# Rolle's Theorem 1

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.

• Is the function $f:[-1,1]\rightarrow\mathbb{R}$ or $f:[0,\pi]\rightarrow\mathbb{R}$? What is XoE and conXo? – Michael Nov 14 '19 at 22:37
• Your English is fine. What I don't get is: f'(Xo)=-f(Xo)conXo. What I guessed didn't work... – Roman Hric Nov 14 '19 at 22:37

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 $$a and $$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 $$0 and $$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 $$f(x)$$.

• According to the OP's question, "function f" is given and we are looking for $x_0$ that does the job. We are not looking for a function that does the job (if so, sin x would be a good choice, but that's not the case). – Nikolaos Skout Dec 27 '19 at 22:01
• I see and have updated my answer. – Axion004 Dec 27 '19 at 23:54
• Too long for a comment. Please check the answer I posted, including the comment. – Nikolaos Skout Dec 28 '19 at 9:33

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)