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<title>Second order differentiability of the intermediate-point function in Cauchy’s mean-value theorem: Second order differentiability of the intermediate-point function in Cauchy’s mean-value theorem</title>
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<h1>Second order differentiability of the intermediate-point function in Cauchy’s mean-value theorem</h1>
<p class="authors">
<span class="author">Beatrix-Mihaela POP\(^\ast \) Dorel I. DUCA\(^{\ast \ast }\)</span>
</p>
<p class="date">February 20, 2015.</p>
</div>
<p>\(^\ast \)<i class="it">Babeş-Bolyai</i> University of Cluj-Napoca, Faculty of Mathematics and Computer Science, M. Kogălniceanu, no. 1, Cluj-Napoca, Romania, e-mail: <span class="tt">pop_mbeatrix@yahoo.com</span>. </p>
<p>\(^{\ast \ast }\)<i class="it">Babeş-Bolyai</i> University of Cluj-Napoca, Faculty of Mathematics and Computer Science, M. Kogălniceanu, no. 1, Cluj-Napoca, Romania, e-mail: <span class="tt">dduca@math.ubbcluj.ro; dorelduca@yahoo.com</span>. </p>
<p>Dedicated to prof. I. Păvăloiu on the occasion of his 75th anniversary </p>

<div class="abstract"><p> If the functions \(f,g:I\rightarrow \mathbb {R}\) are differentiable on the interval \(I\subseteq \mathbb {R}\), \(a\in I,\) then there exists a function \(\overline{c}:I\rightarrow I\) such that</p>
<div class="displaymath" id="a0000000002">
  \begin{equation*}  \left[ f\left( x\right) -f\left( a\right) \right] g^{\left( 1\right) }\left( \overline{c}\left( x\right) \right) =\left[ g\left( x\right) -g\left( a\right) \right] f^{\left( 1\right) }\left( \overline{c}\left( x\right) \right) ,\text{ for }x\in I. \end{equation*}
</div>
<p>In this paper we study the differentiability of the function \(\overline{c}\), when</p>
<div class="displaymath" id="a0000000003">
  \begin{equation*}  f^{\left( k\right) }\left( a\right) g^{\left( 1\right) }\left( a\right) =f^{\left( 1\right) }\left( a\right) g^{\left( k\right) }\left( a\right) ,\text{ for all }k\in \{ 1,...,n-1\}  \end{equation*}
</div>
<p>and</p>
<div class="displaymath" id="a0000000004">
  \begin{equation*}  f^{\left( n\right) }\left( a\right) g^{\left( 1\right) }\left( a\right) \neq f^{\left( 1\right) }\left( a\right) g^{\left( n\right) }\left( a\right). \end{equation*}
</div>
<p><b class="bf">MSC.</b> 26A24 </p>
<p><b class="bf">Keywords.</b> Cauchy’s theorem, intermediate point, mean-value theorem. </p>
</div>
<h1 id="a0000000005">1 Introduction</h1>
<p>The mean value theorem is a cornerstone of the differential calculus. <br />Cauchy’s theorem is one of the generalizations of the mean value theorem. </p>
<p>The purpose of this note is to extend the results by D.I. Duca and O. Pop <span class="cite">
	[
	<a href="#Dorel-Ovidiu-Lagrange" >1</a>
	]
</span> concerning the mean value theorem to Cauchy’s theorem. Also, the results can be considered to extend the results by D.I. Duca and O. Pop from <span class="cite">
	[
	<a href="#Dorel-Ovidiu-Cauchy" >2</a>
	]
</span>. </p>
<p>Cauchy’s theorem is usually presented in the following form: </p>
<p><div class="theorem_thmwrapper " id="TCauchy-c">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">1</span>
  </div>
  <div class="theorem_thmcontent">
  <p>\((\)A.L. Cauchy\()\) Let \(I\) be an interval in \(\mathbb {R}\), and let\(\  a\) be a point of \(I.\) If the functions \(f,g:I\rightarrow \mathbb {R} \) are differentiable on \(I\), then for each \(x\in I\backslash \{ a\} \) there exists a point \(c_{x},\) from the interval with extremities \(x\) and \(a\), such that</p>
<div class="displaymath" id="a0000000006">
  \begin{equation*}  \left[ f\left( x\right) -f\left( a\right) \right] g^{\left( 1\right) }\left( c_{x}\right) =\left[ g\left( x\right) -g\left( a\right) \right] f^{\left( 1\right) }\left( c_{x}\right) . \end{equation*}
</div>
<p>If, in addition, \(g^{\left( 1\right) }\left( x\right) \neq 0,\) for all \(x\in \mathrm{int}I,\) then \(g\left( x\right) \neq g\left( a\right) ,\) for all \(x\in I\backslash \{ a\} ,\) and</p>
<div class="equation" id="Cauchy-c-1">
<p>
  <div class="equation_content">
    \begin{equation}  \frac{f\left( x\right) -f\left( a\right) }{g\left( x\right) -g\left( a\right) }=\frac{f^{\left( 1\right) }\left( c_{x}\right) }{g^{\left( 1\right) }\left( c_{x}\right) },\quad \text{ for all }x\in I\backslash \{ a\} .\medskip \label{Cauchy-c-1} \end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>

  </div>
</div> </p>
<p>If the function \(f^{\left( 1\right) }/g^{\left( 1\right) }\) is injective on \(I,\  \)then for each \(x\in I\backslash \{ a\} \) there exists a unique point \(c_{x}\), from the interval with the extremities \(x\) and \(a\), such that \(\left( \text{\ref{Cauchy-c-1}}\right) \) holds. In this case, we can define the function \(c:I\backslash \{ a\} \rightarrow I\backslash \{ a\} \) by</p>
<div class="equation" id="Cauchy-c-2">
<p>
  <div class="equation_content">
