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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
findminimum&maximumvalueoffunctionf(x)=sin2x+sinx12,πxπ
Commented by mr 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(x)=(sin2xsinx+14)14=(sinx12)214fmax=14atsinx=12,i.e.x=π6,5π6fmin=52atsinx=1,i.e.x=π2
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
Maximumvalue=0.25noteatmaximumvauef(x)=0minimumvalue=2.5

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