The function is of the form
f(x) = ax-2^xf(x)=ax−2x
and so
f^'(x) = a-(ln 2) 2^x
Thus f^'(b) = 0 implies
0 = a-(ln 2) 2^b
2^b = a/ln 2 implies b = log_2(a/ln 2) = ln(a/ln2)/ln 2
To complete the answer we need to determine the value of a explicitly.
a = root{3}{6+root{3}{6+root{3}{6+...}}} implies
a^3 = 6+root{3}{6+root{3}{6+root{3}{6+...}}} = 6+a implies
a^3-a-6 = 0 implies
a^3-2a^2 +2a^2-4a +3a-6 = 0 implies
(a^2+2a+3)(a-2) = 0
Since a^2+2a+3=0 can not have real roots, we must have a = 2
Thus
b = ln(2/ln 2)/ln 2 = frac{ln 2 - ln(ln2)}{ln 2} implies
b = 1-ln(ln 2)/ln 2
