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Revision difference : least and greatest value of function |
| Version 6 |
Version 5 |
| \begin{thmplain} |
\begin{thmplain} |
| \, If the real function $f$ is |
\, If the real function $f$ is |
| \begin{enumerate} |
\begin{enumerate} |
| \item continuous on the closed interval\, $[a,\,b]$\, and |
\item continuous on the closed interval\, $[a,\,b]$\, and |
| \item differentiable on the open interval\, $(a,\,b)$, |
\item differentiable on the open interval\, $(a,\,b)$, |
| \end{enumerate} |
\end{enumerate} |
| then the function has on the interval\, $[a,\,b]$\, a least value and a greatest value.\, These are always got in the end of the interval or in the zero of the derivative. |
then the function has on the interval\, $[a,\,b]$\, a least value and a greatest value.\, These are always got in the end of the interval or in the zero of the derivative. |
| \end{thmplain} |
\end{thmplain} |
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| \textbf{Remark 1.}\, If the preconditions of the theorem are fulfilled by a function $f$, then one needs only to determine the values of $f$ in the end points $a$ and $b$ of the interval and in the zeros of the derivative $f'$ inside the interval; then the least and the greatest value are found among those values. |
\textbf{Remark 1.}\, If the preconditions of the theorem are fulfilled by a function $f$, then one needs only to determine the values of $f$ in the end points $a$ and $b$ of the interval and in the zeros of the derivative $f'$ inside the interval; then the least and the greatest value are found among those values. |
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| \textbf{Remark 2.}\, Note that the theorem does not require anything of the derivative $f'$ in the points $a$ and $b$; one needs not even the right-sided derivative in $a$ or the left-sided derivative in $b$.\, Thus e.g. the function\, $f:\,x \mapsto \sqrt{1-x^2}$,\, fulfilling the conditions of the theorem on the interval\, $[-1,\,1]$\, but not having such one-sided derivatives, gains its least value in the end-point\, $x = -1$\, and its greatest value in the zero\, $x = \frac{1}{2}$\, of the derivative. |
\textbf{Remark 2.}\, Note that the theorem does not require anything of the derivative $f'$ in the points $a$ and $b$; one needs not even the right-sided derivative in $a$ or the left-sided derivative in $b$.\, Thus e.g. the function\, $f:\,x \mapsto \sqrt{1-x^2}$,\, fulfilling the conditions of the theorem on the interval\, $[-1,\,1]$\, but not having such one-sided derivatives, gains its least value in the end-point\, $x = -1$\, and its greatest value in the zero\, $x = \frac{1}{2}$\, of the derivative. |
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