Предмет: Алгебра, автор: Accel12

Определите множество значений функции f(x)=\frac{2x + 3}{(x+2)^2} посредством введения параметра


Accel12: Сколько нужно баллов давать, чтобы люди начали решать что-то сложнее 2+2?)

Ответы

Автор ответа: bb573878
2

Ответ:

Объяснение:

\displaystyle\\f(x)=\frac{2x+3}{(x+2)^2}

ищем все значения а, при которых уравнение

\displaystyle\\\frac{2x+3}{(x+2)^2} =a

имеет решения

\displaystyle\\2x+3=a(x^2+4x+4)\\\\2x+3=ax^2+4ax+4a\\\\ax^2+x(4a-2)+4a-3=0

если а=0 уравнение имеет решение х=-1,5

если а≠0 уравнение является квадратным

чтобы оно имело корни, дискриминант должен

быть неотрицательным

\displaystyle\\D=(4a-2)^2-4a(4a-3)=16a^2-16a+4-16a^2+12a=\\\\=-4a+4\geq 0\\\\4a\leq 4\\\\a\leq 1

Мы нашли множество значений   \displaystyle\\E(f)=(-\infty; 1]

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