a: \(f\left(x\right)=\dfrac{x^2+4}{x}\)
=>\(f\left(x\right)=x+\dfrac{4}{x}\)
=>\(f'\left(x\right)=1-\dfrac{4}{x^2}\)
Đặt f'(x)=0
=>\(-\dfrac{4}{x^2}+1=0\)
=>\(x^2=4\)
=>\(\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
mà 1<=x<=3
nên x=2
\(f\left(2\right)=\dfrac{2^2+4}{2}=\dfrac{4+4}{2}=4\)
\(f\left(1\right)=\dfrac{1^2+4}{1}=\dfrac{1+4}{1}=5;f\left(3\right)=\dfrac{3^2+4}{3}=\dfrac{13}{3}\)
Vì 4<13/3<5
nên \(f\left(x\right)_{min\left[1;3\right]}=4;f\left(x\right)_{max\left[1;3\right]}=5\)
b: \(y=x-\dfrac{1}{x}\)
=>\(y'=1+\dfrac{1}{x^2}\)
Đặt y'=0
=>\(1+\dfrac{1}{x^2}=0\)
=>\(x^2=-1\)(vô lý)
\(y\left(1\right)=1-\dfrac{1}{1}=1-1=0;y\left(3\right)=3-\dfrac{1}{3}=\dfrac{8}{3}\)
=>\(y_{min\left[1;3\right]}=0;y_{max\left[1;3\right]}=\dfrac{8}{3}\)
