Question

Khảo sát sự biến thiên của

1)y=f(x)=\(\dfrac{1}{1-x}\)trên \(\left(1;+\infty\right)\)

2)y=f(x)=\(\sqrt{x+4}+\sqrt{x+1}\)trên \(\left(4;+\infty\right)\)

3)y=f(x)=\(\left|2x-4\right|+x\) trên \(\left(-\infty;2\right)\)

Answer 1

1)\(\forall x1,x2\in\left(1,+\infty\right),x1\ne x2\)

\(f\left(x1\right)-f\left(x2\right)=\dfrac{1}{1-x1}-\dfrac{1}{1-x2}=\dfrac{1-x2-1+x1}{\left(1-x1\right)\left(1-x2\right)}=\dfrac{x1-x2}{\left(1-x1\right)\left(1-x2\right)}\)

\(\dfrac{f\left(x1\right)-f\left(x2\right)}{x1-x2}=\dfrac{\dfrac{x1-x2}{\left(1-x1\right)\left(1-x2\right)}}{x1-x2}=\dfrac{1}{\left(1-x1\right)\left(1-x2\right)}\)

vì \(x1,x2\in\left(1;+\infty\right)\)nên \(\left\{{}\begin{matrix}x1>1\\x2>1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1-x1< 0\\1-x2< 0\end{matrix}\right.\) \(\Rightarrow\dfrac{1}{\left(1-x1\right)\left(1-x2\right)}>0\)

Vậy hàm số đồng biến trên \(\left(1;+\infty\right)\)