Semi sorted binary search
I spend some more time refining the solutions. Also found a relevant question that asks to find the min/max of the array here.
Problem Context
Searching a value in an array can be up to O(n) in time complexity - you have to iterate through the array and check for each element if it's equal the search value. But - if the array is sorted, this is reduced to O(log(n)) due to a strategy called Binary Search. Binary search allows us at each time cut half of the options to consider. You check the middle element in the array, if the search value is bigger you can discard the lower half, if it's lower you can discard the upper half, and now you do the same for the remaining array - until you either find the value or end up with an empty/1-element array - in that case the element is not in the array.
Here there is a problem that the array is swapped - the question is can we still reduce the time complexity to O(log(n)) ?
The key insight here is that even though the array itself is not perfectly sorted, at least half of it IS - and so we can always tell something about that half, and reduce our scope. This is true for every sub-division we make of the array - at least half of it will be sorted (and in some cases, both halves).
If we are looking for the min/max value, we basically don't care about the part that is sorted, because the pivot value will be in the other half (though the mid value should be included, as it can be the min/max value) - so we can discard the sorted part.
If we are looking for a specific value - we can automatically check if our value is in the range of the sorted half, if it is we discard the non-sorted half, and focus on the sorted. If it's not we focus on the non-sorted - and do the same.
Find Min value
Here is a solution in python - here we check if the right half is sorted. If it is, we reduce our scope to the left half. If not, we reduce our scope to the right half (note that in that case we can safely discard the mid point and move beyond it).
def findMin(arr):
    start = 0
    end = len(arr) - 1
    while(start &lt; end):
        mid = (start + end) // 2
        rightSorted = arr[mid] &lt; arr[end]
        if rightSorted:
            end = mid
        else:
            start = mid + 1
    return arr[start]

The decision to check the right is not arbitrary - it is necessary in case we search for a min value. If we instead search the left, and find out it is sorted - we could not confidently discard that half, as the array might be perfectly sorted - in that case the 0th element is the minimum. So we choose the right in order to be completely sure we still have the min value in the sub array.
Find Max value
In python this is actually very easy - you just need to go one element before the minimum - and even if the min is in 0 position, python interprets -1 as the last element of the array. So you could just use the same code as above, but with return arr[start-1].
Alternatively, we could reverse the min function above:
def findMax(arr):
    start = 0
    end = len(arr) - 1
    while(start &lt; end):
        mid = (start + end) // 2
        leftSorted = arr[mid] &gt; arr[start]
        if leftSorted:
            start = mid
        else:
            end = mid - 1
    return arr[start]

Finding a specific value
If we want to check if a specific value exists in the array - the same key insight hold.
def findVal(arr, val):  
    start = 0
    end = len(arr) - 1    
    while(start &lt; end):
        mid = (start + end) // 2
        leftSorted = arr[start] &lt;= arr[mid]
        if leftSorted:
            if val &gt;= arr[start] and val &lt;= arr[mid]:
                end = mid
            else:
                start = mid + 1
        else:
            if val &gt;= arr[mid] and val &lt;= arr[end]:
                start = mid
            else:
                end = mid - 1
    return val == arr[start]

A few differences:

we have to check if our value is in the sorted half, and if not discard it.

we have to watch out for the case of 2 elements in an ordered array. The issue here is that if we are not careful, we can end up in an endless loop - because the mid point will be either the 1st or the 2nd, depending if we are using a floor or a ceil operation. Since the default python // operation is a floor, mid will be equal to start, and we have to check the left side in order to either choose the 1st or discard it. [if we use ceil, we should check the right side]