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I need your help. I don't know how to translate exactly the math problem I have been given for homework, since English is not my mother language, so I would really appreciate it if you didn't judge me.
Function $f:[0, \pi] \rightarrow \Bbb R$ is given, which is continuous on $[0, \pi]$ and differentiable on $(0, \pi)$. Prove that there is a $x_0 \in (0, \pi)$ such as $f'(x_0)=-f(x_0) cosx_0$.
I hope someone will understand what I mean. Thank you a lot.
Rolle's Theorem
Suppose $f(x)$ is a function which satisfies the following:
(i) $f(x)$ is continuous on the closed interval $[a,b]$.
(ii) $f(x)$ is differentiable on the open interval (a,b).
(iii) $f(a)=f(b)$
Then there is a number $c$ such that $a<c<b$ and $f′(c)=0$. Or, in other words $f(x)$ has a critical point in $(a,b)$.
We are given $f(x)$, and we know that $f(x)$ is continuous on $[0,\pi]$ and differentiable on $(0,\pi)$. We need to find a suitable $x_0\in(0,\pi)$ so that $f'(x_0)= −f(x_0)\cos (x_0)$.
My original analysis was to find $f(x)$ by solving
$$f'(x)=-f(x) \cos(x)$$
which forms
$$f(x)=ce^{-\sin(x)}$$
so if we let $c=1$
$$f'(x)=-e^{-\sin{x}}\cos(x)=-f(x)\cos(x)$$
Then, $f(0)=c=f(\pi)$. Furthermore, if $x_0={\pi}/{2}$ then $x_0\in(0,\pi)$ and
$$f'\left(\frac{\pi}{2}\right)=-ce^{-\large{\sin\left(\pi/2\right)}}\cos\left(\frac{\pi}{2}\right)=0$$ therefore by Rolle's Theorem there is an $x_0$ such that $0<x_0<\pi$ and $f′(x_0)=0$ which is $x_0=\pi/2$.
However, this analysis is incorrect as we cannot assume that $f(x)=e^{-\sin(x)}$. There are an infinite amount of other choices for $f(x)$ where $f(0)\neq f(\pi)$. Two such examples are $f(x)=2x$ and $f(x)=\cos(x)$. Since we cannot guarantee that $f(x)$ satisfies $f(0)=f(\pi)$, it follows that we cannot apply Rolle's Theorem since we don't know that $f(0)=f(\pi)$ for all suitable functions $f(x)$. Therefore, the the problem is not well defined and we cannot guarantee that there exists a $x_0\in(0,\pi)$ such that $f'(x_0)=0$ for every possible $f(x)$.
Claim Problem, as stated, is not well defined. Proof For function $e^x$ there is no $x_0\in (0,\pi)$ as stated.
Proof of the OP's question, under the condition $f(0)=f(\pi)$ (which i suppose was forgotten).
We want to prove that there exists root of the equation $f'(x)+f(x)\cos x=0$, equivalently, of $(f(x)e^{\sin x})'=0$. So take $g(x):=f(x)e^{\sin x}$, which satisfies Rolle's theorem's conditions and therefore there is a root of its derivative, as wanted.
Comment Quantifiers exists and for all play a major role in math, and appear more often as one would expect. Treatment is different in each case. For example take the expression $p(x)\equiv ax^2+bx+c=0$. If we need to prove that there exists $x$ such that $p(x)=0$, then one gets the root given by the known formulas. If we have for all $x$ that $p(x)=0$, then $a=b=c=0$.
Back to the original problem. It might seem tempting to see there exists $x$ such that $g'(x)=0$ as for all $x$ $g'(x)=0$. This would lead to a new problem, having nothing to do with the original question, of finding $g$. In our case, one would get $f(x)=ce^{\sin x}$, while $f$ is already given (although we don't know its formula).
Consider a function $f:[0, \pi] \rightarrow \Bbb R$ given by f(x) = sin(x)
f is continuous on [0,π]
f is differentiable on (0,π)
f(0)=0=f(π)
Thus , by Rolle's theorem there is a X° such that f'(X°) =0
Implies Cos(X°)=0 , this implies X°=π/2
Hence X° =π/2 Satisfies f'(X°)= - f(X°)Cos(X°)
Cos(π/2) = -Sin(π/2)Cos(π/2)
