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

1. $f'(x)=6x^2-30x+36=6(x-2)(x-3)$, critical points $x=2,3$. 2. Evaluate at critical points and endpoints: $f(1)=24$, $f(2)=29$, $f(3)=28$, $f(5)=56$. 3. The largest value is $56$ at $x=5$. 4. Trap $29$ is only the local-maximum value, not the absolute maximum. _Source: syllabus Mathematics Ch 6 "Application of Derivatives", p.26_