    \begin{equation}  c\left( x\right) =c_{x},\quad \text{ for all }x\in I\backslash \{ a\} . \label{Cauchy-c-2} \end{equation}
  </div>
  <span class="equation_label">2</span>
</p>
</div>
<p>The function \(c\) has the property that</p>
<div class="equation" id="Cauchy-c-3">
<p>
  <div class="equation_content">
    \begin{equation}  \frac{f\left( x\right) -f\left( a\right) }{g\left( x\right) -g\left( a\right) }=\frac{f^{\left( 1\right) }\left( c\left( x\right) \right) }{g^{\left( 1\right) }\left( c\left( x\right) \right) },\quad \text{ for all }x\in I\backslash \{ a\} .\medskip \label{Cauchy-c-3} \end{equation}
  </div>
  <span class="equation_label">3</span>
</p>
</div>
<p>If the function \(f^{\left( 1\right) }/g^{\left( 1\right) }\) is not injective on \(I,\  \)then for some \(x\in I\backslash \{ a\} \) there exist several points \(c_{x}\), from the interval with the extremities \(x\) and \(a\), such that \(\left( \text{\ref{Cauchy-c-1}}\right) \) is true. If for each \(x\in I\backslash \{ a\} \) we choose one \(c_{x}\) from the interval with the extremities \(x\) and \(a\) which satisfies \(\left( \text{\ref{Cauchy-c-1}}\right) ,\) then we can also define the function \(c:I\backslash \{ a\} \rightarrow I\backslash \{ a\} \) by formula \(\left( \text{\ref{Cauchy-c-2}}\right) \). This function \(c\) satisfies \(\left( \text{\ref{Cauchy-c-3}}\right) \), too \((\)see <span class="cite">
	[
	<a href="#Dorel-Ovidiu-Cauchy" >2</a>
	]
</span>\().\) </p>
<p>Consequently, the following statement is true. </p>
<p><div class="theorem_thmwrapper " id="a0000000007">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(I\) be an interval in \(\mathbb {R}\), and let\(\  a\) be a point of \(I.\) If the functions \(f,g:I\rightarrow \mathbb {R}\) are differentiable on \(I\) and \(g^{\left( 1\right) }\left( x\right) \neq 0,\) for all \(x\in I\backslash \{ a\} , \) then there exists a function \(c:I\backslash \{ a\} \rightarrow I\backslash \{ a\} \) such that \((\)<a href="#Cauchy-c-3">3</a>\()\) is true. </p>
<p>Furthermore, if, in addition, \(f^{\left( 1\right) }/g^{\left( 1\right) }\) is injective, then the function \(c\) is unique. </p>

  </div>
</div> </p>
<p>If \(x\in I\backslash \{ a\} \) tends to \(a,\) because \(\left\vert c\left( x\right) -a\right\vert \leq \left\vert x-a\right\vert ,\) we have</p>
<div class="displaymath" id="a0000000008">
  \begin{equation*}  \underset {x\rightarrow a}{\lim }\text{ }c\left( x\right) =a. \end{equation*}
</div>
<p>Then the function \(\overline{c}:I\rightarrow I\) defined by</p>
<div class="equation" id="Cauchy-c-4">
<p>
  <div class="equation_content">
    \begin{equation}  \overline{c}\left( x\right) =\left\{  \begin{array}{ll} c\left( x\right) , &  \text{if }x\in I\backslash \{ a\}  \\ a, &  \text{if }x=a\end{array}\right. \label{Cauchy-c-4} \end{equation}
  </div>
  <span class="equation_label">4</span>
</p>
</div>
<p>is continuous at \(x=a.\) </p>
<p>The purpose of this paper is to establish under which circumstances the function \(\overline{c}\) is twice differentiable at the point \(x=a\) and to compute its derivatives \(\overline{c}^{\left( 1\right) }\left( a\right) \) and \(\overline{c}^{\left( 2\right) }\left( a\right) .\) Do the derivatives \(\overline{c}^{\left( 1\right) }\left( a\right) \) and \(\overline{c}^{\left( 2\right) }\left( a\right) \) depend upon the functions \(f\) and \(g\)? If there exist several functions \(\overline{c}\) which satisfy \(\left( \text{\ref{Cauchy-c-3}}\right) ,\) do the derivatives of the function \(\overline{c}\) at \(x=a\) depend upon the function \(\overline{c}\) we choose? </p>
<p>Since for \(x\in I\backslash \{ a\} ,\)</p>
<div class="displaymath" id="a0000000009">
  \begin{equation*}  \frac{\overline{c}\left( x\right) -\overline{c}\left( a\right) }{x-a}=\frac{c\left( x\right) -a}{x-a}, \end{equation*}
</div>
<p>if we denote by</p>
<div class="displaymath" id="a0000000010">
  \begin{equation*}  \theta \left( x\right) =\frac{c\left( x\right) -a}{x-a}, \end{equation*}
</div>
<p>then</p>
<div class="displaymath" id="a0000000011">
  \begin{equation*}  \theta \left( x\right) \in ]0,1[ \end{equation*}
</div>
<p>and</p>
<div class="displaymath" id="a0000000012">
  \begin{equation*}  c\left( x\right) =a+\left( x-a\right) \theta \left( x\right) \end{equation*}
</div>
<p>and hence</p>
<div class="displaymath" id="a0000000013">
  \begin{equation*}  \left[ f\left( x\right) -f\left( a\right) \right] g^{\left( 1\right) }\left( a+\left( x-a\right) \theta \left( x\right) \right) =\left[ g\left( x\right) -g\left( a\right) \right] f^{\left( 1\right) }\left( a+\left( x-a\right) \theta \left( x\right) \right) . \end{equation*}
</div>
<p>Consequently, the following statement is true. </p>
<p><div class="theorem_thmwrapper " id="TCauchy-theta">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3</span>
  </div>
  <div class="theorem_thmcontent">
