Hi
Question:
Show that only takes two values. Show this, and decide these values.
Is this a correct/complete solution?
My solution:
, so is a constant function on
Then take any , say .
. This is the only value f takes on the interval since f is constant on this interval.
In the same way, we show that for , .
Thus the values are
Thx for your input
But no one said that f was constant on . What is true is that the derivative of f is 0 on that set. That tells us that f is constant on any interval on which it is differentiable. Thus, f is constant on and on but not necessarily the same constant on those two intervals.
Here is another solution using calculus. (Since this topic is started in the Calculus section, I presume it is okay to use calculus. )
Hence is constant over any interval on which it is continuous. Since it is continuous over and over determine the constant value of in each of these two invervals. For example, for we have and so for all Do the same thing for the negative interval.