## mathcalculus Group Title HELPPPPPPP!!!! locate all critical points (both types) of h(x)= radical 8-6*x^2 The critical point(s): one year ago one year ago

1. mathcalculus Group Title

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

2. mathcalculus Group Title

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

3. mathcalculus Group Title

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

4. zepdrix Group Title

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?

5. mathcalculus Group Title

yes

6. zepdrix Group Title

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 $$\large f$$.

7. mathcalculus Group Title

okay.

8. mathcalculus Group Title

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

9. zepdrix Group Title

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).

10. zepdrix Group Title

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.

11. mathcalculus Group Title

yes i found 2 radical 2

12. mathcalculus Group Title

13. zepdrix Group Title

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

14. mathcalculus Group Title

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

15. mathcalculus Group Title

how?

16. mathcalculus Group Title

so x is not 0?

17. zepdrix Group Title

$\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}$

18. mathcalculus Group Title

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

19. zepdrix Group Title

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.

20. mathcalculus Group Title

okay... then

21. mathcalculus Group Title

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

22. mathcalculus Group Title

have you used webwork?

23. mathcalculus Group Title

im not sure how to plug in the answer

24. mathcalculus Group Title

(0,

25. zepdrix Group Title

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

26. mathcalculus Group Title

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

27. zepdrix Group Title

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

28. mathcalculus Group Title

the critical points are:

29. mathcalculus Group Title

ive tried every way but it's "Wrong"

30. mathcalculus Group Title

yes sqrt( )

31. zepdrix Group Title

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

32. mathcalculus Group Title

Operands of '*' are not of compatible types

33. mathcalculus Group Title

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

34. mathcalculus Group Title

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

35. mathcalculus Group Title

(0, sqrt8)?

36. zepdrix Group Title

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))

37. zepdrix Group Title

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

38. mathcalculus Group Title

not me, thats what webworks says..

39. zepdrix Group Title

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

40. mathcalculus Group Title

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 .)

41. mathcalculus Group Title

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

42. zepdrix Group Title

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)

43. mathcalculus Group Title

correctttt :)

44. mathcalculus Group Title

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

45. mathcalculus Group Title

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

46. mathcalculus Group Title

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

47. zepdrix Group Title

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

48. mathcalculus Group Title

49. zepdrix Group Title

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

50. mathcalculus Group Title

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

51. mathcalculus Group Title

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 .

52. zepdrix Group Title

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

53. mathcalculus Group Title

thank you so much! :)

54. mathcalculus Group Title

thank you thank you! ^_^

55. mathcalculus Group Title

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