Let I be an interval & f:I-->R be a continuous 1-to-1 function.Prove that f is either strictly increasing or

Question

strictly decreasing.

Answer 1

I don't know what level of math you are taking, but what you suggest can be easily done in calculus.

First, you take a derivative of the function f(x) and let it be f'(x)
Then, you take a second derivative of the function (x) and let it be f''(x)

Where f''(x)=0 is where the slope of the change becomes zero. As a consequence, it can either flaten out and go the same way or it changes direction as to either increasing or decreasing. If there is no solution to this equation, then it won't change directions at all. So your answer is simple.

If it has the solution, then it could flaten out for a while and head to different direction, or flat out and go to the same direction. Then, you have to define an interval between the zeros and the rest, plug the x value into the first and see if the answer is positive or negative. Postive means the function is increasing, and negative means the function is decreasing. You'll have to do this for all intervals and evaluate if they are consistant.

Answer 2

amplify f(x) by capacity of defining h(x) on [0,a million] as = 0 if x = 0 or a million, and f(x) in any different case. h(x) is non-end on [0,a million]. considering [0,a million] is compact (closed and bounded), so is extremely like [0,a million] under h, so it has a max and a min. If the max and min are the two 0, then h and f are the two consistent = 0, so it incredibly is a max and min complete by capacity of f. If the max or min at the instant are not 0, the max or min happens in (0,a million), so it incredibly is complete by capacity of f. enable g = -x(x-a million). That has zeroes at 0 and a million, so no minimum in (0,a million), and a max in between at a million/2.