Proving that a function is a subset of an interval for specific x-values

Question

I got the following function $f: x \rightarrow x^3-3x^2+2x$ (dealing only with real numbers here).

Now I have to prove that $$\{f(x):x\in [1,2]\} \subseteq [-1/2, 0]$$

I have not really learned a strategy to do so. Any hints?

Answer 1

Hint: Find extrema of your function by solving $f'(x)=0$. Also look if your funciton is increasing or decreasing in for $x \in [1,2]$. Then calculate $f(1)$ and $f(2)$. Can you complete it from here?

$f'(x)=3x^2-6x+2=0 \implies x_{1/2}=\frac{6\pm \sqrt{6^2-4\cdot 3\cdot 2}}{2\cdot 3}=1\pm\frac{\sqrt{3}}{3}$. Only $1+\sqrt{3}/3 \in [1,2]$. What kind of extrema is this (use the second derivative test, it should be a minima,

calculate $f(1+\sqrt{3}/3)$)? $f(1)=0$ and $f(2)=0$