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anonymous

  • 4 years ago

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

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  1. Mertsj
    • 4 years ago
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    Try an indirect proof....assume that it is negative and show that it leads to a contradiction.

  2. nenadmatematika
    • 4 years ago
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    for every x in R y=x^2 stays equal to zero or bigger than zero

  3. myininaya
    • 4 years ago
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    what do you mean you don't know how to write x^2>0?

  4. myininaya
    • 4 years ago
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    and yes because one counterexample for all complex would be if x=i i^2=-1<0

  5. nenadmatematika
    • 4 years ago
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    correct....just look out because it could be equal to zero when x=0

  6. myininaya
    • 4 years ago
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    yep yep

  7. myininaya
    • 4 years ago
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    so the best thing to say is x^2>=0 for all reals

  8. Zarkon
    • 4 years ago
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    show thats its minumm value is 0

  9. myininaya
    • 4 years ago
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    which nenad already said

  10. Zarkon
    • 4 years ago
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    I butchered the crap out of that word....minimum ;)

  11. myininaya
    • 4 years ago
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    zarkon you don't have to know how to spell

  12. myininaya
    • 4 years ago
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    you just have to get it a little right

  13. Zarkon
    • 4 years ago
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    good...because I can't

  14. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    f'=2x=0 => x=0 which means x=0 is a critical number f''=2 which means it is concave up

  16. myininaya
    • 4 years ago
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    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
    • 4 years ago
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    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
    • 4 years ago
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    Thank you very much!

  19. myininaya
    • 4 years ago
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    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
    • 4 years ago
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    let me know if i made a mistake somewhere

  21. myininaya
    • 4 years ago
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    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
    • 4 years ago
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    "For which values for a and b is f increasing?" wait i'm confused

  23. anonymous
    • 4 years ago
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    I'm sorry I'll rephrase it

  24. myininaya
    • 4 years ago
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    f'=0 when x=a ,x=b

  25. myininaya
    • 4 years ago
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    => f is not increasing at x=a,x=b

  26. Zarkon
    • 4 years ago
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    that is not necessarly true

  27. myininaya
    • 4 years ago
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    f'=(x-a)(x-b)^2?

  28. myininaya
    • 4 years ago
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    f'=0 when x=a f'=0 when x=b

  29. anonymous
    • 4 years ago
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    I'm sorry it's in Swedish but the picture might help.

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  30. myininaya
    • 4 years ago
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    zarkon i want a counterexample from you

  31. anonymous
    • 4 years ago
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    Thats correct, x=a and x=b gives the zeros for the derivative

  32. myininaya
    • 4 years ago
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    but we are to find values a and b so that f is increasing

  33. myininaya
    • 4 years ago
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    but f'=(x-a)(x-b)^2

  34. myininaya
    • 4 years ago
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    f'=0 when x=a,b

  35. Zarkon
    • 4 years ago
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    \[f(x)=x^3\]

  36. myininaya
    • 4 years ago
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    oh fine

  37. anonymous
    • 4 years ago
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    Yes I know. And you get extra credit for pointing that out in the solution.

  38. myininaya
    • 4 years ago
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    no zarkon a>0 b>0

  39. myininaya
    • 4 years ago
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    0<a<b

  40. Zarkon
    • 4 years ago
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    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
    • 4 years ago
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    but a, b cannot equal zero

  42. Zarkon
    • 4 years ago
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    x-a can be

  43. Zarkon
    • 4 years ago
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    I was just pointing out that just because the derivative is zero does not mean that the function is not increasing.

  44. myininaya
    • 4 years ago
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    right i get that part you are correct i didn't think about x^3

  45. asnaseer
    • 4 years ago
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    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
    • 4 years ago
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    and other functions like x^3

  47. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    but i wonder if i can think of one example where b>a>0

  49. anonymous
    • 4 years ago
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    But the thing I couldn't solve was: Give a good motivation why (x-b)^2 is always positive

  50. Zarkon
    • 4 years ago
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    the function is increasing over b the soluituion should just be x>a

  51. anonymous
    • 4 years ago
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    It's not increasing if x=b or x=a

  52. Zarkon
    • 4 years ago
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    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
    • 4 years ago
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    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
    • 4 years ago
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    i should say open interval

  55. Zarkon
    • 4 years ago
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    we ar given the derivative...graph the original function

  56. myininaya
    • 4 years ago
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    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
    • 4 years ago
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    this is never true for b>a>0 \[\frac{-a}{2}(\frac{a}{2}-b)^2>0\]

  58. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    Thanks for your help Myininaya

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

  61. anonymous
    • 4 years ago
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    Sorry sorry, I didn't think realize that. But that's the correct answer.

  62. Zarkon
    • 4 years ago
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    it is not

  63. asnaseer
    • 4 years ago
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    myininaya - your original inequalities were correct - they lead to the correct solution of x > a

  64. myininaya
    • 4 years ago
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    but -a/2<0 since a>0

  65. myininaya
    • 4 years ago
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    i don't see a solution to the problem

  66. asnaseer
    • 4 years ago
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    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
    • 4 years ago
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    therefore function is increasing for x > a

  68. myininaya
    • 4 years ago
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    but want f'>0 always we wanted f to be always increasing and that gets messed up on my f'(a/2)

  69. myininaya
    • 4 years ago
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    oh we wanted to know where f is increasing?

  70. myininaya
    • 4 years ago
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    or did we want to choose a and b so that f is increasing everywhere?

  71. myininaya
    • 4 years ago
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    first problem is x>a but second problem has no solution

  72. Zarkon
    • 4 years ago
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    it is not possible to pick an a and b such that f is increasing everywhere

  73. anonymous
    • 4 years ago
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    This is an analyzing problem. We should find when f is increasing and when it's zero

  74. anonymous
    • 4 years ago
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    We don't have to do that

  75. anonymous
    • 4 years ago
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    Just state that if x > a, f is increasing. as long as it isn't = a or b

  76. anonymous
    • 4 years ago
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    But since it can't be > a and = a. x > a and x != b is the solution given in the answer sheet.

  77. asnaseer
    • 4 years ago
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    I think your answer sheet is wrong, it should just be x > a as Zarkon stated earlier

  78. anonymous
    • 4 years ago
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    if x=b then f' = 0 which means not increasing

  79. asnaseer
    • 4 years ago
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    |dw:1328403876728:dw|

  80. asnaseer
    • 4 years ago
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    the function can still be increasing through a point of inflection

  81. anonymous
    • 4 years ago
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    Not at the point

  82. Zarkon
    • 4 years ago
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    points do not increase and decrease...functions do over intervals

  83. asnaseer
    • 4 years ago
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    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
    • 4 years ago
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    "The rate of change at a particular moment. Same as the value of the derivative at a particular point."

  85. anonymous
    • 4 years ago
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    I'm sorry I understand what you mean

  86. anonymous
    • 4 years ago
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    talk about killing an ant with an h bomb

  87. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    yes satellite

  89. anonymous
    • 4 years ago
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    oh i see now that i read the posts

  90. anonymous
    • 4 years ago
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    there was some confusion over what happens at b which reminds me

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