  <p> Let \(I\) be an interval in \(\mathbb {R}\), and let\(\  a\) be a point of \(I.\) If the functions \(f,g:I\rightarrow \mathbb {R}\) are differentiable on \(I\), and \(g^{\left( 1\right) }\left( x\right) \neq 0,\) for all \(x\in \mathrm{int}I,\) then </p>
<div class="displaymath" id="a0000000014">
  \begin{equation*}  g\left( x\right) \neq g\left( a\right) ,\text{ for all }x\in I\backslash \{ a\} , \end{equation*}
</div>
<p>and there exists a function \(\theta :I\backslash \{ a\} \rightarrow ]0,1[\) such that</p>
<div class="displaymath" id="a0000000015">
  \begin{equation*}  \frac{f\left( x\right) -f\left( a\right) }{g\left( x\right) -g\left( a\right) }=\frac{f^{\left( 1\right) }\left( a+\left( x-a\right) \theta \left( x\right) \right) }{g^{\left( 1\right) }\left( a+\left( x-a\right) \theta \left( x\right) \right) },\text{ for all }x\in I\backslash \{ a\} . \end{equation*}
</div>
<p>Furthermore, if, in addition, the function \(f^{\left( 1\right) }/g^{\left( 1\right) }\) is injective, then the function \(\theta \) is unique.</p>

  </div>
</div> </p>
<p>Obviously, the function \(\overline{c}:I\rightarrow I\) defined by \(\left( \text{\ref{Cauchy-c-4}}\right) \) is differentiable at \(x=a\) if and only if the function \(\theta :I\backslash \{ a\} \rightarrow ]0,1[\) defined by</p>
<div class="displaymath" id="a0000000016">
  \begin{equation*}  \theta \left( x\right) =\frac{\overline{c}\left( x\right) -\overline{c}\left( a\right) }{x-a}=\frac{c\left( x\right) -a}{x-a},\text{ for all }x\in I\backslash \{ a\}  \end{equation*}
</div>
<p>has limit at the point \(x=a.\) Moreover, if the function \(\overline{c}\) is differentiable at \(x=a,\) then</p>
<div class="displaymath" id="a0000000017">
  \begin{equation*}  \overline{c}^{\left( 1\right) }\left( a\right) =\underset {x\rightarrow a}{\lim }\text{ }\theta \left( x\right) . \end{equation*}
</div>
<p>In <span class="cite">
	[
	<a href="#Dorel-Ovidiu-Cauchy" >2</a>
	]
</span>, one proves the following theorem </p>
<p><div class="theorem_thmwrapper " id="Duca-Pop-Cauchy">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">4</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(I\) be an interval in \(\mathbb {R}\), let \(a\) be a point of \(I\) and let \(f,g:I\rightarrow \mathbb {R}\) be functions which satisfy the following conditions: </p>
<p>\(\left( a\right) \) the functions \(f\) and \(g\) are \(n\geq 2\) times differentiable on \(I,\) </p>
<p>\(\left( b\right) \) the functions \(f^{\left( n\right) }\) and \(g^{\left( n\right) }\) are continuous on \(I,\) </p>
<p>\(\left( c\right) \) \(f^{\left( 1\right) }\left( a\right) g^{\left( k\right) }\left( a\right) =f^{\left( k\right) }\left( a\right) g^{\left( 1\right) }\left( a\right) ,\) for all \(k\in \{ 1,...,n-1\} ,\) </p>
<p>\(\left( d\right) \) \(f^{\left( 1\right) }\left( a\right) g^{\left( n\right) }\left( a\right) \neq f^{\left( n\right) }\left( a\right) g^{\left( 1\right) }\left( a\right) .\) </p>
<p>Then the following statements are true: </p>
<p>\(1^{0}\) If \(\theta :I\backslash \{ a\} \rightarrow ]0,1[\) is a function such that</p>
<div class="displaymath" id="a0000000018">
  \begin{equation*}  \left[ f\left( x\right) -f\left( a\right) \right] g^{\left( 1\right) }\left( a+\left( x-a\right) \theta \left( x\right) \right) =\left[ g\left( x\right) -g\left( a\right) \right] f^{\left( 1\right) }\left( a+\left( x-a\right) \theta \left( x\right) \right) , \end{equation*}
</div>
<p>for all \(x\in I\backslash \{ a\} ,\) then there exists the limit</p>
<div class="displaymath" id="a0000000019">
  \begin{equation*}  \underset {x\rightarrow a}{\lim }\text{ }\theta \left( x\right) =\left\{  \begin{array}{ll} \frac{1}{2}, &  \text{if }n=2 \\ \frac{1}{\sqrt[n-1]{n}}, &  \text{if }n\geq 3.\end{array}\right. \end{equation*}
</div>
<p>\(2^{0}\) If \(c:I\backslash \{ a\} \rightarrow I\) is a function such that</p>
<div class="displaymath" id="a0000000020">
  \begin{equation*}  \left[ f\left( x\right) -f\left( a\right) \right] g^{\left( 1\right) }\left( c\left( x\right) \right) =\left[ g\left( x\right) -g\left( a\right) \right] f^{\left( 1\right) }\left( c\left( x\right) \right) , \end{equation*}
</div>
<p>for all \(x\in I\backslash \{ a\} ,\) then the function \(\overline{c}:I\rightarrow I\) defined by \(\left( \text{\ref{Cauchy-c-4}}\right) \) is differentiable at \(x=a\) and</p>
<div class="displaymath" id="a0000000021">
  \begin{equation*}  \overline{c}^{\left( 1\right) }\left( a\right) =\left\{  \begin{array}{ll} \frac{1}{2}, &  \text{if }n=2 \\ \frac{1}{\sqrt[n-1]{n}}, &  \text{if }n\geq 3.\end{array}\right. \end{equation*}
</div>

  </div>
</div> </p>
<p>Related to the higher order differentiability of the intermediate point function in Cauchy’s mean-value theorem one know the following result (see <span class="cite">
	[
	<a href="#Dorel-Beatrix-Remeti" >3</a>
	]
</span>) </p>
