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amy0799

  • one year ago

If f(x)=x^2-x and g(x)=1/x, and h(x) is defined with the expressions below, arrange the values of h'(1) from smallest to largest. 1. h(x) = f(x)-g(x) 2. h(x) = f(x)g(x) 3. h(x) = f(x)/g(x)

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  1. amy0799
    • one year ago
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    do you know how to do this? @IrishBoy123

  2. dinamix
    • one year ago
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    h'(1) = f'(1)-g'(1) u must find f'(x) and g'(x)

  3. dinamix
    • one year ago
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    @amy0799

  4. IrishBoy123
    • one year ago
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    where are you stuck?

  5. dinamix
    • one year ago
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    for 2- h'(1) =f'(1)g(1)+g'(1)f(1) 3- h'(1) =( f'(1) g(1) - f(1)g'(1) )/ (g(1))^2 this the answer @IrishBoy123 right lol

  6. amy0799
    • one year ago
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    h(x) = f(x)-(x) =\[x^{2}-x-\frac{ 1 }{ x }\] \[=2x-1-\frac{ 1 }{ x^{2} }\] Is this correct?

  7. IrishBoy123
    • one year ago
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    not quite last term?

  8. amy0799
    • one year ago
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    the last term is wrong?

  9. IrishBoy123
    • one year ago
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    \[-\frac{ 1 }{ x } = - x^{-1}\]

  10. IrishBoy123
    • one year ago
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    differentiate that again

  11. amy0799
    • one year ago
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    \[2x-1-\frac{ 1 }{x^{-2} }\]

  12. dinamix
    • one year ago
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    2x-1-(-1/x^2) = f'(x)-g'(x)

  13. dinamix
    • one year ago
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    @amy0799 its like my answer see it

  14. IrishBoy123
    • one year ago
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    \[\frac{d}{dx} (\frac{1}{x}) = \frac{d}{dx}(x^{-1}) = -x^{-2} = -\frac{1}{x^2}\] the sign changes :p

  15. amy0799
    • one year ago
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    y wouldn't it be x^-2?

  16. IrishBoy123
    • one year ago
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    \[2x-1-(-1/x^2) = 2x-1+1/x^2\]

  17. dinamix
    • one year ago
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    x^-2 = 1/x^2 u are right but we dont need it @amy0799

  18. IrishBoy123
    • one year ago
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    you had it right with our first go except you got the sign wrong on last term

  19. dinamix
    • one year ago
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    me ? @IrishBoy123 u spoke with me

  20. dinamix
    • one year ago
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    lol

  21. IrishBoy123
    • one year ago
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    no @dinamix :p i was addressing @amy0799

  22. amy0799
    • one year ago
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    h(x)=f(x)g(x) \[=x^{2}-x(-\frac{ 1 }{ x^{2} })+\frac{ 1 }{ x }(2x-1)\] is this right so far?

  23. dinamix
    • one year ago
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    @amy0799 yup this

  24. amy0799
    • one year ago
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    \[\frac{ -x-1 }{ x }+\frac{ 2x-1 }{ x }\]

  25. amy0799
    • one year ago
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    would this simplify to \[\frac{ -x-2 }{ x }?\]

  26. dinamix
    • one year ago
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    -x+1/x @amy0799

  27. dinamix
    • one year ago
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    not -x-1/x

  28. IrishBoy123
    • one year ago
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    i'm afraid i am not following what you have done so i shall leave it to you and @dinamix

  29. dinamix
    • one year ago
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    ok @IrishBoy123 i help him ty anyway for coming

  30. amy0799
    • one year ago
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    -1+(-1)=-2 I dont understand how you got -1

  31. dinamix
    • one year ago
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    \[(x^2-x)(-1/x^2) = -1+\frac{ 1 }{ x } = \frac{ -x+1 }{ x}\]

  32. dinamix
    • one year ago
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    u made mistake @amy0799 u find it ?

