Bài tập 5.22 trang 203 SBT Toán 11

a) Ta có:

\(\begin{array}{l}
f\prime \left( x \right) = {x^6} - 9{x^3} + 8\\
 \Rightarrow f\prime \left( x \right) > 0 \Leftrightarrow {x^6} - 9{x^3} + 8 > 0\\
 \Leftrightarrow \left( {{x^3} - 1} \right)\left( {{x^3} - 8} \right) > 0\\
 \Leftrightarrow \left[ \begin{array}{l}
{x^3} < 1\\
{x^3} > 8
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x < 1\\
x > 2
\end{array} \right.
\end{array}\)

b) Ta có:

\({g'\left( x \right) = \frac{{\left( {2x - 5} \right)\left( {x - 2} \right) - \left( {{x^2} - 5x + 4} \right)}}{{{{\left( {x - 2} \right)}^2}}} = \frac{{{x^2} - 4x + 6}}{{{{\left( {x - 2} \right)}^2}}}}\)

Suy ra

\({g'\left( x \right) \le 0 \Leftrightarrow \frac{{{x^2} - 4x + 6}}{{{{\left( {x - 2} \right)}^2}}} \le 0 \Rightarrow {x^2} - 4x + 6 \le 0 \Leftrightarrow {{\left( {x - 2} \right)}^2} + 2 \le 0\left( {vn} \right)}\)