<p><div class="theorem_thmwrapper " id="TCauchyRemeti">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">5</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(I\) be an interval in \(\mathbb {R}\), let \(a\) be a point of \(I\) and let \(f,g:I\rightarrow \mathbb {R}\) be functions which satisfy the following conditions: </p>
<p>\(\left( a\right) \) the functions \(f\) and \(g\) are \(3\) times differentiable on \(I,\) </p>
<p>\(\left( b\right) \) the functions \(f^{\left( 3\right) }\) and \(g^{\left( 3\right) }\) are continuous at \(x=a,\) </p>
<p>\((c)\) \(\  g^{(1)}(x)\neq 0,\) for all \(x\in \) \(\mathrm{int}\) \(I\), </p>
<p>\((d)\) \(\  f^{(1)}(a)\) \(g^{(2)}(a)\neq f^{(2)}(a)\) \(g^{(1)}(a).\) </p>
<p>Then the following statements are true: </p>
<p>\(1^{0}\) There exists a real number \(\delta {\gt}0\) such that: </p>
<p>\(\  \  i)\) \(\  \  ]a-\delta ,a+\delta \lbrack \subseteq I;\) </p>
<p>\(\  \  ii)\) \(\  \  f^{(1)}(x)g^{(2)}(x)\neq f^{(2)}(x)g^{(1)}(x),\) for all \(x\in ]a-\delta ,a+\delta \lbrack ;\) </p>
<p>\(\  \  iii)\) \(\  \)the function \(f^{\left( 1\right) }/g^{\left( 1\right) }\) is injective on \(]a-\delta ,a+\delta \lbrack .\) </p>
<p>\(2^{0}\) There exists a unique function \(c:]a-\delta ,a+\delta \lbrack \setminus \{ a\} \rightarrow ]a-\delta ,a+\delta \lbrack \setminus \{ a\} \) such that\(\medskip \)</p>
<div class="equation" id="Cf8a">
<p>
  <div class="equation_content">
    \begin{equation}  \frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f^{(1)}(c(x))}{g^{(1)}(c(x))},\medskip \label{Cf8a} \end{equation}
  </div>
  <span class="equation_label">5</span>
</p>
</div>
<p>for all \(x\in ]a-\delta ,a+\delta \lbrack \setminus \{ a\} .\) </p>
<p>\(3^{0}\) There exists a unique function \(\theta :]a-\delta ,a+\delta \lbrack \setminus \{ a\} \rightarrow ]0,1[\) such that\(\medskip \)</p>
<div class="equation" id="Cf9a">
<p>
  <div class="equation_content">
    \begin{equation}  \frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f^{(1)}(a+(x-a)\theta (x))}{g^{(1)}(a+(x-a)\theta (x))},\medskip \label{Cf9a} \end{equation}
  </div>
  <span class="equation_label">6</span>
</p>
</div>
<p>for all\(\  x\in I\backslash \{ a\} .\) </p>
<p>\(4^{0}\) The function \(\theta \) has limit at \(x=a\) and\(\medskip \)</p>
<div class="displaymath" id="a0000000022">
  \begin{equation*}  \underset {x\rightarrow a}{\lim }\text{ }\theta (x)=\tfrac {1}{2}.\medskip \end{equation*}
</div>
<p>\(5^{0}\) The function \(\overline{\theta }:]a-\delta ,a+\delta \lbrack \rightarrow ]0,1[\) defined by\(\medskip \)</p>
<div class="displaymath" id="a0000000023">
  \begin{equation*}  \overline{\theta }\left( x\right) =\left\{  \begin{array}{ll} \theta \left( x\right) , &  \text{if }x\in ]a-\delta ,a+\delta \lbrack \backslash \{ a\} \medskip \\ 1/2, &  \text{if }x=a,\end{array}\right. \medskip \end{equation*}
</div>
<p>is differentiable at \(x=a\) and\(\medskip \)</p>
<div class="displaymath" id="a0000000024">
  \begin{equation*}  \overline{\theta }^{\left( 1\right) }\left( a\right) =\frac{f^{\left( 1\right) }\left( a\right) g^{\left( 3\right) }\left( a\right) -f^{\left( 3\right) }\left( a\right) g^{\left( 1\right) }\left( a\right) }{24\left[ f^{\left( 1\right) }\left( a\right) g^{\left( 2\right) }\left( a\right) -f^{\left( 2\right) }\left( a\right) g^{\left( 1\right) }\left( a\right) \right] }.\medskip \end{equation*}
</div>
<p>\(6^{0}\) The function \(\overline{c}:]a-\delta ,a+\delta \lbrack \rightarrow ]a-\delta ,a+\delta \lbrack \) defined by\(\medskip \)</p>
<div class="displaymath" id="a0000000025">
  \begin{equation*}  \overline{c}\left( x\right) =\left\{  \begin{array}{ll} c\left( x\right) , &  \text{if }x\in ]a-\delta ,a+\delta \lbrack \backslash \{ a\} \medskip \\ a, &  \text{if }x=a,\end{array}\right. \medskip \end{equation*}
</div>
<p>is second order differentiable at the point \(x=a\) and\(\medskip \)</p>
<div class="displaymath" id="a0000000026">
  \begin{equation*}  \overline{c}^{\left( 1\right) }\left( a\right) =\frac{1}{2},\text{ \  }c^{\left( 2\right) }\left( a\right) =\frac{f^{\left( 1\right) }\left( a\right) g^{\left( 3\right) }\left( a\right) -f^{\left( 3\right) }\left( a\right) g^{\left( 1\right) }\left( a\right) }{12\left[ f^{\left( 1\right) }\left( a\right) g^{\left( 2\right) }\left( a\right) -f^{\left( 2\right) }\left( a\right) g^{\left( 1\right) }\left( a\right) \right] }.\medskip \end{equation*}
</div>

  </div>
</div> </p>
<p>There are no known result related to the higher order differentiability of the intermediate-point function in Cauchy’s mean-value theorem if \(f^{(1)}(a)g^{(2)}(a)\)<br />\(=f^{(2)}(a)g^{\left( 1\right) }\left( a\right) \) </p>
<p>In this paper, we establish under which circumstances the intermediate-point function in Cauchy’s mean-value theorem is twice differentiable at the point \(x=a\) and to compute its derivatives, \(\overline{c}^{\left( 1\right) }\left( a\right) \  \)and  \(c^{\left( 2\right) }\left( a\right) \), if</p>
<div class="displaymath" id="a0000000027">
