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    MHF Contributor arbolis's Avatar
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    Analysis simple problem

    How can I prove (or show) that f has a unique positive root, if f(x)=4sin(x)+1-x?
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    Quote Originally Posted by arbolis View Post
    How can I prove (or show) that f has a unique positive root, if f(x)=4sin(x)+1-x?
    Intermediate value theorem. Find a a so that f(a) < 0 and a b so that f(b)>0 then the function must take a zero in that interval.
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    MHF Contributor arbolis's Avatar
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    Find a so that and a so that then the function must take a zero in that interval.
    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 f is strictly increasing or decreasing in the little interval, then it cross only once the x axis.
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    Quote Originally Posted by arbolis View Post
    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 f is strictly increasing or decreasing in the little interval, then it cross only once the x axis.
    To establish uniqueness of the positive root I think you 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 .......
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    MHF Contributor arbolis's Avatar
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    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 -4\leq 4sin(x)\leq 4 \Rightarrow -3-x \leq 4sin(x)+1-x \leq 5-x. From this, we know that our function f is bounded by the 2 lines. Equaling them to 0, we find that x=-3 for the first and x=5 for the second. This implies that f has at least a zero between x=-3 and x=5, but as we only consider positive roots in the exercise, we can say that f has at least a zero on [0,5].
    To be rigorous, I don't see anything else than finding the derivative of f and to study its sign over [0,5]. But before this, I will make a verification. As I assumed f had a zero on [0,5], I can verify it multiplying f(0) by f(5). If the result is negative, then we're right assuming it has at least a zero on [0,5]. So f(0)*f(5)=1*(4sin(5)-4) which is negative.
    Now f'(x)=4cos(x)-1.
     f'(x)=0 when x=arcos(\frac{1}{4}).
    When x<arcos(\frac{1}{4}) \approx 1,32, f'(x)>0 and when  x>arcos(\frac{1}{4}), f'(x)<0. Therefore we conclude that f is increasing on [0,arcos(\frac{1}{4})] and decreasing on [arcos(\frac{1}{4}),5], which implies that f cross only once the x axis somewhere on [0,5].

    Edit : I didn't understand well
    you 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
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