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