A) 75
B) 89
C) 125
D) 139
Correct Answer: D
Solution :
Let \[f(x)=2{{x}^{3}}-24x+107\] At \[x=-3,\ f(-3)=2{{(-3)}^{3}}-24(-3)+107=125\] At \[x=3,\ \ f(3)=2{{(3)}^{3}}-24(3)+107=89\] For maxima or minima, \[{f}'\,(x)=6{{x}^{2}}-24=0\] \[\Rightarrow x=2,\ \ -2\] So at \[x=2,\ f(2)=2{{(2)}^{3}}-24(2)+107=75\] at \[x=-2,\ \ f(-2)=2{{(-2)}^{3}}-24(-2)+107=139\] Thus the maximum value of the given function in [? 3, 3] is 139.You need to login to perform this action.
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