I'm having some issues regarding multiplicities in terms of determining and proving how many roots there are for a function.
Take for instance: $x^3 - 3x + 2$.
We can see that $\ f'(x) > 0, \ x\in (-\infty, -1) \cup (1, \infty)$ and that $\ f'(x) < 0, \ x\in(-1, 1)$. Since $f(x)$ is strictly increasing or decreasing on those intervals there can be at most one root in each interval. Upon further analysis it is clear that since $f(-3) = -16$ and $f(-1) = 4$ so by the Intermediate Value Theorem, $f(x)$ has a real root for $x \in (-3, -1)$.
However if we look at $f(-1) = 4$ and $f(1) = 0$ we can't say whether there is a root here. Likewise for $x \in (1, \infty)$. My question is that I am not sure what to say about the number of roots at the multiplicity and how to prove such existence.
Clearly $f(1) = 1 - 3 + 2 = 0$ is a root. Not sure how to go on from there.
hint
$$f (x)=x^3-x-2x+2$$
$$=x (x-1)(x+1)-2 (x-1)$$
$$=(x-1)(x^2+x-2) $$
$$=(x-1)(x^2-x+2x-2) $$
$$=(x-1)^2 (x+2)$$