Here's the question you clicked on:
alrightatmaths
f(x) = x^3 + 8x - 19 Show that the equation f(x)=0 has only one real root.
I can only recommend the Newton-Method or the Method of Bisection for this problem.
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' ?
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?
Or that it's always positive.
That is just what I wanted to say ;) So f is always increasing...