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Question Number 123337 by aurpeyz last updated on 25/Nov/20

find the maximum and minimum values of 2cos2x−cos4x in the range 0<x<π. sketch the curve

$${find}\:{the}\:{maximum}\:{and}\:{minimum} \\ $$ $${values}\:{of}\:\mathrm{2}{cos}\mathrm{2}{x}−{cos}\mathrm{4}{x}\:{in}\:{the}\:{range} \\ $$ $$\mathrm{0}<{x}<\pi.\:{sketch}\:{the}\:{curve} \\ $$

Commented bybemath last updated on 25/Nov/20

max = 1.5 ; min = −3 range ⇒ −3≤y≤1.5

$${max}\:=\:\mathrm{1}.\mathrm{5}\:;\:{min}\:=\:−\mathrm{3}\: \\ $$ $${range}\:\Rightarrow\:−\mathrm{3}\leqslant{y}\leqslant\mathrm{1}.\mathrm{5} \\ $$

Commented bybemath last updated on 25/Nov/20

Commented byaurpeyz last updated on 25/Nov/20

thanks

$${thanks} \\ $$

Commented byaurpeyz last updated on 25/Nov/20

pls explain how you got min=−3. i tried so hard but couldnt

$${pls}\:{explain}\:{how}\:{you}\:{got}\:{min}=−\mathrm{3}.\: \\ $$ $${i}\:{tried}\:{so}\:{hard}\:{but}\:{couldnt} \\ $$

Commented bybemath last updated on 25/Nov/20

let ∅(x)=2cos 2x−cos 4x ∅(x)=2cos 2x−(2cos^2 2x−1) ∅(x)=−2cos^2 2x+2cos 2x+1 ∅(x) = −2(cos^2 2x−cos 2x)+1 ∅(x)=−2[(cos 2x−(1/2))^2 −(1/4)]+1 ∅(x)=−2(cos 2x−(1/2))^2 +(3/2) → { ((max when cos 2x=(1/2))),((min when cos 2x=−1)) :}

$${let}\:\emptyset\left({x}\right)=\mathrm{2cos}\:\mathrm{2}{x}−\mathrm{cos}\:\mathrm{4}{x}\: \\ $$ $$\emptyset\left({x}\right)=\mathrm{2cos}\:\mathrm{2}{x}−\left(\mathrm{2cos}\:^{\mathrm{2}} \mathrm{2}{x}−\mathrm{1}\right) \\ $$ $$\emptyset\left({x}\right)=−\mathrm{2cos}\:^{\mathrm{2}} \mathrm{2}{x}+\mathrm{2cos}\:\mathrm{2}{x}+\mathrm{1} \\ $$ $$\emptyset\left({x}\right)\:=\:−\mathrm{2}\left(\mathrm{cos}\:^{\mathrm{2}} \mathrm{2}{x}−\mathrm{cos}\:\mathrm{2}{x}\right)+\mathrm{1} \\ $$ $$\emptyset\left({x}\right)=−\mathrm{2}\left[\left(\mathrm{cos}\:\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right]+\mathrm{1} \\ $$ $$\emptyset\left({x}\right)=−\mathrm{2}\left(\mathrm{cos}\:\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}} \\ $$ $$\rightarrow\begin{cases}{{max}\:{when}\:\mathrm{cos}\:\mathrm{2}{x}=\frac{\mathrm{1}}{\mathrm{2}}}\\{{min}\:{when}\:\mathrm{cos}\:\mathrm{2}{x}=−\mathrm{1}}\end{cases} \\ $$

Commented byMJS_new last updated on 25/Nov/20

cos 2x =−1+2cos^2 x cos 4x =−1+2cos^2 2x = =−1+2(−1+2cos^2 x)^2 = =1−8cos^2 x +8cos^4 x ⇒2cos 2x −cos 4x = =−3+12cos^2 x −8cos^4 x =f(x) we can always use t=tan (x/2) ⇔ x=2arctan t ⇒ sin x =((2t)/(t^2 +1))∧cos x =−((t^2 −1)/(t^2 +1)) ⇒ −3+12cos^2 x −8cos^4 x = =((t^8 +20t^6 −90t^4 +20t^2 +1)/((t^2 +1)^4 ))=g(t) (dg/dt)=0 −32((t(t^6 −15t^4 +15t^2 −1))/((t^2 +1)^5 ))=0 t(t^6 −15t^4 +15t^2 −1)=0 t_1 =0 t^6 −15t^4 +15t^2 −1=0 trying factors of −1 ⇒ t_(2, 3) =±1 t^4 −14t^2 +1=0 ⇒ t^2 =7±4(√3) ⇒ t=±(√(7±4(√3)))=±(2±(√3)) now we have t∈{−2−(√3), −1, −2+(√3), 0, 2−(√3), 1, 2+(√3)} now t=tan (x/2) ⇒ x=2nπ+2arctan t for 0<x<π we get (x=0∨x=π ⇒ f(x)=1) x=(π/6) ⇒ f(x)=(3/2) x=(π/2) ⇒ f(x)=−3 x=((5π)/6) ⇒ f(x)=(3/2) ⇒ minimum is −3 and maximum is (3/2)

