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monroe17

  • 3 years ago

The problem to use the intermediate value theorem to show that there is a root of the given equation in the specified interval. sin(x)= x^2-x, (1,2) Can someone see if this is correct? f(x) = sin(x) - x^2 + x, then f is continuous since sin(x) and x^2 - x are continuous and their composition is also continuous. sin(x) = x^2 - x which is equivalent to showing that f(x) = 0. f(1) = 0.841 and f(2) = -1.091, 0 s lying between f(1) and f(2). Since f is continuous, there is a c in (1,2) such that f(c) = 0 there is a root to the equation sin(x) = x^2 - x

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  1. monroe17
    • 3 years ago
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    @jim_thompson5910 can you check my work for me? :)

  2. ivanmlerner
    • 3 years ago
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    I'm sure it is right, but I'm not this jim.

  3. monroe17
    • 3 years ago
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    LOL! okay :) is there a specific format the answer needs to be in?

  4. Zarkon
    • 3 years ago
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    I wouldn't use 'their composition'

  5. Zarkon
    • 3 years ago
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    'their difference' would be better

  6. jim_thompson5910
    • 3 years ago
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    You are correct, all polynomials are continuous Sine is continuous, so sin(x) - x^2 + x is continuous f(1) = 0.841 and f(2) = -1.091 So because of the sign change and because f(x) is continuous, there is a number c such that f(c) = 0 since f(1) < 0 < f(2)

  7. monroe17
    • 3 years ago
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    Thank you guys!

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