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

The correct answer is option (2) → 89

$y=2x^3-24x+107$

$\frac{dy}{dx}=0⇒6x^2-24=0$

$x=±2$

checking at $x=1,2,3$ as $-2∉[1,3]$

$y(1)=85,y(2)=75,y(3)=89$

$y(3)=89$ → max. value