4: \(y=\frac{3x-1}{x-3}\)
=>y'=\(\frac{\left(3x-1\right)^{\prime}\left(x-3\right)-\left(3x-1\right)\left(x-3\right)^{\prime}}{\left(x-3\right)^2}=\frac{3\left(x-3\right)-\left(3x-1\right)}{\left(x-3\right)^2}=\frac{3x-9-3x+1}{\left(x-3\right)^2}=\frac{-8}{\left(x-3\right)^2}\)
Đặt y'=0
=>x∈∅
\(f\left(0\right)=\frac{3\cdot0-1}{0-3}=\frac{-1}{-3}=\frac13;f\left(2\right)=\frac{3\cdot2-1}{2-3}=\frac{5-1}{-1}=-4\)
=>\(f\left(x\right)_{\max\left\lbrack0;2\right\rbrack}\) =1/3 khi x=0
5: \(y=\frac{x-1}{x+1}\)
=>y'=\(\frac{\left(x-1\right)^{\prime}\left(x+1\right)-\left(x-1\right)\left(x+1\right)^{\prime}}{\left(x+1\right)^2}=\frac{x+1-x+1}{\left(x+1\right)^2}=\frac{2}{\left(x+1\right)^2}\)
Đặt y'=0
=>x∈∅
\(f\left(0\right)=\frac{0-1}{0+1}=-1\)
\(f\left(3\right)=\frac{3-1}{3+1}=\frac24=\frac12\)
=>\(f\left(x\right)_{\min\left\lbrack0;3\right\rbrack}=-1\) khi x=0
