How can I prove (or show) that f has a unique positive root, if ?
Yeah I thought about that, but I think that the function could take more than 1 zero, say in a very little interval close the another zero. I know it won't happen in this example, but it would be possible in another one. In other words it wouldn't be a rigorous proof of the uniqueness of a positive zero. During the night I thought about finding more or less where is the zero and then chose a close interval to it and derivative the function to check its sign. For showing that the function is strictly increasing or decreasing in the little interval, then it cross only once the x axis.
I could spend some time today on the problem. Finally it wasn't that easy, but ain't hard neither. As I didn't have a calculator, it was too long, because without the calculator it's more complicated.
First, we notice that . From this, we know that our function is bounded by the 2 lines. Equaling them to 0, we find that for the first and for the second. This implies that has at least a zero between and , but as we only consider positive roots in the exercise, we can say that f has at least a zero on .
To be rigorous, I don't see anything else than finding the derivative of and to study its sign over . But before this, I will make a verification. As I assumed had a zero on , I can verify it multiplying by . If the result is negative, then we're right assuming it has at least a zero on . So which is negative.
Now .
when .
When , and when , . Therefore we conclude that is increasing on and decreasing on , which implies that cross only once the x axis somewhere on .
Edit : I didn't understand wellyou could consider the coordinates of the turning points of the function f(x) = 4 sin(x) + 1 - x for x > a, where a > 0 and f(a) = 0 for the first time