Asosiy tarkibga o'tish

3.5-bo'lim : Grafikalash funktsiyalari

4. Quyidagi funksiya grafigida kamida 4 ta tartiblangan juft nuqtalardan iborat jadval tuzing va funksiya grafigini chizish uchun jadvaldagi tartiblangan juftlardan foydalaning.

\[f\left( x \right) = \left\{ {\begin{array}{*{20}{l}}{10 - 2x}&{{\mbox{if }}x < 2}\\{{x^2} + 2}&{{\mbox{if }}x \ge 2}\end{array}} \right.\]
Yechim

Bu erda biz qismlarga bo'lingan funktsiyaga egamiz, shuning uchun ballar jadvalimizni tuzishda \(x\) ning har bir diapazonidan ball tanlashimiz kerak. Shuningdek, biz \(x = 2\) “kesish nuqtasini” yuqori tenglamaga qo‘shishni xohlaymiz, shuning uchun u amal qiladigan \(x\) diapazonining oxirida qayerda bo‘lishini bilamiz.

Mana biz ushbu muammoni hal qilishda foydalanadigan fikrlar jadvali.

\(x\) \(10 - 2x\) \(\left( {x,y} \right)\)
-1 12 \(\left( { - 1,12} \right)\)
1 8 \(\left( {1,8} \right)\)
2 6 \(\left( {2,6} \right)\)


\(x\) \({x^2} + 2\) \(\left( {x,y} \right)\)
2 6 \(\left( {2,6} \right)\)
3 11 \(\left( {3,11} \right)\)
5 27 \(\left( {5,27} \right)\)
2-qadam

Bu funksiyaning eskizi.

The graph of the piecewise function. To the left of \(x = 2\) it is the straight line \(10 - 2x\) falling from the upper left down to \(\left( {2,6} \right)\) and to the right of \(x = 2\) it is the parabola \({x^2} + 2\) increasing from \(\left( {2,6} \right)\) up to the upper right. The two pieces meet at \(\left( {2,6} \right)\) so there is no break in the graph. The points from the table are marked and labeled and they are \(\left( { - 1,12} \right)\), \(\left( {1,8} \right)\), \(\left( {2,6} \right)\), \(\left( {3,11} \right)\) and \(\left( {5,27} \right)\).

E'tibor bering, bu holda ikkita bo'lak \(\left( {2,6} \right)\) nuqtasida "uchrashdi". Ba'zida bu muammolar bilan sodir bo'ladi, ba'zida esa bo'lmaydi. Shuning uchun biz har doim ikkala tenglamani "kesish nuqtasida" baholaymiz. Buni qilmasdan, biz qismlar bir-biriga mos keladimi yoki yo'qligini bilmaymiz.