a: \(f\left(x\right)=\dfrac{x^2-8x}{x+1}\)
=>\(f'\left(x\right)=\dfrac{\left(x^2-8x\right)'\cdot\left(x+1\right)-\left(x^2-8x\right)\left(x+1\right)'}{\left(x+1\right)^2}\)
=>\(f'\left(x\right)=\dfrac{\left(2x-8\right)\left(x+1\right)-\left(x^2-8x\right)}{\left(x+1\right)^2}\)
=>\(f'\left(x\right)=\dfrac{2x^2+2x-8x-8-x^2+8x}{\left(x+1\right)^2}=\dfrac{x^2+2x-8}{\left(x+1\right)^2}\)
Đặt f'(x)=0
=>\(\dfrac{x^2+2x-8}{\left(x+1\right)^2}=0\)
=>\(x^2+2x-8=0\)
=>(x+4)(x-2)=0
=>\(\left[{}\begin{matrix}x=-4\left(loại\right)\\x=2\left(nhận\right)\end{matrix}\right.\)
\(f\left(1\right)=\dfrac{1^2-8\cdot1}{1+1}=\dfrac{-7}{2};f\left(2\right)=\dfrac{2^2-8\cdot2}{2+1}=\dfrac{-4}{3};f\left(3\right)=\dfrac{3^2-8\cdot3}{3+1}=\dfrac{-15}{4}\)
Vì -15/4<-7/2<-4/3
nên \(f\left(x\right)_{min\left[1;3\right]}=-\dfrac{15}{4};f\left(x\right)_{max\left[1;3\right]}=-\dfrac{4}{3}\)
b: \(y=x+\dfrac{4}{x}\)
=>\(y'=1-\dfrac{4}{x^2}\)
Đặt y'=0
=>\(1-\dfrac{4}{x^2}=0\)
=>\(\dfrac{4}{x^2}=1\)
=>\(x^2=4\)
=>\(\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
mà \(x\in\left[-5;-1\right]\)
nên x=-2
\(y\left(-2\right)=-2+\dfrac{4}{-2}=-2-2=-4;y\left(-5\right)=-5+\dfrac{4}{-5}=-\dfrac{29}{5}\)
\(y\left(-1\right)=-1+\dfrac{4}{-1}=-1-4=-5\)
Vì -29/5<-5<-4
nên \(y_{min\left[-5;-1\right]}=-\dfrac{29}{5};y_{max\left[-5;-1\right]}=-4\)
