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The maximum value of 2x3−24x+1072x^3 - 24x + 107 in the interval [1,3][1, 3] is :

Solution

✅ Correct Option: 2

f′(x)=6x2−24=0⇒x=±2f'(x)=6x^2-24=0 \Rightarrow x=\pm 2. Critical point in [1,3][1,3] is x=2x=2.

f(1)=2−24+107=85f(1)=2-24+107=85, f(2)=16−48+107=75f(2)=16-48+107=75, f(3)=54−72+107=89f(3)=54-72+107=89.

Maximum is 8989 at x=3x=3.

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