Khảo sát tính tăng giảm của hàm số: \(y=f(x)=\frac{1}{x}\)
-  Hàm số \(f(x)=\frac{1}{x}\) có miền xác định \(D=R\setminus \left\{ {0} \right\}=(-\infty,0)\cup (0,+\infty)\)
-  Giả sử \(x_{1},x_{2} \in (-\infty,0)\) và \(x_{1}<x_{2}\).Ta xét:
\(f(x_{1})-f(x_{2})=\frac{1}{x_{1}}-\frac{1}{x_{2}}=\frac{x_{2}-x_{1}}{x_{1}x_{2}}>0 \Rightarrow  f(x_{1})>f(x_{2})\).
-  Vậy hàm số \(f(x)=\frac{1}{x}\) giảm trên \((-\infty,0)\).
-  Giả sử \(x_{1}, x_{2}\in (0,+\infty)\) và \(x_{1}<x_{2}\). Ta xét:
\( f(x_{1})-f(x_{2})=\frac{1}{x_{1}}-\frac{1}{x_{2}}=\frac{x_{2}-x_{1}}{x_{1}x_{2}}>0 \Rightarrow  f(x_{1})>f(x_{2})\).
-  Vậy hàm số \(f(x)=\frac{1}{x}\) giảm trên \((0,+\infty)\). 
-  Tóm lại hàm số \(f(x)=\frac{1}{x}\) giảm trên \(R\)

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