Consider the function f(x) = 3x - 6 x + 3 2x - 3x defined on the interval [0, ] . The sum of the lengths of the sub-intervals where f(x) is strictly increasing is:

Step-by-step solution

Let f(x) = 3x - 6 x + 3 2x - 3x . Differentiating with respect to x , we get: f'(x) = 3 - 6 x + 6 2x - 3 3x Using the multiple-angle identities 2x = 2 ^2 x - 1 and 3x = 4 ^3 x - 3 x , we can express f'(x) entirely in terms of x : f'(x) = 3 - 6 x + 6(2 ^2 x - 1) - 3(4 ^3 x - 3 x) f'(x) = 3 - 6 x + 12 ^2 x - 6 - 12 ^3 x + 9 x f'(x) = -12 ^3 x + 12 ^2 x + 3 x - 3 Factorizing by grouping terms: f'(x) = -12 ^2 x( x - 1) + 3( x - 1) f'(x) = (3 - 12 ^2 x)( x - 1) f'(x) = 3(1 - 4 ^2 x)( x - 1) For f(x) to be strictly incre