Tính x1.x2.
Giải thích
A
Ta có \(f'\left( x \right) = {\left( {x - {x^2}} \right)^\prime }{e^{x - {x^2}}} = \left( {1 - 2x} \right).{e^{x - {x^2}}}\);
\(f''\left( x \right) = - 2{e^{x - {x^2}}} + {\left( {1 - 2x} \right)^2}{e^{x - {x^2}}}\).
Có f"(x) = 0 \( \Leftrightarrow - 2{e^{x - {x^2}}} + {\left( {1 - 2x} \right)^2}{e^{x - {x^2}}} = 0\)\( \Leftrightarrow \left[ {{{\left( {1 - 2x} \right)}^2} - 2} \right]{e^{x - {x^2}}} = 0\)\( \Leftrightarrow {\left( {1 - 2x} \right)^2} - 2 = 0\)
\( \Leftrightarrow \left[ \begin{array}{l}1 - 2x = \sqrt 2 \\1 - 2x = - \sqrt 2 \end{array} \right.\)\( \Leftrightarrow \left[ \begin{array}{l}x = \frac{{1 - \sqrt 2 }}{2}\\x = \frac{{1 + \sqrt 2 }}{2}\end{array} \right.\). Do đó \({x_1}{x_2} = - \frac{1}{4}\).