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

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

zepdrix
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0yes i found 2 radical 2

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

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

zepdrix
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0to find the critical points.. dont we find the derivative first?

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

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

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

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

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

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

zepdrix
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0Operands of '*' are not of compatible types

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

zepdrix
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0not me, thats what webworks says..

zepdrix
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0i understand the 0 and sqrt 8 order pair...

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

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

zepdrix
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0thank you so much! :)

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

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