If $f(x) = 2x^3 - 15x^2+36x+1,x ∈ [1,5]$, then the absolute minimum value of $f(x)$ is:

24

The correct answer is Option (3) → 24

$f(x)=2x^{3}-15x^{2}+36x+1,\ x\in[1,5]$

$f'(x)=6x^{2}-30x+36=6(x-2)(x-3)\ \Rightarrow\ \text{critical points }x=2,3\in[1,5]$

Evaluate at $x=1,2,3,5$:

$f(1)=2-15+36+1=24$

$f(2)=16-60+72+1=29$

$f(3)=54-135+108+1=28$

$f(5)=250-375+180+1=56$

Absolute minimum value $=24$ (at $x=1$)