  \begin{equation*}  f^{(1)}(a)g^{(k)}(a)=f^{(k)}(a)g^{\left( 1\right) }\left( a\right) ,\text{ for all }k\in \{ 1,...,n-1\}  \end{equation*}
</div>
<p>and</p>
<div class="displaymath" id="a0000000028">
  \begin{equation*}  f^{(1)}(a)g^{(n)}(a)\neq f^{(n}(a)g^{\left( 1\right) }\left( a\right) . \end{equation*}
</div>
<h1 id="a0000000029">2 Main result</h1>
<p>The following statement is true. </p>
<p><div class="theorem_thmwrapper " id="a0000000030">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">6</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(I\) be an interval in \(\mathbb {R}\), let \(a\) be a point of \(I\) and let \(f,g:I\rightarrow \mathbb {R}\) be functions which satisfy the following conditions: </p>
<p>\(\left( a\right) \) the functions \(f\) and \(g\) are \(n+1\geq 3\) times differentiable on \(I,\) </p>
<p>\(\left( b\right) \) the functions \(f^{\left( n+1\right) }\) and \(g^{\left( n+1\right) }\) are continuous at \(x=a,\) </p>
<p>\(\left( c\right) \) \(f^{\left( 1\right) }\left( a\right) g^{\left( k\right) }\left( a\right) =f^{\left( k\right) }\left( a\right) g^{\left( 1\right) }\left( a\right) ,\) for all \(k\in \{ 1,...,n-1\} ,\) </p>
<p>\(\left( d\right) \) \(f^{\left( 1\right) }\left( a\right) g^{\left( n\right) }\left( a\right) \neq f^{\left( n\right) }\left( a\right) g^{\left( 1\right) }\left( a\right) .\) </p>
<p>\(1^{0}\) If \(\theta :I\backslash \{ a\} \rightarrow ]0,1[\) is a function with property that</p>
<div class="equation" id="Cf-9b">
<p>
  <div class="equation_content">
    \begin{equation}  \left[ f\left( x\right) -f\left( a\right) \right] g^{\left( 1\right) }\left( a+\left( x-a\right) \theta \left( x\right) \right) \label{Cf-9b} =\left[ g\left( x\right) -g\left( a\right) \right] f^{\left( 1\right) }\left( a+\left( x-a\right) \theta \left( x\right) \right) , \end{equation}
  </div>
  <span class="equation_label">7</span>
</p>
</div>
<p>for all \(x\in I\backslash \{ a\} ,\) then the function \(\overline{\theta }:I\rightarrow ]0,1[\) defined by</p>
<div class="displaymath" id="a0000000031">
  \begin{equation*}  \overline{\theta }\left( x\right) =\left\{  \begin{array}{ll} \theta \left( x\right) , &  \text{if }x\in I\backslash \{ a\}  \\ \frac{1}{\sqrt[n-1]{n}}, &  \text{if }x=a,\end{array}\right. \end{equation*}
</div>
<p>is differentiable at \(x=a\) and</p>
<div class="displaymath" id="a0000000032">
  \begin{equation*}  \overline{\theta }^{\left( 1\right) }\left( x\right) =\frac{1}{\left( n-1\right) \sqrt[n-1]{n^{2}}D_{1,n}}\left[ \frac{2-\sqrt[n-1]{n}}{2}D_{2,n}+\frac{n\sqrt[n-1]{n}-n-1}{\left( n+1\right) }D_{1,n+1}\right] , \end{equation*}
</div>
<p>where</p>
<div class="displaymath" id="a0000000033">
  \begin{equation*}  D_{k,j}=f^{\left( k\right) }\left( a\right) g^{\left( j\right) }\left( a\right) -f^{\left( j\right) }\left( a\right) g^{\left( k\right) }\left( a\right) ,\text{ for all }k,j\in \{ 1,...,n+1\} . \end{equation*}
</div>
<p>\(2^{0}\) If \(c:I\backslash \{ a\} \rightarrow I\) is a function with property that</p>
<div class="displaymath" id="a0000000034">
  \begin{equation*}  \left[ f\left( x\right) -f\left( a\right) \right] g^{\left( 1\right) }\left( c\left( x\right) \right) =\left[ g\left( x\right) -g\left( a\right) \right] f^{\left( 1\right) }\left( c\left( x\right) \right) , \end{equation*}
</div>
<p>for all \(x\in I\backslash \{ a\} ,\) then the function \(\overline{c}:I\rightarrow I\) defined by</p>
<div class="displaymath" id="a0000000035">
  \begin{equation*}  \overline{c}\left( x\right) =\left\{  \begin{array}{ll} c\left( x\right) , &  \text{if }x\in I\backslash \{ a\}  \\ a, &  \text{if }x=a,\end{array}\right. \end{equation*}
</div>
<p>is second order differentiable at \(x=a\) and</p>
<div class="displaymath" id="a0000000036">
  \begin{equation*}  \overline{c}^{\left( 1\right) }\left( a\right) =\frac{1}{\sqrt[n-1]{n}}, \end{equation*}
</div>
<div class="displaymath" id="a0000000037">
  \begin{equation*}  \overline{c}^{\left( 2\right) }\left( a\right) =\frac{2}{\left( n-1\right) \sqrt[n-1]{n^{2}}D_{1,n}}\left[ \frac{2-\sqrt[n-1]{n}}{2}D_{2,n}+\frac{n\sqrt[n-1]{n}-n-1}{\left( n+1\right) }D_{1,n+1}\right] . \end{equation*}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000038">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> By Taylor’s formula, for each \(x\in I\backslash \{ a\} \) there exist two real numbers \(\widehat{\theta }_{f,x},\) \(\widehat{\theta }_{g,x}\in ]0,1[\) such that</p>
<div class="equation" id="Cauchy-1-2">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( x\right) \label{Cauchy-1-2} =\underset {i=0}{\overset {n}{\sum }}\text{ }\frac{f^{\left( i\right) }\left( a\right) }{i!}\left( x-a\right) ^{i}+\frac{f^{\left( n+1\right) }\left( A_{f}\left( x\right) \right) }{\left( n+1\right) !}\left( x-a\right) ^{n+1},\medskip \end{equation}
  </div>
  <span class="equation_label">8</span>
</p>
</div>
<p>and</p>
<div class="equation" id="Cauchy-1-3">
<p>
  <div class="equation_content">
