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

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Spacelimbus
 3 years ago
Best ResponseYou've already chosen the best response.0I can only recommend the NewtonMethod or the Method of Bisection for this problem.

ZeHanz
 3 years ago
Best ResponseYou've already chosen the best response.1It is a polynomial of the 3rd degree. Look at the endbehaviour 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' ?

alrightatmaths
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah, 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?

alrightatmaths
 3 years ago
Best ResponseYou've already chosen the best response.0Or that it's always positive.

ZeHanz
 3 years ago
Best ResponseYou've already chosen the best response.1That is just what I wanted to say ;) So f is always increasing...
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