## anonymous 4 years ago Prove that x^2 is always positive. I understand why, but I don't know how to write out.

1. Mertsj

Try an indirect proof....assume that it is negative and show that it leads to a contradiction.

for every x in R y=x^2 stays equal to zero or bigger than zero

3. myininaya

what do you mean you don't know how to write x^2>0?

4. myininaya

and yes because one counterexample for all complex would be if x=i i^2=-1<0

correct....just look out because it could be equal to zero when x=0

6. myininaya

yep yep

7. myininaya

so the best thing to say is x^2>=0 for all reals

8. Zarkon

show thats its minumm value is 0

9. myininaya

10. Zarkon

I butchered the crap out of that word....minimum ;)

11. myininaya

zarkon you don't have to know how to spell

12. myininaya

you just have to get it a little right

13. Zarkon

good...because I can't

14. anonymous

Here is the problem: f'(x) = (x-a)(x-b)^2 0 < a < b "For which values for a and b is f increasing?" And I can't motivate why (x-b)^2 is always positive. Am I over thinking this?

15. myininaya

f'=2x=0 => x=0 which means x=0 is a critical number f''=2 which means it is concave up

16. myininaya

so a and b are critical numbers $f''=(x-b)^2+(x-a) \cdot 2(x-b)=(x-b)[x-b+2(x-a))]=(x-b)[3x-b-2a]$ $f''(a)=(a-b)[3a-b-2a]=(a-b)[a-b]=(a-b)^2=(b-a)^2>0$ $f''(b)=0 \text{ inconclusive }$

17. myininaya

b>a so b is not a ever so (b-a)^2 is always positive of course we are pretending we have real a and real b

18. anonymous

Thank you very much!

19. myininaya

but it seems like the question you asking now is different from the one you are asking earlier $f'(x)=(x-a)(x-b)^2$ we want f'>0 $(x-a)(x-b)^2>0$ f' is zero when x=a or x=b |-----|---------|---- 0 a b chose number btw 0 and a like a/2 we want the following to be greater than 0 $f'(\frac{a}{2})=(\frac{-a}{2})(\frac{a}{2}-b)^2$ we want to choose a number btw a and b now like (a+b)/2 again we want the following to be greater than 0 $f'(\frac{a+b}{2})=\frac{-a+b}{2}(\frac{a-b}{2})^2=-\frac{a-b}{2}(\frac{a-b}{2})^2=-(\frac{a-b}{2})^3$ now we choose a number after b like b+a we also want this to be bigger than 0 $f'(b+a)=(b)(a)^2=a^2b$ no now we need to think of a and b such that all of these inequalities are satisfy

20. myininaya

let me know if i made a mistake somewhere

21. myininaya

so we have the following three inequalities: $\frac{-a}{2}(\frac{a}{2}-b)^2>0, -(\frac{a-b}{2})^3>0 , a^2b>0$

22. myininaya

"For which values for a and b is f increasing?" wait i'm confused

23. anonymous

I'm sorry I'll rephrase it

24. myininaya

f'=0 when x=a ,x=b

25. myininaya

=> f is not increasing at x=a,x=b

26. Zarkon

that is not necessarly true

27. myininaya

f'=(x-a)(x-b)^2?

28. myininaya

f'=0 when x=a f'=0 when x=b

29. anonymous

I'm sorry it's in Swedish but the picture might help.

30. myininaya

zarkon i want a counterexample from you

31. anonymous

Thats correct, x=a and x=b gives the zeros for the derivative

32. myininaya

but we are to find values a and b so that f is increasing

33. myininaya

but f'=(x-a)(x-b)^2

34. myininaya

f'=0 when x=a,b

35. Zarkon

$f(x)=x^3$

36. myininaya

oh fine

37. anonymous

Yes I know. And you get extra credit for pointing that out in the solution.

38. myininaya

no zarkon a>0 b>0

39. myininaya

0<a<b

40. Zarkon

def: a function $$f$$ is increasing on and interval $$I$$ if for any $$x,y\in I$$ with $$x<y$$ we have that $$f(x)<f(y)$$ $x^3$ satisfies this definition on$$\mathbb{R}$$

41. myininaya

but a, b cannot equal zero

42. Zarkon

x-a can be

43. Zarkon

I was just pointing out that just because the derivative is zero does not mean that the function is not increasing.

