\] 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 \(\large f\).
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
yeah I determined one critical point, but what about y?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
I'm saying that there is another critical point,\[\large \sqrt{86x^2}\]When this thing equals zero, we're dividing by zero, so f'(x) is undefined, DNE. \[\large 0=\sqrt{86x^2}\] Solve for x to find your other critical point(s).
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
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.
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
yes i found 2 radical 2
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
and i wrote the answer like this... (0,2 radical 2)
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Hmm I think you end up with \(\large x=\pm \dfrac{2}{\sqrt3}\)
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
\[\large 0=86x^2\]\[\large 8=6x^2\]\[\large \frac{8}{6}=x^2 \qquad \rightarrow \qquad \frac{4}{3}=x^2\]\[\large x=\pm \sqrt{\frac{4}{3}}\qquad \rightarrow \qquad x=\pm \frac{2}{\sqrt3}\]
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
to find the critical points.. dont we find the derivative first?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Yes, this was our derivative,\[\large h'(x)=\frac{6x}{\sqrt{86x^2}}\]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.
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
we solve the denominator = to 0 but what about the radical we ignore it?
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
have you used webwork?
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
im not sure how to plug in the answer
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
The square root? I squared both sides. 0^2 = 0
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
no the answer; how do we write it if there a positive and a negative 2 radical 3.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
I've used webassign, not webworks. Hmm Is there a panel on the left somewhere where you can use tools like sqrt?
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
the critical points are:
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
ive tried every way but it's "Wrong"
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Write it as three separate critical points: (0, sqrt(8)) (2/sqrt(3),0) (2/sqrt(3),0) Does that work maybe? D:
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
Operands of '*' are not of compatible types
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
because its between of (0, sqrt(8)) (2/sqrt(3),0) (2/sqrt(3),0)
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
hold on im so confused.... lol how did you get (0,8)
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
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))
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
"Operands of '*' are not of compatible types" I don't understand what you're saying :o
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
not me, thats what webworks says..
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Does it tell you to list them as ordered pairs? Usually you just list critical points as x=
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
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 .)
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
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
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
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)
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
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
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
i understand the 0 and sqrt 8 order pair...
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
why couldnt i write 0, 2 radical 2 as a critical point?
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
Scroll up to see my steps. I don't want to write them out again. Set the denominator equal to zero, solve for x.
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
nvm but now... how about the +& 2 radical 3.
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
We have an \(\large x^2\). When we take the root of a variable, we get two solutions, the positive and negative root.
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
i understand how you made the cal. but why did you ignore the radical and wrote it as 86x^2
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
yes ,right. we have a postive and negative. i understand that but why did you ignore the radical and wrote it as 86x^2? when the original one had a radical .
 one year ago

zepdrixBest ResponseYou've already chosen the best response.0
I didn't ignore the radical.\[\large 0=\sqrt{86x^2}\]Step 1, square both sides.
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
thank you so much! :)
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
thank you thank you! ^_^
 one year ago

mathcalculusBest ResponseYou've already chosen the best response.0
i'll keep that the radicals need a + and  in mind.
 one year ago
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