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A-function-f-is-define-by-f-3-2sinx-for-0-x-360-find-the-range-of-f-

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Question Number 23785 by tawa tawa last updated on 06/Nov/17
A function f is define by f : → 3 − 2sinx, for 0 ≤ x ≤ 360 find the range of f
Afunctionfisdefinebyf:→3−2sinx,for0⩽x⩽360findtherangeoff
Answered by mrW1 last updated on 06/Nov/17
sin x∈[−1,+1] f=3−2sin x ∈[1,5]
sinx∈[−1,+1]f=3−2sinx∈[1,5]
Commented by tawa tawa last updated on 06/Nov/17
How please, explain more.
Howplease,explainmore.
Commented by Joel577 last updated on 06/Nov/17
The range of function sin x is [−1,1] Hence the minimum value of f(x) = 3 − 2sin x is 1 and the max is 5
Therangeoffunctionsinxis[−1,1]Hencetheminimumvalueoff(x)=3−2sinxis1andthemaxis5
Commented by tawa tawa last updated on 06/Nov/17
God bless you sir. But i have not really understand.
Godblessyousir.Butihavenotreallyunderstand.
Commented by mrW1 last updated on 06/Nov/17
f(x)=3−2sin x is minimum if sin x=1, and the minimum is 3−2=1. f(x)=3−2sin x is maximum if sin x=−1, and the maximum is 3+2=5. ⇒1≤f(x)≤5
f(x)=3−2sinxisminimumifsinx=1,andtheminimumis3−2=1.f(x)=3−2sinxismaximumifsinx=−1,andthemaximumis3+2=5.⇒1⩽f(x)⩽5
Commented by tawa tawa last updated on 06/Nov/17
Wow, now i understand. God bless you sir.
Wow,nowiunderstand.Godblessyousir.
Answered by FilupES last updated on 09/Nov/17
f(x)=3−2sin(x), 0≤x≤360 −1≤sin(x)≤1, ∀x g=2sin(x) min(g)=−2 max(g)=2 f(x)=3−g min(f)=3−max(g)=1 max(f)=3−min(g)=5 ∴ 1≤f(x)≤5
f(x)=3−2sin(x),0⩽x⩽360−1⩽sin(x)⩽1,∀xg=2sin(x)min(g)=−2max(g)=2f(x)=3−gmin(f)=3−max(g)=1max(f)=3−min(g)=5∴1⩽f(x)⩽5
Commented by tawa tawa last updated on 11/Nov/17
God bless you sir
Godblessyousir

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