Question-159046

Question Number 159046 by mnjuly1970 last updated on 12/Nov/21

Commented by cortano last updated on 13/Nov/21

f(x)=((3(1−cos^2 x)^2 +2)/(6cos^2 x+1)) f(x)=((3(cos^4 x−2cos^2 x+1)+2)/(6cos^2 x+1)) f(x)=((3cos^4 x−6cos^2 x+5)/(6cos^2 x+1)) let cos^2 x=u ∧ 0≤u≤1 ⇒y=((3u^2 −6u+5)/(6u+1)) when u=0⇒y=5 when u=1⇒y=(2/7) R_f : (2/7)≤f(x)≤5

$${f}\left({x}\right)=\frac{\mathrm{3}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)^{\mathrm{2}} +\mathrm{2}}{\mathrm{6cos}\:^{\mathrm{2}} {x}+\mathrm{1}} \\ $$$${f}\left({x}\right)=\frac{\mathrm{3}\left(\mathrm{cos}\:^{\mathrm{4}} {x}−\mathrm{2cos}\:^{\mathrm{2}} {x}+\mathrm{1}\right)+\mathrm{2}}{\mathrm{6cos}\:^{\mathrm{2}} {x}+\mathrm{1}} \\ $$$${f}\left({x}\right)=\frac{\mathrm{3cos}\:^{\mathrm{4}} {x}−\mathrm{6cos}\:^{\mathrm{2}} {x}+\mathrm{5}}{\mathrm{6cos}\:^{\mathrm{2}} {x}+\mathrm{1}} \\ $$$$\:{let}\:\mathrm{cos}\:^{\mathrm{2}} {x}={u}\:\wedge\:\mathrm{0}\leqslant{u}\leqslant\mathrm{1} \\ $$$$\Rightarrow{y}=\frac{\mathrm{3}{u}^{\mathrm{2}} −\mathrm{6}{u}+\mathrm{5}}{\mathrm{6}{u}+\mathrm{1}} \\ $$$${when}\:{u}=\mathrm{0}\Rightarrow{y}=\mathrm{5} \\ $$$${when}\:{u}=\mathrm{1}\Rightarrow{y}=\frac{\mathrm{2}}{\mathrm{7}} \\ $$$${R}_{{f}} \::\:\frac{\mathrm{2}}{\mathrm{7}}\leqslant{f}\left({x}\right)\leqslant\mathrm{5}\: \\ $$

Commented by mnjuly1970 last updated on 13/Nov/21

$${thank}\:{you}\:{so}\:{much}\:{master} \\ $$

Answered by MJS_new last updated on 12/Nov/21

$$\left[\frac{\mathrm{2}}{\mathrm{7}};\:\mathrm{5}\right] \\ $$

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Question-159046

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