  33. dinamix
    • one year ago
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    f(x)g(x) = 1 must find it like @amy0799 so check and tell me

  34. amy0799
    • one year ago
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    \[(x^{2}-x)-\frac{ 1 }{ x ^{2} }=-\frac{ x ^{2} -x+1}{ x ^{2} }\]

  35. amy0799
    • one year ago
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    is this how you set it up?

  36. dinamix
    • one year ago
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    \[\frac{ -x^2+x }{ x^2 } \]

  37. dinamix
    • one year ago
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    its like that @amy0799

  38. amy0799
    • one year ago
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    oh ok

  39. dinamix
    • one year ago
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    so what u find f(x)g(x) = i want your answer

  40. amy0799
    • one year ago
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    h(x)=f(x)/g(x) \[=\frac{ \frac{ 1 }{ x(2x-1)-(x ^{2}-x)-\frac{ 1 }{ x ^{2} } } }{ (\frac{ 1 }{ x })^{2} }\]

  41. amy0799
    • one year ago
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    is this setup right?

  42. amy0799
    • one year ago
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    i messed up, it's 1/x

  43. dinamix
    • one year ago
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    u make some mistake

  44. dinamix
    • one year ago
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    \[\frac{ \ ( \frac{ 1 }{ x})( 2x-1) -[\frac{ -1 }{ x^2} (x^2-x)]} { \frac{ -1 }{ x^2 } }\]

  45. dinamix
    • one year ago
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    @amy0799

  46. dinamix
    • one year ago
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    its not - 1 only 1

  47. dinamix
    • one year ago
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    in demonst

  48. IrishBoy123
    • one year ago
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    where did this come from?!?!?! https://gyazo.com/b501b2ad7a4061eab75c1cf7645b4f67 \[h(x) = \frac{f(x)}{g(x)} = \frac{x^2 - x}{\frac{1}{x}} = x(x^2 - x)\] now, expand and differentiate. that's all there is to it and did we ever come up with an answer to pt 1?

  49. dinamix
    • one year ago
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    \[\frac{ (2x-1+x-1)x^2 }{ x }\]

  50. dinamix
    • one year ago
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    we calcul h'(x)

  51. dinamix
    • one year ago
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    by using f'(x) and g'(x) thats only @IrishBoy123

  52. IrishBoy123
    • one year ago
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    and what exactly is this all about?!?!?!?! https://gyazo.com/d2c0f8f760ddd3e17b317893a135989d \[h(x) = f(x)g(x) = (x^2 - x). \left(\frac{1}{x}\right) = x - 1\] \[h'(x) = 1\] done.

  53. IrishBoy123
    • one year ago
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    if you want to collate your solutions, i am happy to come back later and check them @freckles

  54. dinamix
    • one year ago
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    we have 2 method @amy0799

  55. amy0799
    • one year ago
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    \[\frac{ (3x-2)x ^{2} }{ x }\] can this simplify further?

  56. dinamix
    • one year ago
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    yup

  57. dinamix
    • one year ago
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    h'(x) for the last one

  58. dinamix
    • one year ago
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    h'(x) for function f(x)/g(x) = (3x^2-2x)

  59. dinamix
    • one year ago
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    i mean h'(x) =(3x^2-2x) not f(x)/g(x) did u understand

  60. dinamix
    • one year ago
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    all this time we calcul h'(x) not h(x) @amy0799

  61. dinamix
    • one year ago
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    did u understand mate what we are do

  62. amy0799
    • one year ago
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    woudn't it be \[\frac{ (3x ^{3}-2x ^{2}) }{ x }?\]

  63. dinamix
    • one year ago
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    x^2/x = x right @amy0799 and its same

  64. dinamix
    • one year ago
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    and u have devide by x

  65. amy0799
    • one year ago
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    oh ok. so 3x^2-2x is the answer?

  66. dinamix
    • one year ago
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    yup

  67. amy0799
    • one year ago
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    the question also asked arrange the values of h'(1) from smallest to largest. do i plug in 1 to x?

