Задача №1005
Условие
Даны функции: а) \(f(x)=\frac{x-2}{x+1}\); б) \(\varphi(x)=\frac{|x-2|}{x+1}\). Найти \(f(0)\), \(f(1)\), \(f(2)\), \(f(-2)\), \(f\left(-\frac{1}{2}\right)\); \(f(\sqrt{2})\), \(\left|f\left(\frac{1}{2}\right)\right|\), \(\varphi(0)\), \(\varphi(1)\), \(\varphi(2)\), \(\varphi(-2)\), \(\varphi(4)\). Существует ли \(f(-1)\), \(\varphi(-1)\)?
Решение
\[
\begin{aligned}
& f(0)=\frac{0-2}{0+1}=-2;\;f(1)=\frac{1-2}{1+1}=-\frac{1}{2};\\
& f(2)=\frac{2-2}{2+1}=0;\;f(-2)=\frac{-2-2}{-2+1}=4;\\
& f\left(-\frac{1}{2}\right)=\frac{-\frac{1}{2}-2}{-\frac{1}{2}+1}=-5;\;\left|f\left(\frac{1}{2}\right)\right|=\left|\frac{\frac{1}{2}-2}{\frac{1}{2}+1}\right|=1;\\
& f(\sqrt{2})=\frac{\sqrt{2}-2}{\sqrt{2}+1}=\frac{(\sqrt{2}-2)(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=4-3\sqrt{2}.
\end{aligned}
\]
\[
\begin{aligned}
& \varphi(0)=\frac{|0-2|}{0+1}=2;\;\varphi(1)=\frac{|1-2|}{1+1}=\frac{1}{2};\\
& \varphi(2)=\frac{|2-2|}{2+1}=0;\; \varphi(-2)=\frac{|-2-2|}{-2+1}=-4;\; \varphi(4)=\frac{|4-2|}{4+1}=\frac{2}{5}.
\end{aligned}
\]
Так как при \(x=-1\) имеем \(x+1=0\), то значения \(f(-1)\), \(\varphi(-1)\) не определены.
Ответ:
\(f(0)=-2\), \(f(1)=-\frac{1}{2}\), \(f(2)=0\), \(f(-2)=4\), \(f\left(-\frac{1}{2}\right)=-5\); \(f(\sqrt{2})=4-3\sqrt{2}\), \(\left|f\left(\frac{1}{2}\right)\right|=1\), \(\varphi(0)=2\), \(\varphi(1)=\frac{1}{2}\), \(\varphi(2)=0\), \(\varphi(-2)=-4\), \(\varphi(4)=\frac{2}{5}\).