Eugene1000
21.10.2020 17:40

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. Дедлайн завтра/послезавтра ​

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Ответ:
eshkeree1
16.04.2020 20:03

We learn the concept of a friend from elementary school when we are in school. At school, our teachers introduce us to the power and challenges of friendship. Children of this age retain the notion of a friend, become real friends in life, and consider their playmate as their friend. When he fights with his friends, he defends them. At school, she shared with her friend a piece of bread that her mother had put in her bag to eat when she was hungry. Yes, this was the case 10-15 years ago. And I was amazed to see a girl in the 4th grade look at her friend, who was playing a game, and say, "Go away, I'm not playing with you, your clothes are bad." That is, the child's consciousness is formed from an early age. What makes it so? Aren't they our future? If you are guilty, I would put my family first. Because he does what he sees in his family on the street. At the beginning of the meal, when parents and siblings ask each other for lessons, not about things that can be an example for each other, be friends with high-achieving students, do not be friends with a diaper, he wears bad clothes and does not study well. Yes, he wears bad clothes, he doesn't do well, but he probably has a high moral character, loyalty to a friend, which is not found in everyone. Why don't we consider it a human quality, a person may not say in vain that beauty is a cloth, but the cloth will end, but what about friendship, where is the friendship? This is how a child's outlook on life is shaped by the idea that he should make friends with people who are famous and well-dressed.

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Ответ:
9SawHoffman
21.02.2022 18:03

Объяснение:

а) х=2 это вертикальная асимптота. Это точка разрыва, т. е. это будет та точка, в которой знаменатель равен 0, т.к. на 0 делить нельзя. Следовательно

2·2+b=0;     b=-4

y=3 - это горизонтальная асимптота. К этому значению стремится предел функции. Тогда

\lim_{x \to \infty} \frac{ax+11}{2x-4} =3

Применяя правило Лопиталя, будем иметь

\frac{(ax+11)'}{(2x-4)'} =3\\\frac{a}{2} =3\\a=6

b)

i)

\frac{6x+11}{2x-4}= \frac{6x+11}{2(x-2)}=\frac{3x+5.5}{x-2}=\frac{3x+5.5}{x-2}= \frac{3x-6+11.5}{x-2}= \frac{3x-6}{x-2}+\frac{11.5}{x-2}=3+\frac{11.5}{x-2}

Как видим, к требуемому виду функция не приводится, т.к. 3≠-2

ii) В точках пересечения с осью у абцисса равна 0. Подставляем в уравнение, находим у:

y=\frac{6\cdot0+11}{2\cdot0-4}= -2.75

A(0;-2.75) - точка пересечения с осью у

В точках пересечения с осью х ордината равна 0. Решаем уравнение

\frac{6x+11}{2x-4}=0\\ 6x-4=0\\x=\frac{2}{3}

B(\frac{2}{3} ;0) - точка пересечения  с осью х.

iii) Дополнительно исследуем функцию в точке разрыва

\lim_{x \to 2-} \frac{6x+11}{2x-4}= -\infty\\ \lim_{x \to 2+} \frac{6x+11}{2x-4}= +\infty

Схематически строим график


Дробно-линейная функция задана уравнением: f(x)=(ax+11)/(2x+b) a) Асимптоты функции имеют уравнения
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