Find first missing number in an array containing consecutive numbers

Given a sorted array with no duplicates find the first number that is missing in the sequence.
For an input array where no number is missing, return 1 + largest element in the array.

Example 1: [1, 2, 3, 4, 7] 
The first missing number is 5.

Example 2: [-1, 0, 1, 2, 3, 4, 5, 6, 7, 9] 
The first missing number is 8.

Example 3: [1, 2, 3, 4, 5, 6] 
There is no missing number. So return 7.

We are aware of binary search technique to find a number in sorted array. But is there a way to extend it to find the missing number? The idea here is to locate the first missing index and we can do that using binary search. A number is missing in the left for an array x, if x[0] + index < x[index], otherwise it's missing in the right. The binary search will eventually lead to to consecutive indices where the number is missing or to the end of the array. In the latter case the result is the largest number in the array + 1.

int find_first_missing_number(int x[], int n) {

    int a = x[0];
    int lb = 0;
    int ub = n - 1;

    // Narrow down the range where the number may be located
    while(ub - lb > 1) {
        int mid = lb +(ub-lb)/2;

        if (x[mid] == a + mid) {
            lb = mid;
        } else if (x[mid] > a + mid) {
            ub = mid;
        }
    }

    // At this point lb + 1 should be equal to ub and
    // if the number NOT missing then x[lb] + 1 should be 
    // equal to x[ub]
    // if the number is missing then x[lb] + 1 is the first
    // missing number
    if (x[lb] + 1 == x[ub])
        return x[ub] + 1;

    return x[lb] + 1;
}