1. ## Analysis simple problem

How can I prove (or show) that f has a unique positive root, if $f(x)=4sin(x)+1-x$?

2. Originally Posted by arbolis
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.

3. 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.

4. Originally Posted by arbolis
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 .......

5. 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, $f'(x)>0$ 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