f′(x)=6x2−24=0⇒x=±2. Critical point in [1,3] is x=2.
f(1)=2−24+107=85, f(2)=16−48+107=75, f(3)=54−72+107=89.
Maximum is 89 at x=3.
The maximum value of 2x3−24x+107 in the interval [1,3] is :
Held on 22 May 2023 · Verified 13 Jul 2026.
87
89
90
85
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