Question Number 12704 by @ANTARES_VY last updated on 29/Apr/17
![y=sin2x−x (x∈[0;𝛑]) find the range of the function.](https://www.tinkutara.com/question/Q12704.png)
$$\boldsymbol{\mathrm{y}}=\boldsymbol{\mathrm{sin}}\mathrm{2}\boldsymbol{\mathrm{x}}−\boldsymbol{\mathrm{x}}\:\:\left(\boldsymbol{\mathrm{x}}\in\left[\mathrm{0};\boldsymbol{\pi}\right]\right)\:\:\boldsymbol{\mathrm{find}}\:\:\boldsymbol{\mathrm{the}} \\ $$$$\boldsymbol{\mathrm{range}}\:\:\boldsymbol{\mathrm{of}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{function}}. \\ $$
Answered by mrW1 last updated on 29/Apr/17
![y=sin 2x−x y′=2cos 2x−1 for y to be local max. or min.: y′=0⇒2cos 2x−1=0⇒cos 2x=(1/2) ⇒2x=(π/3),((5π)/3) ⇒x=(π/6),((5π)/6) at x=π: y=sin (2π)−π=−π at x=((5π)/6): y=sin (((5π)/3))−((5π)/6)=−((√3)/2)−((5π)/6)=−((3(√3)+5π)/6)<−π at x=0: y=sin (0)+0=0 at x=(π/6): y=sin ((π/3))−(π/6)=((√3)/2)−(π/6)=((3(√3)−π)/6)>0 for x∈[0,π] y∈[−((3(√3)+5π)/6),((3(√3)−π)/6)]](https://www.tinkutara.com/question/Q12716.png)
$${y}=\mathrm{sin}\:\mathrm{2}{x}−{x} \\ $$$${y}'=\mathrm{2cos}\:\mathrm{2}{x}−\mathrm{1} \\ $$$$ \\ $$$${for}\:{y}\:{to}\:{be}\:{local}\:{max}.\:{or}\:{min}.: \\ $$$${y}'=\mathrm{0}\Rightarrow\mathrm{2cos}\:\mathrm{2}{x}−\mathrm{1}=\mathrm{0}\Rightarrow\mathrm{cos}\:\mathrm{2}{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}{x}=\frac{\pi}{\mathrm{3}},\frac{\mathrm{5}\pi}{\mathrm{3}} \\ $$$$\Rightarrow{x}=\frac{\pi}{\mathrm{6}},\frac{\mathrm{5}\pi}{\mathrm{6}} \\ $$$$ \\ $$$${at}\:{x}=\pi: \\ $$$${y}=\mathrm{sin}\:\left(\mathrm{2}\pi\right)−\pi=−\pi \\ $$$${at}\:{x}=\frac{\mathrm{5}\pi}{\mathrm{6}}: \\ $$$${y}=\mathrm{sin}\:\left(\frac{\mathrm{5}\pi}{\mathrm{3}}\right)−\frac{\mathrm{5}\pi}{\mathrm{6}}=−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\mathrm{5}\pi}{\mathrm{6}}=−\frac{\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{5}\pi}{\mathrm{6}}<−\pi \\ $$$$ \\ $$$${at}\:{x}=\mathrm{0}: \\ $$$${y}=\mathrm{sin}\:\left(\mathrm{0}\right)+\mathrm{0}=\mathrm{0} \\ $$$${at}\:{x}=\frac{\pi}{\mathrm{6}}: \\ $$$${y}=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}\right)−\frac{\pi}{\mathrm{6}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}−\frac{\pi}{\mathrm{6}}=\frac{\mathrm{3}\sqrt{\mathrm{3}}−\pi}{\mathrm{6}}>\mathrm{0} \\ $$$$ \\ $$$${for}\:{x}\in\left[\mathrm{0},\pi\right] \\ $$$${y}\in\left[−\frac{\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{5}\pi}{\mathrm{6}},\frac{\mathrm{3}\sqrt{\mathrm{3}}−\pi}{\mathrm{6}}\right] \\ $$