Prove that $x^3 -3x^2 +6 = 0$ has only one real root 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? 
 A: 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.
A: Sketching the graph of the function might be a good start. 
The roots of the first derivative 
$$
f'(x)=3x(x-2)
$$
gives you the monotone intervals of $f(x)=x^3-3x^2+6$:
$$
(-\infty,0),\quad  (0,2),\quad (2,\infty)
$$
You the tell the monotonicity in these interval by the signs of the derivative. Checking the values of $f$ at the end points and using intermediate value theorem would give you a proof. 

To see how your "intuition" is wrong, you can just google "x^3-3x^2+6":

A: let $f$ be defined by $f(x) = x^3 - 3x^2 + 6$ basic descartes rule of sign tells there is one negative root and either two or zero positive roots. we need to rule out that there are no positive roots. now division by $(x-3)$ shows that $f(x) \ge 9 \mbox{ for} x \ge 3$ 
to make further progress, i need to use some calculus. the critical numbers are $0$ and $2.$ so the global minimum of $f(x), 0 \le x < \infty = f(2) = 2 > 0$. 
therefore, $f$ has no positive zero and only one negative zero.
