Forf(x)=(x−1)2/3 , the mean value theorem is applicable tof(x) in the interval
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a
[2,4]
b
[0,2]
c
[−2,2]
d
Any finite interval
answer is A.
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Detailed Solution
Sincef'(x) does not exist atx=1 , ( f'(x)=23(x−1)−1/3) therefore, mean value theorem is not applicable to the given function in any interval which contains 1 as an interior point. SO, in this case the required interval is[2,4] .