Question

Answer 1

\(y=\dfrac{x^2+3}{x-1}\)

\(y'=\dfrac{x^2-2x-3}{\left(x-1\right)^2}\)

\(y'=0\Leftrightarrow\dfrac{x^2-2x-3}{\left(x-1\right)^2}\Leftrightarrow\left\{{}\begin{matrix}x_1=-1\\x_2=3\end{matrix}\right.\)

Ta có: \(y\left(2\right)=7;y\left(3\right)=6;y\left(4\right)=\dfrac{19}{3}\)

Suy ra: \(Miny=Min\left\{y\left(2\right),y\left(3\right),y\left(4\right)\right\}=y\left(3\right)=6\)

         \(x\in\left[2;4\right]\)