Advanced Mean Value Theorem - Part 1

We solve the problem using two different functions. One for the case $f$ doesn’t have any zeros in $\left( 0,1 \right)$ and another one in the opposite case.

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Case 1 : Suppose that for the given function $f$ it holds $f\left( x \right)\ne 0$ , $\forall x\in \left( 0,\ 1 \right)$ .
Then, the function ${\color{#D61F06}{g\left( x \right)=\dfrac{1}{f\left( x \right)}-x}}$ can be defined on $\left[ 0,\ 1 \right]$ , $g$ is continuous within the interval $\left[ 0,\ 1 \right]$ , differentiable within the interval $\left( 0,\ 1 \right)$ , with ${g}'\left( x \right)=-\dfrac{{f}'\left( x \right)}{{{f}^{2}}\left( x \right)}-1$ and $g\left( 0 \right)=g\left( 1 \right)=1$ . Hence, by Rolle’s theorem , there exists a $\xi\in \left( 0,\ 1 \right)$ such that
${g}'\left( \xi \right)=0\Rightarrow -\dfrac{{f}'\left( \xi \right)} {{{f}^{2}}\left( \xi \right)}-1=0\Rightarrow {f}'\left( \xi \right)+{{f}^{2}}\left( \xi \right)=0$

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Case 2 : Suppose the given function $f$ has at least one zero ${{x}_{0}}\in \left( 0,\ 1 \right)$ .
Define ${\color{#20A900}{h\left( x \right)=f\left( x \right)\cdot {{e}^{F\left( x \right)}}}}$ on the interval $\left[ 0,\ 1 \right]$ , where $F\left( x \right)=\int\limits_{0}^{x}{f\left( t \right)}dt$ , $\forall x\in \left[ 0,\ 1 \right]$ .

By the Fundamental Theorem of Calculus $F$ is differentiable with ${F}'\left( x \right)=f\left( x \right)$ . Consequently, $h$ is differentiable (thus, continuous) within the interval $\left[ 0,\ 1 \right]$ , with $\begin{aligned} {h}'\left( x \right) & ={f}'\left( x \right)\cdot {{e}^{F\left( x \right)}}+f\left( x \right)\cdot {{\left( {{e}^{F\left( x \right)}} \right)}^{\prime }} \\ & ={f}'\left( x \right)\cdot {{e}^{F\left( x \right)}}+f\left( x \right)\cdot {F}'\left( x \right){{e}^{F\left( x \right)}} \\ & ={f}'\left( x \right)\cdot {{e}^{F\left( x \right)}}+{{f}^{2}}\left( x \right)\cdot {{e}^{F\left( x \right)}} \\ & =\left( {f}'\left( x \right)+{{f}^{2}}\left( x \right) \right)\cdot {{e}^{F\left( x \right)}} \\ \end{aligned}$

Moreover, $h\left( 0 \right)=1$ , $h\left( 1 \right)=\frac{1}{2}{{e}^{F\left( 1 \right)}}>0$ and $h\left( {{x}_{0}} \right)=f\left( {{x}_{0}} \right)\cdot {{e}^{F\left( {{x}_{0}} \right)}}=0$ .
By the extreme value theorem , $h$ has a minimum value which, being non positive, is evidently attained at some $\xi \in \left( 0,\ 1 \right)$ .

By Fermat’s theorem for Extrema (Interior Extremum Theorem) it holds

${h}'\left( \xi \right)=0\Rightarrow \left( {f}'\left( \xi \right)+{{f}^{2}}\left( \xi \right) \right)\cdot {{e}^{F\left( \xi \right)}}=0\Rightarrow {f}'\left( \xi \right)+{{f}^{2}}\left( \xi \right)=0$

Therefore, whichever case we are given, there exists a $\xi \in \left( 0,\ 1 \right)$ such that ${f}'\left( \xi \right)+{{f}^{2}}\left( \xi \right)=0$ .

Let $a$ be the smallest positive root of $f(x)=\frac12$ . From the given information, there is such an $a$ , and $a\le1$ . Also, by the intermediate value theorem, $f(x)>\frac12$ for $x \in (0,a)$ .

Define $g(x)=\frac{1}{f(x)}$ on the interval $[0,a]$ , so that $g(0)=1$ , $g(a)=2$ and $g(x)$ inherits differentiability. Then, by the mean value theorem, there exists a $\xi \in (0,a)$ such that $g'(\xi)=1$ .

But $g'(\xi)=-\frac{f'(\xi)}{f(\xi)^2}$ so this implies that $f'(\xi)+f(\xi)^2=0$ as required.

If we solve this first-order, separable ODE for $y = f(\xi)$ we get:

$f'(\xi) = -f^{2}(\xi), f(0)=1, f(1)=\frac{1}{2}$ ,

or $\frac{dy}{y^2} = -d\xi$ ;

or $-\frac{1}{y} = -\xi + C$

which both of the above boundary conditions yield $C=-1$ , or $\boxed{y = f(\xi) = \frac{1}{\xi + 1}}.$ This function is indeed continuous and differentiable for all $0 \le \xi \le 1 \Rightarrow$ there will always be $\xi \in (0,1)$ that satisfies the original ODE above.

Since $f(x)$ is differentiable for all real numbers, it must be continuous for all real numbers. Therefore, it satisfies the requirements for the Mean Value Theorem, meaning that there exists some $c \in (0, 1)$ such that $f'(c) = \frac{f(1) - f(0)}{1 - 0} = -\frac{1}{2}$ (since $c$ is in the interval $(0, 1)$ then we can just denote it as $\xi$ ). If we assume $f^2(\xi) + f'(\xi) = 0 \rightarrow f'(\xi) = -f^2(\xi)$ , then $-f^2(\xi) = -\frac{1}{2}$ or $f(\xi) = \frac{\sqrt2}{2}$ . Because $f(x)$ is continuous on the interval $(0, 1)$ and has a range of $(\frac{1}{2}, 1)$ , regardless of our choice of $f(x)$ , there exists a $\xi$ that satisifies $f^2(\xi) + f'(\xi) = 0$ . If the value we calculated for $f(\xi)$ did not fall in the range of $f(x)$ for the domain $(0, 1)$ , the property would not apply.