f(x) = 3x log2 x
f(2) = 3(2)(1) = 6
f(1) = 3(1)(0) = 0
so the line joining the endpoints has slope = 6
MVT states there is a c between 1 and 2 such that f'(c) = 6
f'(x) = 3log2 x + 3x/(x ln 2) = 3(log2 x + 1/ln2) = 3/ln2 (lnx + 1)
so, we set f'(c) = 6 and solve for c:
6 = 3/ln2(ln c + 1)
ln c + 1 = 2ln2
ln c = 2ln2 - 1
c = e^(2ln2-1) = 4/e = 1.47
which is in the interval [1,2]
use the mean value theorem to find the c's on the open interval (a,b) such that fprime(c)= (f(b)-f(a))/(b-a)
f(x)= 3xlog(base 2)x , [1,2]
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