    \begin{equation}  g\left( x\right) \label{Cauchy-1-3} =\underset {i=0}{\overset {n}{\sum }}\text{ }\frac{g^{\left( i\right) }\left( a\right) }{i!}\left( x-a\right) ^{i}+\frac{g^{\left( n+1\right) }\left( A_{g}\left( x\right) \right) }{\left( n+1\right) !}\left( x-a\right) ^{n+1}, \end{equation}
  </div>
  <span class="equation_label">9</span>
</p>
</div>
<p>where</p>
<div class="displaymath" id="a0000000039">
  \begin{equation*}  A_{f}\left( x\right) =a+\left( x-a\right) \widehat{\theta }_{f,x}\text{ \   and \  \  }A_{g}\left( x\right) =a+\left( x-a\right) \widehat{\theta }_{g,x}.\medskip \end{equation*}
</div>
<p>On the other hand, by Taylor’s formula applied to the functions \(f^{\left( 1\right) }\) and \(g^{\left( 1\right) }\), for each \(x\in I\backslash \{ a\} \) there exist two real numbers \(\widetilde{\theta }_{f,x},\) \(\widetilde{\theta }_{g,x}\in ]0,1[\) such that\(\medskip \)</p>
<div class="displaymath" id="Cauchy-1-4">
  \begin{align} & f^{\left( 1\right) }\left( a+\left( x-a\right) \theta \left( x\right) \right) = \label{Cauchy-1-4}\\ & =\underset {k=1}{\overset {n}{\sum }}\text{ }\frac{f^{\left( k\right) }\left( a\right) }{\left( k-1\right) !}\left( x-a\right) ^{k-1}\theta ^{k-1}\left( x\right) +\frac{f^{\left( n+1\right) }\left( B_{f}\left( x\right) \right) }{n!}\left( x-a\right) ^{n}\theta ^{n}\left( x\right) ,\nonumber \end{align}
</div>
<p>and</p>
<div class="displaymath" id="Cauchy-1-5">
  \begin{align} & g^{\left( 1\right) }\left( a+\left( x-a\right) \theta \left( x\right) \right) = \label{Cauchy-1-5}\\ & =\underset {k=1}{\overset {n}{\sum }}\text{ }\frac{g^{\left( k\right) }\left( a\right) }{\left( k-1\right) !}\left( x-a\right) ^{k-1}\theta ^{k-1}\left( x\right) +\frac{g^{\left( n+1\right) }\left( B_{g}\left( x\right) \right) }{n!}\left( x-a\right) ^{n}\theta ^{n}\left( x\right) ,\nonumber \end{align}
</div>
<p>where</p>
<div class="displaymath" id="a0000000040">
  \begin{equation*}  B_{f}\left( x\right) =a+\left( x-a\right) \widetilde{\theta }_{f,x}\text{  \  and \  \  }B_{g}\left( x\right) =a+\left( x-a\right) \widetilde{\theta }_{g,x}.\medskip \end{equation*}
</div>
<p>Substituting \(\left( \text{\ref{Cauchy-1-2}}\right) -\left( \text{\ref{Cauchy-1-5}}\right) \) in \(\left( \text{\ref{Cf-9b}}\right) ,\) we obtain that,\(\medskip \)</p>
<div class="displaymath" id="Cauchy-1-9">
  \begin{align} & \underset {i=1}{\overset {n-1}{\sum }}\underset {k=i+1}{\overset {n}{\sum }}\text{ }D_{i,k}\left[ \tfrac {\theta ^{k-1}\left( x\right) }{i!\left( k-1\right) !}-\tfrac {\theta ^{i-1}\left( x\right) }{k!\left( i-1\right) !}\right] \left( x-a\right) ^{k+i-1} \label{Cauchy-1-9}\\ & \quad +\underset {i=1}{\overset {n}{\sum }}\text{ }\tfrac {f^{\left( i\right) }\left( a\right) g^{\left( n+1\right) }\left( B_{g}\left( x\right) \right) -f^{\left( n+1\right) }\left( B_{f}\left( x\right) \right) g^{\left( i\right) }\left( a\right) }{i!n!}\left( x-a\right) ^{n+i}\theta ^{n}\left( x\right) \nonumber \\ & \quad +\underset {k=1}{\overset {n}{\sum }}\text{ }\tfrac {f^{\left( n+1\right) }\left( A_{f}\left( x\right) \right) g^{\left( k\right) }\left( a\right) -f^{\left( k\right) }\left( a\right) g^{\left( n+1\right) }\left( A_{g}\left( x\right) \right) }{\left( n+1\right) !\left( k-1\right) !}\left( x-a\right) ^{n+k}\theta ^{k-1}\left( x\right) \nonumber \\ & \quad +\tfrac {f^{\left( n+1\right) }\left( A_{f}\left( x\right) \right) g^{\left( n+1\right) }\left( B_{g}\left( x\right) \right) -f^{\left( n+1\right) }\left( B_{f}\left( x\right) \right) g^{\left( n+1\right) }\left( A_{g}\left( x\right) \right) }{n!\left( n+1\right) !}\left( x-a\right) ^{2n+1}\theta ^{n}\left( x\right)\nonumber \\ & =0,\nonumber \end{align}
</div>
<p>for all \(x\in I\backslash \{ a\} .\) </p>
<p>Since</p>
<div class="displaymath" id="a0000000041">
  \begin{equation*}  \frac{f^{\left( 1\right) }\left( a\right) }{g^{\left( 1\right) }\left( a\right) }=\frac{f^{\left( 2\right) }\left( a\right) }{g^{\left( 2\right) }\left( a\right) }=\cdot \cdot \cdot =\frac{f^{\left( n-1\right) }\left( a\right) }{g^{\left( n-1\right) }\left( a\right) } \end{equation*}
</div>
<p>and</p>
<div class="displaymath" id="a0000000042">
  \begin{equation*}  \frac{f^{\left( 1\right) }\left( a\right) }{g^{\left( 1\right) }\left( a\right) }\neq \frac{f^{\left( n\right) }\left( a\right) }{g^{\left( n\right) }\left( a\right) }. \end{equation*}
</div>
<p>we have</p>
<div class="displaymath" id="a0000000043">
  \begin{equation*}  D_{k,j}=0,\text{ for all }k,j\in \{ 1,...,n-1\} , \end{equation*}
</div>
<p>and</p>
<div class="displaymath" id="a0000000044">
  \begin{equation*}  D_{1,n}\neq 0. \end{equation*}
</div>
<p>Then \(\left( \text{\ref{Cauchy-1-9}}\right) \) becomes </p>
<div class="displaymath" id="Cauchy-1-6">
