Continuity - 2-masala yechimi | MathModel.uz

2.9-bo'lim : Davomiylik

2. $f\left( x \right)$ ning grafigi quyida keltirilgan. Ushbu grafik asosida funktsiyaning qayerda uzluksiz ekanligini aniqlang.

$The graph of $f\left( x \right)$ on $ - 12 \le x \le 12$. Coming down from the upper left the graph decreases to an open dot at the bottom of a valley at approximately $\left( { - 8, - 6} \right)$, with a separate closed dot sitting above it at approximately $\left( { - 8, - 3} \right)$. From the open dot the graph increases, crossing the x-axis near $x = - 4$, and rises to a closed dot at approximately $\left( { - 2,3} \right)$ with the graph continuing up off the top of the picture just to the right of that point. The graph picks back up near $\left( {0,1} \right)$, rises to a small peak at approximately $\left( {3,3} \right)$, dips slightly and then runs out to an open dot at $\left( {6,2} \right)$, with a separate closed dot sitting above it at $\left( {6,5} \right)$. From the closed dot the graph decreases through a closed dot at approximately $\left( {10,1} \right)$ and continues on down to the lower right.$

Yechim

Yechimni boshlashdan oldin, $x = a$ da funksiya uzluksiz bo‘lishi uchun $f\left( a \right)$ ham, $\mathop {\lim }\limits_{x \to a} f\left( x \right)$ ham mavjud bo‘lishi va bizda quyidagilar bo‘lishi kerakligini esga oling:

\[\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)\]

Ushbu fikrdan foydalanib, funktsiyaning doimiy emasligi aniq bo'lishi kerak.

Birinchidan, $x = - 8$ da bizda borligiga e'tibor bering,

\[\mathop {\lim }\limits_{x \to - {8^ - }} f\left( x \right) = - 6 = \mathop {\lim }\limits_{x \to - {8^ + }} f\left( x \right)\]

va shuning uchun biz $\mathop {\lim }\limits_{x \to - 8} f\left( x \right) = - 6$ ekanligini ham bilamiz. Biz $f\left( { - 8} \right) = - 3$ ni ham ko'rishimiz mumkin va shuning uchun bizda,

\[ - 6 = \mathop {\lim }\limits_{x \to - 8} f\left( x \right) \ne f\left( { - 8} \right) = - 3\]

Funktsiya va chegara turli qiymatlarga ega bo'lgani uchun biz $f\left( x \right)$ degan xulosaga kelishimiz mumkin uzluksiz at $x = - 8$.

Keling, bizda mavjud $x = - 2$ ni ko'rib chiqaylik,

\[\mathop {\lim }\limits_{x \to - {2^ - }} f\left( x \right) = 3 \ne \infty = \mathop {\lim }\limits_{x \to - {2^ + }} f\left( x \right)\]

va shuning uchun biz $\mathop {\lim }\limits_{x \to \, - 2} f\left( x \right)$ mavjud emasligini ham bilamiz. Shunday qilib, biz $f\left( x \right)$ degan xulosaga kelishimiz mumkin uzluksiz $x = - 2$ da, chunki limit mavjud emas.

Nihoyat, $x = 6$ ni ko'rib chiqaylik. Bu erda biz borligini ko'rishimiz mumkin,

\[\mathop {\lim }\limits_{x \to {6^ - }} f\left( x \right) = 2 \ne 5 = \mathop {\lim }\limits_{x \to {6^ + }} f\left( x \right)\]

va shuning uchun biz \[\mathop {\lim }\limits_{x \to \,6} f\left( x \right)\] mavjud emasligini ham bilamiz. Shunday qilib, yana bir bor, limit mavjud emasligi sababli, biz $f\left( x \right)$ degan xulosaga kelishimiz mumkin uzluksiz at $x = 6$.

Ushbu grafikdagi barcha boshqa nuqtalar funksiya va chegara mavjud bo'ladi va bizda $\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)$ bo'ladi va shuning uchun doimiy bo'ladi.

Xulosa qilib aytganda, ushbu grafik uchun uzilish nuqtalari: $x = - 8$, $x = - 2$ va $x = 6$.