Doug C. answered 08/31/16
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(a) f(x) = x − 3-x = 0.
Since f(0) = −1 and f(1) = 2/3
, by the intermediate value theorem,
there is some solution of f in the interval [0, 1].
Since f(0) = −1 and f(1) = 2/3
, by the intermediate value theorem,
there is some solution of f in the interval [0, 1].
(b) f(x) = x3 + 4.001x2 + 4.002x + 1.101 = 0.
Since f(−3) = −1.896 and f(−2) = 1.101, by the intermediate value
theorem, there is some solution of f in the interval [−3, −2].
Since f(−3) = −1.896 and f(−2) = 1.101, by the intermediate value
theorem, there is some solution of f in the interval [−3, −2].
Could there have been a typo in part b) 4.002x not ^x. I have seen these problems before.
Consider graphing the functions to get an idea of the values to use to apply the intermediate value theorem.
https://www.desmos.com/calculator/j5nws8g1zj