Apply mean value theorem f(x)=7-(6/x) on [1,6]
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
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