Min. of function.

2^{17}-2 ?

130050

modulus inequalities might help.
hint: sum ( -1ⁱ * (x-x_i))<= sum (/(x-x_i)/)

I made a mistake in the final expression: It should be 2^{17}+2-2^{10} = 130050

We have from the triangle inequality, |x-a|+|b-x| ≥ (b-a) for all x in [a,b]

Applying this

|x-2|+\left|2^{16}-x\right| \ge 2^{16}-2 \\ \\ \left|x-2^2\right| + \left|2^{15}-x\right|\ge 2^{15}-4 \\ \\ .\\ .\\ .\\ \left|x-2^8\right| + \left|2^{9}-x\right|\ge 2^{9}-2^8

Thus LHS \ge 2^{16}-2 + 2^{15}-4+...+2^{9}-2^8 = 130050

The minimum will occur at the intersection of the intervals

\left[2,2^{16}\right], \left[2^2, 2^{15}\right],...,\left[2^8, 2^{9}\right]

which is obviously the interval \left[2^8, 2^{9}\right]

f is max at x = infinity

so f wil be minimum wen f'(x)= 0 , i.e wen f is independent of x

f is independent of x for all x in ( 2⁸, 2⁹) as first 8 of the mods will giv x-2^k and other 8 wil giv 2^m-x

so x gets cancelled and hence f is independent of x

so f _min = - ( 2+2²+2³+..2⁸) + (2⁹+ 2¹⁰+...2¹⁶)

= 2⁹(2⁸-1) - 2(2⁸-1)=2(2⁸-1)²

Thanks to bhatt Sir and Qwerty.