Giải bài 5.52 trang 208 - SBT Đại số và Giải tích lớp 11
Tìm \(f'\left( 1 \right),f'\left( 2 \right),f'\left( 3 \right) \) nếu \(f\left( x \right)=\left( x-1 \right){{\left( x-2 \right)}^{2}}{{\left( x-3 \right)}^{3}} \)
Ta có:
\(\begin{align} & y=\left( x-1 \right){{\left( x-2 \right)}^{2}}{{\left( x-3 \right)}^{3}} \\ & \Rightarrow {y}'={{\left( x-2 \right)}^{2}}{{\left( x-3 \right)}^{3}}+\left( x-1 \right)\left[ 2\left( x-2 \right){{\left( x-3 \right)}^{3}}+{{\left( x-2 \right)}^{2}}.3.{{\left( x-3 \right)}^{2}} \right] \\ & ={{\left( x-2 \right)}^{2}}{{\left( x-3 \right)}^{3}}+\left( x-1 \right)\left( x-2 \right){{\left( x-3 \right)}^{2}}\left[ 2\left( x-3 \right)+3\left( x-2 \right) \right] \\ & ={{\left( x-2 \right)}^{2}}{{\left( x-3 \right)}^{3}}+\left( x-1 \right)\left( x-2 \right){{\left( x-3 \right)}^{2}}\left( 5x-12 \right) \\ & =\left( x-2 \right){{\left( x-3 \right)}^{2}}\left( {{x}^{2}}-5x+6+5{{x}^{2}}-17x+12 \right) \\ & =\left( x-2 \right)\left( x-3 \right)^2\left( 6{{x}^{2}}-22x+18 \right) \\ \end{align} \)
Suy ra:
\(f'(1)=(1-2)(1-3)^2(6-22+18)=8\)
\(f'(2)=(2-2)(2-3)^2(6.4-22.2+18)=0\)
\(f'(3)=(3-2)(3-3)^2(6.9-22.3+18)=0\)