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

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

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

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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0It'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?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Hmm 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 \(\large f\).

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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0I'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).

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0As 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.

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

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

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

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

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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.0Yes, 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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0Locate 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 .)

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0now 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
 one year ago
Best ResponseYou've already chosen the best response.0Hmm 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)

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0okay, 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

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

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

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

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

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

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

mathcalculus
 one year ago
Best ResponseYou've already chosen the best response.0yes ,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 .

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

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

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

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