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Question Number 85146 by john santu last updated on 19/Mar/20

find minimum & maximum value   of function   f(x)= −sin^2 x+sin x−(1/2) , −π≤x≤π

$$\mathrm{find}\:\mathrm{minimum}\:\&\:\mathrm{maximum}\:\mathrm{value}\: \\ $$ $$\mathrm{of}\:\mathrm{function}\: \\ $$ $$\mathrm{f}\left(\mathrm{x}\right)=\:−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{sin}\:\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\:,\:−\pi\leqslant\mathrm{x}\leqslant\pi \\ $$

Commented bymr W last updated on 19/Mar/20

f(x)=−(sin^2  x−sin x+(1/4))−(1/4)  =−(sin x−(1/2))^2 −(1/4)  f_(max) =−(1/4) at sin x=(1/2), i.e. x=(π/6),((5π)/6)  f_(min) =−(5/2) at sin x=−1, i.e. x=−(π/2)

$${f}\left({x}\right)=−\left(\mathrm{sin}^{\mathrm{2}} \:{x}−\mathrm{sin}\:{x}+\frac{\mathrm{1}}{\mathrm{4}}\right)−\frac{\mathrm{1}}{\mathrm{4}} \\ $$ $$=−\left(\mathrm{sin}\:{x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}} \\ $$ $${f}_{{max}} =−\frac{\mathrm{1}}{\mathrm{4}}\:{at}\:\mathrm{sin}\:{x}=\frac{\mathrm{1}}{\mathrm{2}},\:{i}.{e}.\:{x}=\frac{\pi}{\mathrm{6}},\frac{\mathrm{5}\pi}{\mathrm{6}} \\ $$ $${f}_{{min}} =−\frac{\mathrm{5}}{\mathrm{2}}\:{at}\:\mathrm{sin}\:{x}=−\mathrm{1},\:{i}.{e}.\:{x}=−\frac{\pi}{\mathrm{2}} \\ $$

Answered by jagoll last updated on 19/Mar/20

Answered by Rio Michael last updated on 19/Mar/20

Maximum value = −0.25  note at maximum vaue  f^′ (x) = 0  minimum value = −2.5

$$\mathrm{Maximum}\:\mathrm{value}\:=\:−\mathrm{0}.\mathrm{25} \\ $$ $$\mathrm{note}\:\mathrm{at}\:\mathrm{maximum}\:\mathrm{vaue}\:\:{f}^{'} \left({x}\right)\:=\:\mathrm{0} \\ $$ $$\mathrm{minimum}\:\mathrm{value}\:=\:−\mathrm{2}.\mathrm{5} \\ $$

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