max. value is 139 at x = −2 and min. value is 89 at x = 3

F^’(x)=6x²-24=0

6(x²-4)=0

6(x²-2²)=0

6(x-2)(x+2)=0

X=2,-2

Now, we shall evaluate the value of f at these points and the end points

F(2)=2(2)³-24(2)+107=75

F(-2)=2(-2)³-24(-2)+107=139

F(-3)=2(-3)³-24(-3)+107=125

F(3)=2(3)³-24(3)+107=89