$$\mathrm{cos}\:\mathrm{2}{x}\:=−\mathrm{1}+\mathrm{2cos}^{\mathrm{2}} \:{x} \\ $$ $$\mathrm{cos}\:\mathrm{4}{x}\:=−\mathrm{1}+\mathrm{2cos}^{\mathrm{2}} \:\mathrm{2}{x}\:= \\ $$ $$\:\:\:\:\:=−\mathrm{1}+\mathrm{2}\left(−\mathrm{1}+\mathrm{2cos}^{\mathrm{2}} \:{x}\right)^{\mathrm{2}} = \\ $$ $$\:\:\:\:\:=\mathrm{1}−\mathrm{8cos}^{\mathrm{2}} \:{x}\:+\mathrm{8cos}^{\mathrm{4}} \:{x} \\ $$ $$\Rightarrow\mathrm{2cos}\:\mathrm{2}{x}\:−\mathrm{cos}\:\mathrm{4}{x}\:= \\ $$ $$\:\:\:\:\:=−\mathrm{3}+\mathrm{12cos}^{\mathrm{2}} \:{x}\:−\mathrm{8cos}^{\mathrm{4}} \:{x}\:={f}\left({x}\right) \\ $$ $$\mathrm{we}\:\mathrm{can}\:\mathrm{always}\:\mathrm{use} \\ $$ $${t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\Leftrightarrow\:{x}=\mathrm{2arctan}\:{t} \\ $$ $$\Rightarrow\:\mathrm{sin}\:{x}\:=\frac{\mathrm{2}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\wedge\mathrm{cos}\:{x}\:=−\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}} \\ $$ $$\Rightarrow\:−\mathrm{3}+\mathrm{12cos}^{\mathrm{2}} \:{x}\:−\mathrm{8cos}^{\mathrm{4}} \:{x}\:= \\ $$ $$=\frac{{t}^{\mathrm{8}} +\mathrm{20}{t}^{\mathrm{6}} −\mathrm{90}{t}^{\mathrm{4}} +\mathrm{20}{t}^{\mathrm{2}} +\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{4}} }={g}\left({t}\right) \\ $$ $$\frac{{dg}}{{dt}}=\mathrm{0} \\ $$ $$−\mathrm{32}\frac{{t}\left({t}^{\mathrm{6}} −\mathrm{15}{t}^{\mathrm{4}} +\mathrm{15}{t}^{\mathrm{2}} −\mathrm{1}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{5}} }=\mathrm{0} \\ $$ $${t}\left({t}^{\mathrm{6}} −\mathrm{15}{t}^{\mathrm{4}} +\mathrm{15}{t}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0} \\ $$ $${t}_{\mathrm{1}} =\mathrm{0} \\ $$ $${t}^{\mathrm{6}} −\mathrm{15}{t}^{\mathrm{4}} +\mathrm{15}{t}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$ $$\mathrm{trying}\:\mathrm{factors}\:\mathrm{of}\:−\mathrm{1}\:\Rightarrow\:{t}_{\mathrm{2},\:\mathrm{3}} =\pm\mathrm{1} \\ $$ $${t}^{\mathrm{4}} −\mathrm{14}{t}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$ $$\Rightarrow\:{t}^{\mathrm{2}} =\mathrm{7}\pm\mathrm{4}\sqrt{\mathrm{3}} \\ $$ $$\Rightarrow\:{t}=\pm\sqrt{\mathrm{7}\pm\mathrm{4}\sqrt{\mathrm{3}}}=\pm\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\right) \\ $$ $$\mathrm{now}\:\mathrm{we}\:\mathrm{have} \\ $$ $${t}\in\left\{−\mathrm{2}−\sqrt{\mathrm{3}},\:−\mathrm{1},\:−\mathrm{2}+\sqrt{\mathrm{3}},\:\mathrm{0},\:\mathrm{2}−\sqrt{\mathrm{3}},\:\mathrm{1},\:\mathrm{2}+\sqrt{\mathrm{3}}\right\} \\ $$ $$\mathrm{now}\:{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\Rightarrow\:{x}=\mathrm{2}{n}\pi+\mathrm{2arctan}\:{t} \\ $$ $$\mathrm{for}\:\mathrm{0}<{x}<\pi\:\mathrm{we}\:\mathrm{get} \\ $$ $$\left({x}=\mathrm{0}\vee{x}=\pi\:\Rightarrow\:{f}\left({x}\right)=\mathrm{1}\right) \\ $$ $${x}=\frac{\pi}{\mathrm{6}}\:\Rightarrow\:{f}\left({x}\right)=\frac{\mathrm{3}}{\mathrm{2}} \\ $$ $${x}=\frac{\pi}{\mathrm{2}}\:\Rightarrow\:{f}\left({x}\right)=−\mathrm{3} \\ $$ $${x}=\frac{\mathrm{5}\pi}{\mathrm{6}}\:\Rightarrow\:{f}\left({x}\right)=\frac{\mathrm{3}}{\mathrm{2}} \\ $$ $$\Rightarrow\:\mathrm{minimum}\:\mathrm{is}\:−\mathrm{3}\:\mathrm{and}\:\mathrm{maximum}\:\mathrm{is}\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$

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