## anonymous one year ago Locate the absolute extrema of the function (if any exist) over the indicated variables.

1. anonymous

2. anonymous

This is how I solved a...

3. anonymous

I found the derivative of the function: x(4-x^2) I set it to zero... x=0 4-x^2=0 or x=2 I then plugged the following back into the original function: 2, 0, -2 And I received: 0, 2, 0 So I concluded that my max = 2 and my min = 0. Assuming that what I did was correct, how could I figure out the exact coordinates of my min and max without looking at a graph?

4. freckles

your derivative is just a little off

5. freckles

also 4-x^2=0 gives you x=2 or x=-2

6. anonymous

"your derivative is just a little off" What should it be? "also 4-x^2=0 gives you x=2 or x=-2" Yeah, that makes sense. I forgot about that rule.

7. freckles

you should have used chain rule

8. freckles

$((4-x^2)^\frac{1}{2})'=\frac{1}{2}(4-x^2)^{\frac{-1}{2}} \cdot (4-x^2)'$

9. anonymous

I used the chain rule. I didn't add the exponent though...and I didn't make 2x negative. Oops! O_o

10. anonymous

Now that I know that. How could I find the critical points for that derivative?

11. freckles

$f'(x)=\frac{-x}{\sqrt{4-x^2}}$ is this what you have for f'?

12. anonymous

Not even close unfortunately

13. freckles

you can find the critical numbers by finding when it is 0 or when it does not exist (these numbers should be in the domain of the original function though) so you still have to solve the equations x=0 and 4-x^2=0 so you are right your critical number are -2,0,2

14. freckles

wait what?

15. freckles

did you not understand how I got... $((4-x^2)^\frac{1}{2})'=\frac{1}{2}(4-x^2)^{\frac{-1}{2}} \cdot (4-x^2)'$

16. anonymous

Yes, I actually solved it wrongly, and some how managed to get the right values. This is how I originally solved it...

17. anonymous

$\sqrt{4-x^2}$$(4-x^2)^\frac{1}{2}$$\frac{1}{2}(4-x^2)(2x)$$x(4-x^2)$ I didn't make the 2x negative nor did I remember to subtract 1 from my exponent (instead I forgot about it completely.

18. anonymous

Then I got... x=0 4-x^2=0 or x=2 , x=-2

19. freckles

Ok yeah but are you understand why this is $f'(x)=\frac{-x}{\sqrt{4-x^2}}$ ?

20. anonymous

I know how you get to $-x(4-x^2)^\frac{1}{2}$ is what you have just a simplified version of that?

21. freckles

well 1/2-1 is -1/2 not 1/2

22. anonymous

Sorry, typo

23. freckles

$-x(4-x^2)^\frac{-1}{2}=\frac{-x}{(4-x^2)^\frac{1}{2}} \text{ by law of exponents }$

24. freckles

$x^{-n}=\frac{1}{x^n}$

25. anonymous

There are so many laws I need to re-memorize.

26. anonymous

But yes I do recognize what you did.

27. freckles

anyways you have the right critical numbers x=-2,0,2 so we need to evaluate f(-2),f(0),f(1),f(2) I put f(1) because we need to evaluate it later anyways

28. anonymous

1?

29. freckles

$f(x)=\sqrt{4-x^2} \\ f(-2)=\sqrt{4-(-2)^2}=\sqrt{4-4}=0 \\ f(0)=? \\ f(1)=? \\ f(2)=?$

30. freckles

question d has 1 as an endpoint...

31. anonymous

Oh okay, I was just looking at a

32. freckles

I just wanted to get all the evaluation done with

33. freckles

we can look at a first ... I want to get through evaluating the function at the critical numbers and any endpoints

34. freckles

anyways can you evaluate those 3 above...

35. anonymous

x=2 would end up being 0 x=-1 would also be 0 x=0 would be 2 x=1 would be sqrt(3)

36. anonymous

That was supposed to say -2 not -1

37. freckles

f(-2)=0 f(0)=2 f(2)=0 f(1)=sqrt(3) great work

38. freckles

so look at the outputs above

39. freckles

0,2,0

40. freckles

what is biggest number there

41. anonymous

2

42. freckles

so that is your absolute max

43. freckles

0,2,0 and of these numbers 0 is the absolute min

44. anonymous

That makes sense, but how would I represent that max as coordinates?

45. freckles

which this makes sense... because the graph is the top part of a circle... |dw:1444500172873:dw|

46. freckles

well you already found the coordinates

47. freckles

f(x)=y means (x,y) is a coordinate on f

48. freckles

for example we said f(2)=0 so that means (2,0) is a coordinate on f

49. anonymous

Oh so max would be 0,2

50. freckles

(0,2) would be max ordered pairs have ( ) around them

51. freckles

|dw:1444500273762:dw|

52. anonymous

so could (-2,0) and (2,0) both be min values?

53. freckles

yep

54. anonymous

Because b has a parentheses instead of a bracket, would it be solved for differently?

55. freckles

now we can look at b we have [-2,0) we only care about f(-2) and f(0) here but we cannot actually include what happens at x=0 f(-2)=0 and f(0)=2 is what we found earlier of 0 and 2 the smallest is 0 so your absolute min is at x=0 and we will not include an absolute max for this question because we cannot include what happens at x=0

56. anonymous

Why can't we include 0?

57. freckles

[-2,0) means we have everything between -2 and 0 including -2 but not including 0

58. freckles

|dw:1444500569732:dw| this function on [-2,0) looks like: |dw:1444500587078:dw| my quarter of a circle was horribly drawn I blame it on the cat licking me

59. anonymous

[ means included ( means to that point but the point isn't included itself

60. anonymous

Correct?

61. freckles

see the absolute max doesn't actually exist there is a hole at x=0 because the domain didn't allow us to include what happens at x=0

62. anonymous

Ah okay

63. anonymous

and for C it would just be solving for 0?

64. freckles

brackets means we include the endpoint parenthesis means we don't include the endpoint

65. freckles

we already looked at a and said the absolute mins were (-2,0) and (2,0) and the absolute max was (0,2) c would be the same answer except the absolute mins would not exist since we have (-2,2) and not [-2,2]

66. anonymous

So only the max would exist in that particular problem.

67. freckles

yep

68. anonymous

Should i just state that it cannot be determined or is there a specific term I should use?

69. freckles

I think the word choice I used was good "does not exist" |dw:1444500908510:dw| this graph on [1,2) looks like: |dw:1444500919987:dw|

70. anonymous

Thank you so much for your help! :D

71. freckles

what did you get for b again

72. anonymous

max: does not exist min: (-2, 0)

73. freckles

great

74. freckles

75. anonymous

just a second

76. anonymous

max: sqrt(3) min: does not exist That may or may not be right.

77. freckles

did you still want the coordinates or will the y value just be okay ?

78. freckles

like you can say the max is sqrt(3) but if your teacher prefers coordinates then you should say (1,sqrt(3))

79. anonymous

Yeah that's what I'll put

80. freckles

ok ok that sounds great to me

81. anonymous

excellent! It's surprising how simple this stuff really is compared to how difficult it seems at times

82. freckles

It is really not that bad... You do have to train yourself back on your algebra though.. The calculus I think is easier than the algebra :p

83. anonymous

I do the worst on trig identities personally