  \begin{align} & \tfrac {1}{\left( n-1\right) !}\underset {k=1}{\overset {n-1}{\sum }}\text{ }\tfrac {D_{k,n}}{\left( k-1\right) !}\left[ \tfrac {\theta ^{n-1}\left( x\right) }{k}-\tfrac {\theta ^{k-1}\left( x\right) }{n}\right] \left( x-a\right) ^{k-1} \label{Cauchy-1-6}\\ & \quad +\text{ }\tfrac {1}{n!}\underset {k=1}{\overset {n}{\sum }}\text{ }\tfrac {f^{\left( k\right) }\left( a\right) g^{\left( n+1\right) }\left( B_{g}\left( x\right) \right) -f^{\left( n+1\right) }\left( B_{f}\left( x\right) \right) g^{\left( k\right) }\left( a\right) }{k!}\left( x-a\right) ^{k} + \nonumber \end{align}
</div>
<div class="displaymath" id="a0000000045">
  \begin{align*} & \quad +\tfrac {1}{\left( n+1\right) !}\text{ }\underset {k=1}{\overset {n}{\sum }}\text{ }\tfrac {f^{\left( n+1\right) }\left( A_{f}\left( x\right) \right) g^{\left( k\right) }\left( a\right) -f^{\left( k\right) }\left( a\right) g^{\left( n+1\right) }\left( A_{g}\left( x\right) \right) }{\left( k-1\right) !}\theta ^{k-1}\left( x\right) \left( x-a\right) ^{k}\nonumber \\ & \quad +\tfrac {f^{\left( n+1\right) }\left( A_{f}\left( x\right) \right) g^{\left( n+1\right) }\left( B_{g}\left( x\right) \right) -f^{\left( n+1\right) }\left( B_{f}\left( x\right) \right) g^{\left( n+1\right) }\left( A_{g}\left( x\right) \right) }{n!\left( n+1\right) !}\theta ^{n}\left( x\right) \left( x-a\right) ^{n+1}\nonumber \\ & =0,\nonumber \end{align*}
</div>
<p>for all \(x\in I\backslash \{ a\} .\) </p>
<p>Taking \(x\rightarrow a,\) we obtain</p>
<div class="displaymath" id="a0000000046">
  \begin{equation*}  \underset {x\rightarrow a}{\lim }\text{ }\left[ \frac{\theta ^{n-1}\left( x\right) }{1}-\frac{1}{n}\right] =0, \end{equation*}
</div>
<p>hence</p>
<div class="displaymath" id="a0000000047">
  \begin{equation*}  \overline{\theta }^{n-1}\left( a\right) =\underset {x\rightarrow a}{\lim }\text{ }\theta ^{n-1}\left( x\right) =\frac{1}{n}. \end{equation*}
</div>
<p>Consequently, the function \(\overline{\theta }\) is continuous at \(x=a.\) </p>
<p>From \(\left( \text{\ref{Cauchy-1-6}}\right) ,\) we deduce that</p>
<div class="displaymath" id="a0000000048">
  \begin{align*} & D_{1,n}\left[ \theta ^{n-1}\left( x\right) -\tfrac {1}{n}\right] +\\ & \quad +\underset {k=2}{\overset {n}{\sum }}\text{ }\tfrac {D_{k,n}}{\left( k-1\right) !}\left[ \tfrac {\theta ^{n-1}\left( x\right) }{k}-\tfrac {\theta ^{k-1}\left( x\right) }{n}\right] \left( x-a\right) ^{k-1}\\ & \quad +\tfrac {\theta ^{n}\left( x\right) }{n}\text{ }\underset {k=1}{\overset {n}{\sum }}\text{ }\tfrac {f^{\left( k\right) }\left( a\right) g^{\left( n+1\right) }\left( B_{g}\left( x\right) \right) -f^{\left( n+1\right) }\left( B_{f}\left( x\right) \right) g^{\left( k\right) }\left( a\right) }{k!}\left( x-a\right) ^{k}\\ & \quad +\tfrac {1}{n\left( n+1\right) }\text{ }\underset {k=1}{\overset {n}{\sum }}\text{ }\tfrac {f^{\left( n+1\right) }\left( A_{f}\left( x\right) \right) g^{\left( k\right) }\left( a\right) -f^{\left( k\right) }\left( a\right) g^{\left( n+1\right) }\left( A_{g}\left( x\right) \right) }{\left( k-1\right) !}\left( x-a\right) ^{k}\theta ^{k-1}\left( x\right) \\ & \quad +\tfrac {f^{\left( n+1\right) }\left( A_{f}\left( x\right) \right) g^{\left( n+1\right) }\left( B_{g}\left( x\right) \right) -f^{\left( n+1\right) }\left( B_{f}\left( x\right) \right) g^{\left( n+1\right) }\left( A_{g}\left( x\right) \right) }{\left( n+1\right) !n}\left( x-a\right) ^{n+1}\theta ^{n}\left( x\right) =\\ & =0, \end{align*}
</div>
<p>for all \(x\in I\backslash \{ a\} .\) </p>
<p>Dividing by \(\left( x-a\right) D_{1,n}\) we obtain</p>
<div class="displaymath" id="Cauchy-1-7">
  \begin{align} & \frac{1}{x-a}\left[ \theta ^{n-1}\left( x\right) -\frac{1}{n}\right] +\frac{D_{2,n}}{D_{1,n}}\left[ \frac{\theta ^{n-1}\left( x\right) }{2}-\frac{\theta \left( x\right) }{n}\right] + \label{Cauchy-1-7}\\ & \quad +\frac{f^{\left( 1\right) }\left( a\right) g^{\left( n+1\right) }\left( B_{g}\left( x\right) \right) -f^{\left( n+1\right) }\left( B_{f}\left( x\right) \right) g^{\left( 1\right) }\left( a\right) }{nD_{1,n}}\theta ^{n}\left( x\right) \nonumber \\ & \quad +\frac{f^{\left( n+1\right) }\left( A_{f}\left( x\right) \right) g^{\left( 1\right) }\left( a\right) -f^{\left( 1\right) }\left( a\right) g^{\left( n+1\right) }\left( A_{g}\left( x\right) \right) }{n\left( n+1\right) D_{1,n}}+\emph{o}\left( x-a\right) =0, \nonumber \end{align}
</div>
<p>for all \(x\in I\backslash \{ a\} .\) </p>
<p>Evidently,\(\bigskip \)</p>
<div class="displaymath" id="a0000000049">
  \begin{equation*}  \frac{1}{x-a}\left[ \theta ^{n-1}\left( x\right) -\frac{1}{n}\right] =\frac{1}{x-a}\left[ \theta \left( x\right) -\frac{1}{\sqrt[n-1]{n}}\right] \underset {k=0}{\overset {n-2}{\sum }}\text{ }\theta ^{n-2-k}\left( x\right) \left( \frac{1}{\sqrt[n-1]{n}}\right) ^{k},\bigskip \end{equation*}
</div>
<p>for all \(x\in I\backslash \{ a\} .\) </p>
<p>Since</p>
<div class="displaymath" id="a0000000050">
  \begin{equation*}  \underset {x\rightarrow a}{\lim }\text{ }\theta \left( x\right) =\frac{1}{\sqrt[n-1]{n}}=\overline{\theta }\left( a\right) ,\bigskip \end{equation*}
</div>
