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anonymous

  • 5 years ago

EVALUATE THE EXPONENTIAL FUNCTION FOR THREE POSITIVE VALUES OF X,THREE NEGATIVE VALUES OF X,AND AT X=0.SHOW YOUR WORK.USE THE RESULTING ORDERED PAIRS TO PLOT THE GRAPH.STATE THE DOMAIN AND RANGE OF THE FUNCTIONATTACH GRAPH AS LISTED AND REQUIRED.F(X)=E^-X-1

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  1. anonymous
    • 5 years ago
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    please help me out in this......

  2. anonymous
    • 5 years ago
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    \[f(x) = e ^{-x}-1\]

  3. anonymous
    • 5 years ago
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    is that the question?

  4. anonymous
    • 5 years ago
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    yes this is the question?

  5. anonymous
    • 5 years ago
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    okay, substitute and three positive values of x, say 1,2,3 and three negative values, say -1,-2,-3 in f(x)

  6. anonymous
    • 5 years ago
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    so for x =1, f(x) = f(1) = (e^-1) -1 use your calculator or google to find the value of e^-1 and subtract 1 from that to get f(1)

  7. anonymous
    • 5 years ago
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    just go to google.com and type e^-1. you will get a value. try it out

  8. anonymous
    • 5 years ago
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    according to google calculator e^-1 is 0.367879441. so (e^-1) -1 = -0.632120559 so at x = 1, f(x) = -0.632120559

  9. anonymous
    • 5 years ago
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    similarly find out values of f(x) for x =2,3 and x = -1,-2,-3

  10. anonymous
    • 5 years ago
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    use those values and make a graph such as this: f(x) | | | | | | | |----------------------->x

  11. anonymous
    • 5 years ago
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    iam new to this..thts why getting problems

  12. anonymous
    • 5 years ago
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    okay I understand. But now I have shown you the procedure. Just follow it. Its easy. You can do it. Did you try typing it in google?

  13. anonymous
    • 5 years ago
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    yes i did but i dint got

  14. anonymous
    • 5 years ago
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    okay. do you have a scientific calculator with you?

  15. anonymous
    • 5 years ago
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    did you type in e^-1 exactly as I typed it? just copy e^-1 and paste it in google. and press enter.

  16. anonymous
    • 5 years ago
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    yeah i did the same

  17. anonymous
    • 5 years ago
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    i got now

  18. anonymous
    • 5 years ago
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    okay what did you get?

  19. anonymous
    • 5 years ago
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    can you help me in another question?

  20. anonymous
    • 5 years ago
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    i got the same which you got

  21. anonymous
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    okay good, just do the same for other numbers and you will get your answers.

  22. anonymous
    • 5 years ago
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    Yes, I can help you. But your new problem separately.

  23. anonymous
    • 5 years ago
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    post your new problem separately.

  24. anonymous
    • 5 years ago
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    ok can i post it here

  25. anonymous
    • 5 years ago
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    no.. post it separately. this post is too long already

  26. anonymous
    • 5 years ago
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    ok

  27. anonymous
    • 5 years ago
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    I dont see it zehera.

  28. anonymous
    • 5 years ago
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    its not coming there

  29. anonymous
    • 5 years ago
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    i think that i have poted it before thts y it not coming once again

  30. anonymous
    • 5 years ago
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    eveluate the exponentaial eqution forthree positive valu f x,and three negative values of x,and at x=0.transform the second expression into the equivalent logarithm equation and evalaute the logarithm equation for thre values of x that are greater than 1,three values of x that are between 0 and1,and at x=1.show your work.use the resulting ordered pairs to plot the graph of each function.Y=5x^-2,X=5y^-2

  31. anonymous
    • 5 years ago
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    is that it?

  32. anonymous
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    yes

  33. anonymous
    • 5 years ago
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    so Y = 5x^-2 is equation 1 and X = 5Y^-2 is equation 2 right?

  34. anonymous
    • 5 years ago
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    yes

  35. anonymous
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    okay what is x^-2 ?

  36. anonymous
    • 5 years ago
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    on google i got this but ith another equation

  37. anonymous
    • 5 years ago
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    im getting positive values but for negative vl\alues iam getting confused

  38. anonymous
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    okay, ;ets make it simpler then. what is x^-2 ?

  39. anonymous
    • 5 years ago
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    ok it would be greatful

  40. anonymous
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    okay. but what is x^-2?

  41. anonymous
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    its given that Y=5X^-2

  42. anonymous
    • 5 years ago
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    no, tell me what x^-2 means. I dont want maths. Just tell me in plain english. What do you think x^-2 means?

