Dùng định nghĩa, tìm khoảng tăng giảm của hàm số:\(f(x)=\frac{3}{x^{2}+1}\).
Miền xác định của hàm số là \(D=R\)
Với  \(\forall x_{1},x_{2} \in\ D, x_{1}< x_{2}\). Ta xét:
\(f(x_{1})-f(x_{2})=\frac{3}{x_{1}^{2}+1}-\frac{3}{x^{2}+1}=\frac{3(x_{2}^{2}-x_{1}^{2})}{(x_{1}^{2}+1)(x_{2}^{2}+1)}=\frac{3(x_{2}+x_{1})(x_{2}-x_{1})}{(x_{1}^{2}+1)(x_{2}^{2}+1)}\)
-  Khi \(x_{1}<x_{2}<0\Rightarrow x_{1}+x_{2}<0\) và \(x_{1}-x_{2}>0\)
\(\Rightarrow f(x_{1})-f(x_{2})=\frac{3(x_{2}+x_{1})(x_{2}-x_{1})}{(x_{1}^{2}+1)(x_{2}^{2}+1)}<0\)
\(\Rightarrow f(x_{1})<f(x_{2})\).
Vậy hàm \(f(x)\) tăng dần trên \((-\infty,0)\).
-  Khi \(0<x_{1}<x_{2}\Rightarrow x_{1}+x_{2}>0\) và \(x_{2}-x_{1}>0\)
\(\Rightarrow f(x_{1})-f(x_{2})=\frac{3(x_{2}+x_{1})(x_{2}-x_{1})}{(x_{1}^{2}+1)(x_{2}^{2}+1)}>0\)
\(\Rightarrow f(x_{1})>f(x_{2})\).
Vậy hàm \(f(x)\) giảm trên \((0,+\infty)\)

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