Posted by Jody on Sunday, June 24, 2012 at 5:25pm.
f(1) = 1
f(6) = 6
[f(6)-f(1)]/(6-1) = 5/5 = 1
f'(x) = 6/x^2
1 = 6/x^2 ==> x = √6
so, since 1 < √6 < 6, the MVT applies
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