Answer :
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
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