<div class="displaymath" id="a0000000051">
  \begin{equation*}  \underset {x\rightarrow a}{\lim }\text{ }A_{f}\left( x\right) =\underset {x\rightarrow a}{\lim }\text{ }A_{g}\left( x\right) =\underset {x\rightarrow a}{\lim }\text{ }B_{f}\left( x\right) =\underset {x\rightarrow a}{\lim }\text{ }B_{g}\left( x\right) =a,\bigskip \end{equation*}
</div>
<div class="displaymath" id="a0000000052">
  \begin{equation*}  \underset {x\rightarrow a}{\lim }\text{ }\left[ f^{\left( 1\right) }\left( a\right) g^{\left( n+1\right) }\left( B_{g}\left( x\right) \right) -f^{\left( n+1\right) }\left( B_{f}\left( x\right) \right) g^{\left( n\right) }\left( a\right) \right] =D_{1,n+1},\bigskip \end{equation*}
</div>
<div class="displaymath" id="a0000000053">
  \begin{equation*}  \underset {x\rightarrow a}{\lim }\text{ }\left[ f^{\left( n+1\right) }\left( A_{f}\left( x\right) \right) g^{\left( 1\right) }\left( a\right) -f^{\left( 1\right) }\left( a\right) g^{\left( n\right) }\left( A_{g}\left( x\right) \right) \right] =-D_{1,n+1},\bigskip \end{equation*}
</div>
<p>from \(\left( \text{\ref{Cauchy-1-7}}\right) ,\) we deduce that there exists the limit\(\bigskip \)</p>
<div class="displaymath" id="a0000000054">
  \begin{equation*}  \underset {x\rightarrow a}{\lim }\text{ }\left( \frac{1}{x-a}\left[ \theta \left( x\right) -\frac{1}{\sqrt[n-1]{n}}\right] \right) =\underset {x\rightarrow a}{\lim }\text{ }\frac{\overline{\theta }\left( x\right) -\overline{\theta }\left( a\right) }{x-a}=\overline{\theta }^{\prime }\left( a\right) ,\bigskip \end{equation*}
</div>
<p>and </p>
<div class="displaymath" id="a0000000055">
  \begin{equation*}  \left( n\! -\! 1\right) \overline{\theta }^{n-2}\left( a\right) \overline{\theta }^{\prime }\left( a\right) \! +\! \frac{D_{2,n}}{D_{1,n}}\! \left[ \frac{\overline{\theta }^{n-1}\left( a\right) }{2}\! -\! \frac{\overline{\theta }\left( a\right) }{n}\right]\!  +\! \frac{D_{1,n+1}}{D_{1,n}}\! \left[ \frac{\overline{\theta }^{n}\left( a\right) }{n}\! -\! \frac{1}{n\left( n+1\right) }\right] \! \! =0.\bigskip \end{equation*}
</div>
<p>Consequently,\(\bigskip \)</p>
<div class="displaymath" id="a0000000056">
  \begin{equation*}  \overline{\theta }^{\prime }\left( a\right) =\frac{1}{\left( n-1\right) \sqrt[n-1]{n^{2}}D_{1,n}}\left[ \frac{2-\sqrt[n-1]{n}}{2}D_{2,n}+\frac{n.\sqrt[n-1]{n^{n}}-n-1}{n\left( n+1\right) }D_{1,n+1}\right] . \end{equation*}
</div>
<p>\(2^{0}\) The statement \(2^{0}\) follows from the statement \(1^{0}.\) <div class="proof_wrapper" id="a0000000057">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="a0000000058">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">7</span>
  </div>
  <div class="remark_thmcontent">
  <p>If \(g=1_{I},\) then</p>
<div class="displaymath" id="a0000000059">
  \begin{equation*}  \overline{\theta }^{\prime }\left( a\right) =\frac{n\sqrt[n-1]{n}-n-1}{n\left( n^{2}-1\right) \sqrt[n-1]{n^{2}}}\frac{f^{\left( n+1\right) }\left( a\right) }{f^{\left( n\right) }\left( a\right) }, \end{equation*}
</div>
<p>and</p>
<div class="displaymath" id="a0000000060">
  \begin{equation*}  \overline{c}^{\prime \prime }\left( a\right) =\frac{2\left( n\sqrt[n-1]{n}-n-1\right) }{n\left( n^{2}-1\right) \sqrt[n-1]{n^{2}}}\frac{f^{\left( n+1\right) }\left( a\right) }{f^{\left( n\right) }\left( a\right) }. \end{equation*}
</div>

  </div>
</div> </p>

<p><small class="footnotesize">  </small></p>
<div class="bibliography">
<h1>Bibliography</h1>
<dl class="bibliography">
  <dt><a name="Dorel-Ovidiu-Lagrange">1</a></dt>
  <dd><p><a href ="http://dx.doi.org/10.7153/mia-12-38"> <i class="sc">D.I. Duca</i> and <i class="sc">O.T. Pop</i>, <i class="itshape">Concerning the intermediate point in the mean value theorem</i>, Mathematical Inequalities &amp; Applications, <b class="bf">12</b> (2009), no. 3, pp.&#160;499–512. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="Dorel-Ovidiu-Cauchy">2</a></dt>
  <dd><p><a href ="http://dx.doi.org/10.7153/mia-09-37"> <i class="sc">D.I. Duca</i> and <i class="sc">O. Pop</i>, <i class="itshape">On the intermediate point in Cauchy’s mean-value theorem</i>, Mathematical Inequalities &amp; Applications, <b class="bf">9</b> (2006), no. 3, pp.&#160;375–389. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
  <dt><a name="Dorel-Beatrix-Remeti">3</a></dt>
  <dd><p><i class="sc">B.-M. Pop</i> and <i class="sc">D.I. Duca</i>, <i class="itshape">The derivability of the intermediate point function in Cauchy’s mean-value theorem</i>, Didactica Mathematica, <b class="bf">32</b> (2014), pp.&#160;87–100 (in Romanian). </p>
</dd>
  <dt><a name="Trif">4</a></dt>
  <dd><p><a href ="http://dx.doi.org/10.7153/jmi-02-15"> <i class="sc">T. Trif</i>, <i class="itshape">Asymptotic behavior of intermediate point in certain mean value theorems</i>, J. Math. Ineq., <b class="bf">2</b> (2008), pp.&#160;151–161. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
</dl>


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