Unbounded Binary Search Example (Find the point where a monotonically increasing function becomes positive first time)

Given a function 'int f(unsigned int x)' which takes a non-negative integer 'x' as input and returns an integer as output. The function is monotonically increasing with respect to the value of x, i.e., the value of f(x+1) is greater than f(x) for every input x. Find the value 'n' where f() becomes positive for the first time. Since f() is monotonically increasing, values of f(n+1), f(n+2),... must be positive and values of f(n-2), f(n-3), … must be negative. 
Find n in O(logn) time, you may assume that f(x) can be evaluated in O(1) time for any input x. 

A simple solution is to start from i equals to 0 and one by one calculate the value of f(i) for 1, 2, 3, 4 … etc until we find a positive f(i). This works but takes O(n) time.
Can we apply Binary Search to find n in O(Logn) time? We can't directly apply Binary Search as we don't have an upper limit or high index. The idea is to do repeated doubling until we find a positive value, i.e., check values of f() for following values until f(i) becomes positive.

 f(0) 
 f(1)
 f(2)
 f(4)
 f(8)
 f(16)
 f(32)
 ....
 ....
 f(high)
Let 'high' be the value of i when f() becomes positive for first time.

Can we apply Binary Search to find n after finding 'high'? We can apply Binary Search now, we can use 'high/2' as low and 'high' as high indexes in binary search. The result n must lie between 'high/2' and 'high'.
The number of steps for finding 'high' is O(Logn). So we can find 'high' in O(Logn) time. What about the time taken by Binary Search between high/2 and high? The value of 'high' must be less than 2*n. The number of elements between high/2 and high must be O(n). Therefore, the time complexity of Binary Search is O(Logn) and the overall time complexity is 2*O(Logn) which is O(Logn). 
 

Output : 

The value n where f() becomes positive first is 12

Time Complexity: O(logn)
Auxiliary Space: O(logn)