Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 87171 by naka3546 last updated on 03/Apr/20

Given f(x) = 2 sin^2 x − sin x + 1 , 0 ≤ x ≤ 2π Find maximum and minumum value of f(x) without differential .

$${Given}\:\:{f}\left({x}\right)\:\:=\:\:\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:{x}\:−\:\mathrm{sin}\:{x}\:+\:\mathrm{1}\:\:,\:\:\mathrm{0}\:\:\leqslant\:{x}\:\leqslant\:\mathrm{2}\pi \\ $$$${Find}\:\:{maximum}\:\:{and}\:\:{minumum}\:\:{value} \\ $$$${of}\:\:{f}\left({x}\right)\:\:{without}\:\:{differential}\:. \\ $$

Commented by john santu last updated on 03/Apr/20

for sin x = 1 ⇒y_1 = 2−1+1 = 2 for sin x = −1 ⇒y_2 = 2+1+1 =4 for sin x = (1/4) ⇒y_3 = (1/8)−(1/4)+1 y_3 = (1/8)+(3/4) = (7/8) max = 4 & min = (7/8)

$$\mathrm{for}\:\mathrm{sin}\:\mathrm{x}\:=\:\mathrm{1}\:\Rightarrow\mathrm{y}_{\mathrm{1}} \:=\:\mathrm{2}−\mathrm{1}+\mathrm{1}\:=\:\mathrm{2} \\ $$$$\mathrm{for}\:\mathrm{sin}\:\mathrm{x}\:=\:−\mathrm{1}\:\Rightarrow\mathrm{y}_{\mathrm{2}} \:=\:\mathrm{2}+\mathrm{1}+\mathrm{1}\:=\mathrm{4} \\ $$$$\mathrm{for}\:\mathrm{sin}\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow\mathrm{y}_{\mathrm{3}} \:=\:\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{1} \\ $$$$\mathrm{y}_{\mathrm{3}} \:=\:\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{3}}{\mathrm{4}}\:=\:\frac{\mathrm{7}}{\mathrm{8}} \\ $$$$\mathrm{max}\:=\:\mathrm{4}\:\&\:\mathrm{min}\:=\:\frac{\mathrm{7}}{\mathrm{8}} \\ $$

Commented by mr W last updated on 03/Apr/20

f(x)=2(sin^2 x−2×(1/4)sin x+(1/4^2 ))+1−2×(1/4^2 ) =2(sin x−(1/4))^2 +(7/8) min.=(7/8) max.=2(−1−(1/4))^2 +(7/8)=4

$${f}\left({x}\right)=\mathrm{2}\left(\mathrm{sin}^{\mathrm{2}} \:{x}−\mathrm{2}×\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:{x}+\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} }\right)+\mathrm{1}−\mathrm{2}×\frac{\mathrm{1}}{\mathrm{4}^{\mathrm{2}} } \\ $$$$=\mathrm{2}\left(\mathrm{sin}\:{x}−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{7}}{\mathrm{8}} \\ $$$${min}.=\frac{\mathrm{7}}{\mathrm{8}} \\ $$$${max}.=\mathrm{2}\left(−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{7}}{\mathrm{8}}=\mathrm{4} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com