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3.5-bo'lim : Grafikalash funktsiyalari

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

\[f\left( x \right) = \left\{ {\begin{array}{*{20}{l}}{5 + x}&{{\mbox{if }}x \ge 1}\\2&{{\mbox{if }} - 2 \le x < 1}\\{1 - {x^2}}&{{\mbox{if }}x < - 2}\end{array}} \right.\]
Yechim

Bu erda biz qismlarga bo'lingan funktsiyaga egamiz, shuning uchun ballar jadvalimizni tuzishda \(x\) ning har bir diapazonidan ball tanlashimiz kerak. Shuningdek, biz \(x = - 2\) va \(x = 1\) “kesish nuqtalarini” mos keladigan tenglamaga ulashni xohlaymiz, shuning uchun ularning har biri tegishli bo‘lgan \(x\) diapazonining oxirida qayerda bo‘lishini bilamiz.

Mana biz ushbu muammoni hal qilishda foydalanadigan fikrlar jadvali.

\(x\) \(1 - {x^2}\) \(\left( {x,y} \right)\)
-4 -15 \(\left( { - 4, - 15} \right)\)
-3 -8 \(\left( { - 3, - 8} \right)\)
-2 -3 \(\left( { - 2, - 3} \right)\)


\(x\) 2 \(\left( {x,y} \right)\)
-2 2 \(\left( { - 2,2} \right)\)
-1 2 \(\left( { - 1,2} \right)\)
1 2 \(\left( {1,2} \right)\)


\(x\) \(5 + x\) \(\left( {x,y} \right)\)
1 6 \(\left( {1,6} \right)\)
2 7 \(\left( {2,7} \right)\)
3 8 \(\left( {3,8} \right)\)
2-qadam

Bu funksiyaning eskizi.

The graph of the piecewise function, which is in three pieces and has a break at each of the two cutoff points. For \(x < - 2\) it is the downward opening parabola \(1 - {x^2}\) coming up from the lower left to an open dot at \(\left( { - 2, - 3} \right)\). For \( - 2 \le x < 1\) it is the horizontal line \(y = 2\) running from a closed dot at \(\left( { - 2,2} \right)\) to an open dot at \(\left( {1,2} \right)\). For \(x \ge 1\) it is the straight line \(5 + x\) starting at a closed dot at \(\left( {1,6} \right)\) and increasing to the upper right. The points from the table are marked and labeled and they are \(\left( { - 4, - 15} \right)\), \(\left( { - 3, - 8} \right)\), \(\left( { - 2,2} \right)\), \(\left( { - 1,2} \right)\), \(\left( {1,6} \right)\), \(\left( {2,7} \right)\) and \(\left( {3,8} \right)\).

Bunday holda, grafikning hech bir qismi kesish nuqtalarida uchrashmadi. Shuning uchun biz har bir tenglamaning qiymatini kesish nuqtalarida hisoblaymiz. Buni qilmasdan, biz ular kesishish nuqtasida uchrashganmi yoki yo'qmi, bilmaymiz.