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alrightatmaths

  • one year ago

f(x) = x^3 + 8x - 19 Show that the equation f(x)=0 has only one real root.

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  1. Spacelimbus
    • one year ago
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    I can only recommend the Newton-Method or the Method of Bisection for this problem.

  2. ZeHanz
    • one year ago
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    It is a polynomial of the 3rd degree. Look at the end-behaviour of the function: The limit of f(x) if x => infinity is also infinity. The limit of f(x) if x => -infinity is also -infinity. f is continuous for every x, so it must be 0 at least one time. Now look at the derivative: f'(x)=3x²+8. What do you know about the possible values of f' ?

  3. alrightatmaths
    • one year ago
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    Yeah, you know it has to have one because there is a change in sign basically. And are you saying that f'(x) will never be zero so there's no turning point?

  4. alrightatmaths
    • one year ago
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    Or that it's always positive.

  5. ZeHanz
    • one year ago
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    That is just what I wanted to say ;) So f is always increasing...

  6. alrightatmaths
    • one year ago
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    Good man!

  7. ZeHanz
    • one year ago
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    YW!

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