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

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1. mathcalculus

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

2. mathcalculus

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

3. mathcalculus

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

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

yes

6. 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 $$\large f$$.

7. mathcalculus

okay.

8. mathcalculus

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

9. 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).

10. 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.

11. mathcalculus

yes i found 2 radical 2

12. mathcalculus

and i wrote the answer like this... (0,2 radical 2)

13. zepdrix

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

14. mathcalculus

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

15. mathcalculus

how?

16. mathcalculus

so x is not 0?

17. 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}$

18. mathcalculus

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

19. 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.

20. mathcalculus

okay... then

21. mathcalculus

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

22. mathcalculus

have you used webwork?

23. mathcalculus

im not sure how to plug in the answer

24. mathcalculus

(0,

25. zepdrix

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

26. mathcalculus

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

27. zepdrix

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

28. mathcalculus

the critical points are:

29. mathcalculus

ive tried every way but it's "Wrong"

30. mathcalculus

yes sqrt( )

31. zepdrix

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

Operands of '*' are not of compatible types

33. mathcalculus

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

34. mathcalculus

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

35. mathcalculus

(0, sqrt8)?

36. 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))

37. zepdrix

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

38. mathcalculus

not me, thats what webworks says..

39. zepdrix

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

40. mathcalculus

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

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

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

correctttt :)

44. mathcalculus

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

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

46. mathcalculus

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

47. 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.

48. mathcalculus

nvm but now... how about the +&- 2 radical 3.

49. zepdrix

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

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

51. mathcalculus

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

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

53. mathcalculus

thank you so much! :)

54. mathcalculus

thank you thank you! ^_^

55. mathcalculus

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

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