Minimum of a function using framing rather than derivatives

Question

Is it possible to get the minimum of the function: $f(x) = 2+x^2(3-x)$ in the intercept$ [1, 4]$ by framing, that is you start by $ 1 \leq x \leq 4$ until you get to the minimum.

Answer 1

Without derivatives: $$2+x^2(3-x)=4x^2-x^3+2-x^2=x^2(4-x)+(4-x)(4+x)-14\geq-14.$$ The equality occurs for $x=4$, which says that we got a minimal value.

Answer 2

Find minimum of $2-x^2(x-3)$, $x \in [1,4]$.

$f(x):=x^2(x-3) \ge 0$ for $x \in [3,4].$

$f(3)=0$; $f(4)= 16$;

Show $f(x)$ is increasing in $[3,4]$;

Let $a > b$, $a,b \in [3,4]$:

$f(a)-f(b)=$

$a^3-b^3 -3(a^2-b^2)=$

$(a-b)(a^2+ab+b^2) -3(a-b)(a+b)=$

$(a-b)(a^2+ab+b^2-3(a+b))\gt$

$(a-b)(3^2+3^2+3^2-3(4+4))=$

$(a-b)(27-24) >0$, hence

$f$ strictly increasing in $[3,4]$.

Maximum at $x=4$: $f(4)=16$.

$\min_{x \in [1,4]} (2-f(x))=$

$2-16=-14.$