  68. dinamix
    • one year ago
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    and this h'(x) for function f(x)/g(x) , u understand what was do

  69. dinamix
    • one year ago
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    we find all h'(x) for this fuction f(x)-g(x) and f(x)g(x) and f(x) /g(x)

  70. dinamix
    • one year ago
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    right ?

  71. amy0799
    • one year ago
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    right, so what does arrange the values of h'(1) from smallest to largest mean?

  72. dinamix
    • one year ago
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    h'(1) = f'(1)-g'(1) = 2(1)-1+1(1^2) = ?

  73. dinamix
    • one year ago
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    thats only

  74. amy0799
    • one year ago
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    =2 right?

  75. dinamix
    • one year ago
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    1- h'(1) = 2 yup

  76. amy0799
    • one year ago
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    for f(x)/g(x) 3(1)^2-2(1)=1

  77. dinamix
    • one year ago
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    yup this

  78. dinamix
    • one year ago
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    and for fuction f(x)g(x) ; h'(1) = ?

  79. amy0799
    • one year ago
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    so from smallest to largest it would be f(x)g(x), f(x)/g(x), f(x)-g(x) f(x)g(x)=1

  80. dinamix
    • one year ago
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    not f(x)g(x) =1 its f'(x)g'(x) = 1

  81. dinamix
    • one year ago
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    @amy0799

  82. dinamix
    • one year ago
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    why u make this mistake ?@amy0799

  83. amy0799
    • one year ago
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    what mistake?

  84. dinamix
    • one year ago
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    look up what i write u

  85. dinamix
    • one year ago
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    not f(x)g(x)= 1 its f'(x)g'(x) = 1

  86. amy0799
    • one year ago
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    oh i forgot to put in the '

  87. dinamix
    • one year ago
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    are u sure u understand now .@amy0799

  88. amy0799
    • one year ago
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    yes so the order goes f'(x)g'(x)=1 f'(x)/g'(x)=1 f'(x)-g'(x)=2 ?

  89. dinamix
    • one year ago
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    f'(1)g'(1)=1 f'(1)/g'(1) =1 f'(1)-g'(1)=2 not like u said

  90. dinamix
    • one year ago
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    @amy0799

  91. amy0799
    • one year ago
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    oh ok thank you. can you help me with another problem?

  92. dinamix
    • one year ago
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    ok but fast

  93. amy0799
    • one year ago
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    if f(x)=x^3+2x^2-5x-1 find \[\lim_{h \rightarrow 0}\frac{ f'(-3+h)-f'(-3) }{ h }\]

  94. dinamix
    • one year ago
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    \[\lim_{h \rightarrow 0}\frac{ f(-3+h)-f(-3) }{ h}\]

  95. dinamix
    • one year ago
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    @amy0799 not like u said its not f'(-3+h) -f'(-3) look good how i write it

  96. amy0799
    • one year ago
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    that's how my question was written

  97. dinamix
    • one year ago
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    made mistake can u check plz cuz i didnt see it my life like what u write ?

  98. dinamix
    • one year ago
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    i am sure my write is true @amy0799

  99. amy0799
    • one year ago
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    you could be right, im not sure, can you help me solve it plz

  100. dinamix
    • one year ago
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    f(-3+h) = (-3+h)^3+2(-3+h)^2-5(-3+h)-1

  101. dinamix
    • one year ago
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    the same thing for f(-3) = @amy0799

  102. amy0799
    • one year ago
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    should i find the derivative of f(x)=x^3+2x^2-5x-1 first?

  103. dinamix
    • one year ago
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    \[\lim_{h \rightarrow 0} \frac{ f(-3+h)-f(-3) }{ h } = f'(-3)\]

  104. dinamix
    • one year ago
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    @amy0799 u can find it i cant help u more u should try

  105. dinamix
    • one year ago
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    cuz i will go to sleep its 1:00 Am

  106. amy0799
    • one year ago
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    ok thank you for all your help!!

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