DerivativeInterp - 5-masala yechimi | MathModel.uz

3.2-bo'lim : Hosilaning talqini

5. Quyida $f\left( x \right)$ funksiyasining grafigi keltirilgan. Bundan $f'\left( x \right)$ hosilasining grafigini chizish uchun foydalaning.

$The graph of $f\left( x \right)$ on $ - 3 \le x \le 6$. Starting at approximately $\left( { - 3,4} \right)$ the graph rises to a peak at approximately $\left( { - 2,7} \right)$, then decreases to a valley at approximately $\left( {1,2} \right)$, then increases to a peak at approximately $\left( {5,12.5} \right)$ and finally decreases a little to approximately $\left( {6,9} \right)$ at the right end.$

Yechim

Funktsiya grafigidan ko'rinib turibdiki, biz $x = - 2$, $x = 1$ va $x = 5$ da gorizontal tangens chiziqlarga ega bo'lamiz. Biz bu erda gorizontal tangenslarga ega bo'lishimiz sababli, bu nuqtalarda hosila nolga teng bo'lishi kerakligini ham bilamiz. Shuning uchun biz quyidagi hosilaviy baholashlarni bilamiz.

\[f'\left( { - 2} \right) = 0\hspace{0.5in}f'\left( 1 \right) = 0\hspace{0.5in}f'\left( 5 \right) = 0\]

2-qadam

Yuqorida topilgan nuqtalar $x$ o'qini funksiya ortib borayotgan va kamayadigan hududlarga ajratadi. Eslatib o'tamiz, agar hosila ijobiy bo'lsa, unda funktsiya ortib boradi va xuddi shunday hosila salbiy bo'lsa, funktsiya kamayadi. Ushbu fikrlardan foydalanib, biz har bir mintaqada hosila belgisini osongina aniqlashimiz mumkin. Shunday qilib,

\[\begin{align*}\hspace{0.25in}x & < - 2 & \hspace{0.25in}f'\left( x \right) > 0\\ - 2 < x & < 1 & \hspace{0.25in}f'\left( x \right) < 0\\ 1 < x & < 5 & \hspace{0.25in}f'\left( x \right) > 0\\ x & > 5 & \hspace{0.25in}f'\left( x \right) < 0\end{align*}\]

3-qadam

$x < - 2$ diapazonida hosila ijobiy bo'lishi va $x = - 2$ da nolga teng bo'lishi kerakligini bilamiz, shuning uchun $x = - 2$ ning chap tomonida hosila kamayib borayotgan bo'lishi kerak.

$ - 2 < x < 1$ diapazonida hosila manfiy bo'lishini va diapazonning so'nggi nuqtalarida u nolga teng bo'lishini bilamiz. Shunday qilib, $x = - 2$ ning o'ng tomonida hosila kamayishi kerak (noldan manfiy raqamga o'tadi). Xuddi shunday, $x = 1$ ning chap tomonida lotin ortishi kerak bo'ladi (salbiy raqamdan nolga o'tadi).

Esda tutingki, biz ushbu diapazonda hamma joyda hosila qanday harakat qilishini bilmaymiz, lekin biz oxirgi nuqtalar yaqinidagi umumiy xatti-harakatlardan foydalanishimiz va hosila qanday ko'rinishi haqida tasavvurga ega bo'lish uchun ikkalasini ulashning eng oddiy usulidan foydalanishimiz mumkin.

Shunga o'xshash mulohazalardan so'ng, hosila $x = 1$ ning o'ng tomoniga (noldan musbat raqamga boradi), $x = 5$ ning chap tomoniga (musbat sondan nolga o'tadi) kamayib borishi va $x = 5$ ning o'ng tomoniga (noldan manfiy raqamga o'tadi) kamayishi kerakligini ko'rishimiz mumkin.

4-qadam

Shunday qilib, bularning barchasini bu erda birlashtirib, hosilaning eskizi. E'tibor bering, agar siz lotinning o'ziga xos qiymatlarini sinab ko'rish va baholashni istasangiz, biz vertikal o'qga a shkalasini qo'shdik. 4-misol ushbu bo'limdan.

$The sketch of the derivative $f'\left( x \right)$. It comes down steeply from the upper left, crosses the x-axis at $x = - 2$, drops to a valley of about -2.5 near $x = - 1$, rises back up and crosses the x-axis at $x = 1$, reaches a peak of about 4 near $x = 3.3$ and then falls, crossing the x-axis at $x = 5$ and continuing down to the lower right.$