Học lớp hướng dẫn giải
\(\begin{array}{l} y = \frac{{{x^2} + 3}}{{x - 1}}\\ y' = \frac{{{x^2} - 2x - 3}}{{{{\left( {x - 1} \right)}^2}}}\\ y' = 0 \Leftrightarrow \frac{{{x^2} - 2x - 3}}{{{{\left( {x - 1} \right)}^2}}} \Leftrightarrow \left[ \begin{array}{l} {x_1} = - 1\\ {x_2} = 3 \end{array} \right. \end{array}\)
Ta có: \(y(2) = 7;\,y(3) = 6;\,y(4) = \frac{{19}}{3}\)
Suy ra: \(\mathop {Min}\limits_{x \in \left[ {2;4} \right]} y = Min\left\{ {y\left( 2 \right),y\left( 3 \right),y\left( 4 \right)} \right\} = y\left( 3 \right) = 6\)
