x^3-4x-2 [2,3]

we have to use the intermediate value thereom. not sure how to do this.

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asked by sonya
  1. x^3-4x-2 [2,3]
    the function is continuous between 2 and 3
    at 2 f(2) = -2
    at 3 f(3) = +13

    The theorem states that for every number N between -2 and +13, there is some x between 2 and 3 such that f(x) = N

    Now they did not give you an N so I will say 5 for example is between -2 and +13

    The theorem says that some x between 2 and 3 will come out with f(x)= 5

    I would have to solve it by trial and error but I know to start between 2 and 5
    x f(x)
    2 -2

    try 2.5 --> 3.63 too small
    try 2.7 --> 6.88 too big
    try 2.6 --> 5.18 closer, a bit big
    try 2.58 -> 4.85 a little small
    try 2.585-> 4.93 well you get the idea

    3 13

    The theorem says that

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    posted by Damon

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