  43. anonymous
    • 5 years ago
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    I DONT KNOW?

  44. anonymous
    • 5 years ago
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    okay. \[x ^{-2} = 1/x ^{2}\]

  45. anonymous
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    ok

  46. anonymous
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    so, \[5x ^{-2} = 5/x ^{2}\]

  47. anonymous
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    that means, \[Y = 5/x ^{2}\]

  48. anonymous
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    so now substitute x = 1,2,3 and -1,-2,-3 and get Y values. what is the value of Y at 0?

  49. anonymous
    • 5 years ago
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    1 i think

  50. anonymous
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    how?

  51. anonymous
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    any value at log0 is 1

  52. anonymous
    • 5 years ago
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    you are just asked to evaluate the first equation. Why are you taking log?

  53. anonymous
    • 5 years ago
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    ok you explain me how to do

  54. anonymous
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    if we take one value as 2 thn y=5/4

  55. anonymous
    • 5 years ago
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    is it right?

  56. anonymous
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    right

  57. anonymous
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    ok like this we have to take the other two values

  58. anonymous
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    right. But the question also asks for x =0 . so what happens if you take another value as 0?

  59. anonymous
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    dont lknow

  60. anonymous
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    just put 0 in the equation. see what you get. y = 5/0. what is anything divided by 0?

  61. anonymous
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    0

  62. anonymous
    • 5 years ago
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    so we have to take three values as 0,1,2

  63. anonymous
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    no, anything multiplied by 0 is 0. anything divided by zero is infinity.

  64. anonymous
    • 5 years ago
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    No, pay attention zehera. the problem asks for three positive values. so you use any three positive values. Similarly use three negative values. then use x = 0 which is neither positive nor negative.

  65. anonymous
    • 5 years ago
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    ok

  66. anonymous
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    so totally you have 7 values, three positive, three negative and one 0.

  67. anonymous
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    ok

  68. anonymous
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    okay. now second equation.

  69. anonymous
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    while taking negative values we get 5/-4

  70. anonymous
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    what does the problem say you should do with the second equation?

  71. anonymous
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    if x = -2, what is x ^2?

  72. anonymous
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    x^2 = x * x. so, if x = -2, x * x = ?

  73. anonymous
    • 5 years ago
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    4

  74. anonymous
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    so then for x = -2, Y = ?

  75. anonymous
    • 5 years ago
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    confused

  76. anonymous
    • 5 years ago
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    x = -2. Y = 5/(-2)^2

  77. anonymous
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    but you just said (-2)^2 = 4 so Y = 5/4.

  78. anonymous
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    understood?

  79. anonymous
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    but here the poer is -2 na

  80. anonymous
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    -2^-2=?

  81. anonymous
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    are you talking about equation 1 or equation 2?

  82. anonymous
    • 5 years ago
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    equation2

  83. anonymous
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    okay i was talking about equation 1

  84. anonymous
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    what does the problem say to do with equation 2?

  85. anonymous
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    same as we did in equation 1 but here we have to take negative values

  86. anonymous
    • 5 years ago
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    No, pay attention. The problem says transform equation 2 to its logarthmic form

  87. anonymous
    • 5 years ago
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    am i right?

  88. anonymous
    • 5 years ago
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    im jst getting confused

  89. anonymous
    • 5 years ago
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    you just explain me

  90. anonymous
    • 5 years ago
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    i will follow it

  91. anonymous
    • 5 years ago
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    There are two parts of the problem. We finished the first part using equation 1. Now the second part of the problem says transform the second equation to logarithm form

  92. anonymous
    • 5 years ago
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    the second equation is X=5y^-2

  93. anonymous
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    so you transform it to logarithms by taking log on both sides. that is log X = log(5y^-2)

  94. anonymous
    • 5 years ago
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    that is all I can explain. You should be able to take it from there.

  95. anonymous
    • 5 years ago
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    ok

  96. anonymous
    • 5 years ago
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    thankyou

  97. anonymous
    • 5 years ago
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    you are welcome.

  98. anonymous
    • 5 years ago
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    you helped me lot

  99. anonymous
    • 5 years ago
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    thankyou very much

  100. anonymous
    • 5 years ago
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    bye:)

  101. anonymous
    • 5 years ago
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    http://en.wikipedia.org/wiki/Logarithms

  102. anonymous
    • 5 years ago
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    you are welcome. try to understand what you are doing and you will get your answers easily. good luck!

  103. anonymous
    • 5 years ago
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    thankyou

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