[Math] Prove that $x^3 -3x^2 +6 = 0$ has only one real root

calculusderivativesnumerical methods

I know that if I take the derivative of
$$x^3 -3x^2 +6 = 0$$

and prove it is always greater than zero, I'll find that this functions is always increasing, and therefore if I find an interval where the function has a sign before it number, and then switches sign after it, that's where the root is.

The exercise asks me to find an interval with 'size' 1. Is there a good method to find such interval, that does not require me to try a lot of values?

Best Answer

Actually the derivative here is not always positive! So you'll need something different to show there are not multiple real roots, such as finding the stationary points and note that the function values there have the same sign.

Usually you wouldn't need to find an explicit interval here -- it is enough to know that because the dominant term is $x^3$ (with an odd degree) the function goes towards $-\infty$ as $x\to-\infty$ and towards $\infty$ as x\to\infty$. So it must take both negative and positive values.

What the exercise asks you to is in effect to approximate the root to a precision of $1$. You could use any numeric root-finding procedure for this -- if simple trial-and-error with small integers doesn't seem to work, Newton-Raphson starting at (arbitrarily) $-1$ would probably quickly reach an approximation that's not more than $1$ from the actual root.

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