For $f(x)=2x^3-15x^2+36x+1$ on $[1,5]$, the absolute minimum value is

1. Critical points from $f'(x)=6(x-2)(x-3)$ are $x=2,3$. 2. Values: $f(1)=24$, $f(2)=29$, $f(3)=28$, $f(5)=56$. 3. The smallest is $24$ at the endpoint $x=1$. 4. Trap $1$ is the constant term, not a computed value. _Source: syllabus Mathematics Ch 6 "Application of Derivatives", p.26_