Question Number 199538 by cherokeesay last updated on 04/Nov/23
Answered by mr W last updated on 05/Nov/23
$$\left.{a}\right) \\ $$$${f}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{1}−{m}\left({x}−\mathrm{1}\right) \\ $$$${f}'\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} −{m}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} −\mathrm{1}−\mathrm{3}{x}^{\mathrm{2}} \left({x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{2}{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{2}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{1}\:\Rightarrow{m}=\mathrm{3} \\ $$$$\Rightarrow{x}=−\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow{m}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\left.{b}\right) \\ $$$${m}=\mathrm{3} \\ $$$${f}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{1}−\mathrm{3}\left({x}−\mathrm{1}\right)=\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}+\mathrm{2}\right) \\ $$$${zeros}\:{at}\:{x}=−\mathrm{2},\:{x}=\mathrm{1} \\ $$$$… \\ $$$$\left.{c}\right) \\ $$$$\int_{−\mathrm{2}} ^{\mathrm{1}} \left({x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{2}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}^{\mathrm{4}} −\mathrm{2}^{\mathrm{4}} \right)−\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{1}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} \right)+\mathrm{2}\left(\mathrm{1}+\mathrm{2}\right)=\frac{\mathrm{27}}{\mathrm{4}} \\ $$
Commented by mr W last updated on 05/Nov/23
Commented by cherokeesay last updated on 05/Nov/23
$${thank}\:{you}\:{so}\:{much}\:{master}\:!!! \\ $$$$ \\ $$