44. myininaya

right i get that part you are correct i didn't think about x^3

45. asnaseer

myininaya - as Zarkon says - f'(x)=0 could be a point of inflection. a function can be increasing/decreasing through a point of inflection

46. myininaya

and other functions like x^3

47. anonymous

The answers says that (x-b)^2 will always be positive. Because of that (x-a) determines the sign of the derivative. x > a and x != a,b Thats is all thats needed

48. myininaya

but i wonder if i can think of one example where b>a>0

49. anonymous

But the thing I couldn't solve was: Give a good motivation why (x-b)^2 is always positive

50. Zarkon

the function is increasing over b the soluituion should just be x>a

51. anonymous

It's not increasing if x=b or x=a

52. Zarkon

if you fix a value of x then f is not increasing(it is just a number)...you need to look over an interval....for some open set containig b the function is increasing...even though the derivative is zero at b

53. asnaseer

look at this as an example: http://www.wolframalpha.com/input/?i=y%3D%28x-1%29%28x-4%29%5E2+for+x%3D0+to+5 the function is increasing from x=4 onwards here 4 is equivalent to point x=b

54. Zarkon

i should say open interval

55. Zarkon

we ar given the derivative...graph the original function

56. myininaya

http://www.wolframalpha.com/input/?i=find+a+and+b+such+that+-a%5E2%2F2*%28a%2F2-b%29%5E2%3E0%2C+-%28a-b%29%5E3%2F8%3E0%2C+a%5E2b%3E0 i put in those three inequalities wolfram said there is no solution to that system

57. Zarkon

this is never true for b>a>0 $\frac{-a}{2}(\frac{a}{2}-b)^2>0$

58. anonymous

The answer doesn't require an interval. Even if x will equal a at some point, it's enough to state that at this point the derivative will be = 0

59. anonymous

60. Zarkon

you gave an interval (2 intervals) as an answer... " x > a and x != a,b Thats is all thats needed"

61. anonymous

Sorry sorry, I didn't think realize that. But that's the correct answer.

62. Zarkon

it is not

63. asnaseer

myininaya - your original inequalities were correct - they lead to the correct solution of x > a

64. myininaya

but -a/2<0 since a>0

65. myininaya

i don't see a solution to the problem

66. asnaseer

you got this for x < a:$f'(\frac{a}{2})=(\frac{-a}{2})(\frac{a}{2}-b)^2$which is always negative, so function cannot be increasing over this region. for x>a and x<b you got:$f'(\frac{a+b}{2})=-(\frac{a-b}{2})^3=-\frac{a-b}{2}(\frac{a-b}{2})^2=\frac{b-a}{2}(\frac{a-b}{2})^2$which is always positive since b > a. and finally for x > b you got:$f'(b+a)=a^2b$which is always positive since 0 < a < b

67. asnaseer

therefore function is increasing for x > a

68. myininaya

but want f'>0 always we wanted f to be always increasing and that gets messed up on my f'(a/2)

69. myininaya

oh we wanted to know where f is increasing?

70. myininaya

or did we want to choose a and b so that f is increasing everywhere?

71. myininaya

first problem is x>a but second problem has no solution

72. Zarkon

it is not possible to pick an a and b such that f is increasing everywhere

73. anonymous

This is an analyzing problem. We should find when f is increasing and when it's zero

74. anonymous

We don't have to do that

75. anonymous

Just state that if x > a, f is increasing. as long as it isn't = a or b

76. anonymous

But since it can't be > a and = a. x > a and x != b is the solution given in the answer sheet.

77. asnaseer

I think your answer sheet is wrong, it should just be x > a as Zarkon stated earlier

78. anonymous

if x=b then f' = 0 which means not increasing

79. asnaseer

|dw:1328403876728:dw|

80. asnaseer

the function can still be increasing through a point of inflection

81. anonymous

Not at the point

82. Zarkon

points do not increase and decrease...functions do over intervals

83. asnaseer

if you pick any specific point, then you cannot say that function is increasing/decreasing, it just has some value there. increasing/decreasing applies to an interval.

84. anonymous

"The rate of change at a particular moment. Same as the value of the derivative at a particular point."

85. anonymous

I'm sorry I understand what you mean

86. anonymous

talk about killing an ant with an h bomb

87. anonymous

not that i read the whole thing, but isn't is just asking for which values of x is $x-a>0$ namely $x>a$? if the function is continuous (and we assume it is since we assume it has a derivative everywhere) then the fact that the derivative may be postitive, then zero then positive means the funcion is increasing over the whole interval

88. myininaya

yes satellite

89. anonymous

oh i see now that i read the posts

90. anonymous

there was some confusion over what happens at b which reminds me