Question Number 175916 by mnjuly1970 last updated on 09/Sep/22
$$ \\ $$$$\:\:\:\:{f}\left({x}\right)=\:\mathrm{2}^{\:\sqrt[{\mathrm{3}}]{\:{sin}\left({x}\right)\:+\sqrt{\mathrm{3}}\:{cos}\left({x}\right)}\:} −\:\mathrm{2}^{\:\sqrt[{\mathrm{3}}]{−{sin}\left({x}\right)\:−\sqrt{\mathrm{3}}\:{cos}\left({x}\right)}} \\ $$$$\:\:\:\:\:\:\:{R}_{\:{f}} \:=? \\ $$
Commented by mnjuly1970 last updated on 09/Sep/22
$$\:{yes}\:{sir}\:{thanks}\:{alot} \\ $$
Answered by mahdipoor last updated on 09/Sep/22
![sinx+(√3)cosx=2(cos60.sinx+sin60.cosx)= 2sin(x+60)=u^3 ⇒2^u −2^(−u) =f ⇒ (df/du)=2^u .ln2+2^(−u) .ln2>0 ⇒f is increment function R_u =D_f =[−2^(1/3) ,+2^(1/3) ] R_f =[f(−2^(1/3) ),f(2^(1/3) )] −f(2^(1/3) )=f(−2^(1/3) )= (1/2^2^(1/3) )−2^2^(1/3) =((1−2^2^(4/3) )/2^2^(1/3) )](https://www.tinkutara.com/question/Q175920.png)
$${sinx}+\sqrt{\mathrm{3}}{cosx}=\mathrm{2}\left({cos}\mathrm{60}.{sinx}+{sin}\mathrm{60}.{cosx}\right)= \\ $$$$\mathrm{2}{sin}\left({x}+\mathrm{60}\right)={u}^{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{2}^{{u}} −\mathrm{2}^{−{u}} ={f}\:\Rightarrow \\ $$$$\frac{{df}}{{du}}=\mathrm{2}^{{u}} .{ln}\mathrm{2}+\mathrm{2}^{−{u}} .{ln}\mathrm{2}>\mathrm{0} \\ $$$$\Rightarrow{f}\:{is}\:{increment}\:{function} \\ $$$${R}_{{u}} ={D}_{{f}} =\left[−\mathrm{2}^{\mathrm{1}/\mathrm{3}} ,+\mathrm{2}^{\mathrm{1}/\mathrm{3}} \right]\: \\ $$$${R}_{{f}} =\left[{f}\left(−\mathrm{2}^{\mathrm{1}/\mathrm{3}} \right),{f}\left(\mathrm{2}^{\mathrm{1}/\mathrm{3}} \right)\right]\: \\ $$$$−{f}\left(\mathrm{2}^{\mathrm{1}/\mathrm{3}} \right)={f}\left(−\mathrm{2}^{\mathrm{1}/\mathrm{3}} \right)= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}^{\mathrm{1}/\mathrm{3}} } }−\mathrm{2}^{\mathrm{2}^{\mathrm{1}/\mathrm{3}} } =\frac{\mathrm{1}−\mathrm{2}^{\mathrm{2}^{\mathrm{4}/\mathrm{3}} } }{\mathrm{2}^{\mathrm{2}^{\mathrm{1}/\mathrm{3}} } } \\ $$
Commented by mnjuly1970 last updated on 09/Sep/22
$${thanks}\:{alot} \\ $$