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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

$$\mathrm{A}\:\mathrm{function}\:\mathrm{f}\:\mathrm{is}\:\mathrm{define}\:\mathrm{by}\:\:\mathrm{f}\::\:\rightarrow\:\mathrm{3}\:−\:\mathrm{2sinx},\:\:\mathrm{for}\:\:\mathrm{0}\:\leqslant\:\mathrm{x}\:\leqslant\:\mathrm{360} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\:\mathrm{f} \\ $$

Answered by mrW1 last updated on 06/Nov/17

sin x∈[−1,+1] f=3−2sin x ∈[1,5]

$$\mathrm{sin}\:{x}\in\left[−\mathrm{1},+\mathrm{1}\right] \\ $$$${f}=\mathrm{3}−\mathrm{2sin}\:{x}\:\in\left[\mathrm{1},\mathrm{5}\right] \\ $$

Commented by tawa tawa last updated on 06/Nov/17

How please, explain more.

$$\mathrm{How}\:\mathrm{please},\:\mathrm{explain}\:\mathrm{more}.\: \\ $$

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

$$\mathrm{The}\:\mathrm{range}\:\mathrm{of}\:\mathrm{function}\:\mathrm{sin}\:{x}\:\mathrm{is}\:\left[−\mathrm{1},\mathrm{1}\right] \\ $$$$\mathrm{Hence}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:{f}\left({x}\right)\:=\:\mathrm{3}\:−\:\mathrm{2sin}\:{x} \\ $$$$\mathrm{is}\:\mathrm{1}\:\mathrm{and}\:\mathrm{the}\:\mathrm{max}\:\mathrm{is}\:\mathrm{5} \\ $$

Commented by tawa tawa last updated on 06/Nov/17

God bless you sir. But i have not really understand.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{But}\:\mathrm{i}\:\mathrm{have}\:\mathrm{not}\:\mathrm{really}\:\mathrm{understand}. \\ $$

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

$${f}\left({x}\right)=\mathrm{3}−\mathrm{2sin}\:{x}\:{is}\:{minimum}\:{if}\:\mathrm{sin}\:{x}=\mathrm{1}, \\ $$$${and}\:{the}\:{minimum}\:{is}\:\mathrm{3}−\mathrm{2}=\mathrm{1}. \\ $$$$ \\ $$$${f}\left({x}\right)=\mathrm{3}−\mathrm{2sin}\:{x}\:{is}\:{maximum}\:{if}\:\mathrm{sin}\:{x}=−\mathrm{1}, \\ $$$${and}\:{the}\:{maximum}\:{is}\:\mathrm{3}+\mathrm{2}=\mathrm{5}. \\ $$$$ \\ $$$$\Rightarrow\mathrm{1}\leqslant{f}\left({x}\right)\leqslant\mathrm{5} \\ $$

Commented by tawa tawa last updated on 06/Nov/17

Wow, now i understand. God bless you sir.

$$\mathrm{Wow},\:\mathrm{now}\:\mathrm{i}\:\mathrm{understand}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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

$${f}\left({x}\right)=\mathrm{3}−\mathrm{2sin}\left({x}\right),\:\:\:\:\:\mathrm{0}\leqslant{x}\leqslant\mathrm{360} \\ $$$$−\mathrm{1}\leqslant\mathrm{sin}\left({x}\right)\leqslant\mathrm{1},\:\:\:\:\forall{x} \\ $$$$\: \\ $$$${g}=\mathrm{2sin}\left({x}\right) \\ $$$$\mathrm{min}\left({g}\right)=−\mathrm{2} \\ $$$$\mathrm{max}\left({g}\right)=\mathrm{2} \\ $$$$\: \\ $$$${f}\left({x}\right)=\mathrm{3}−{g} \\ $$$$\mathrm{min}\left({f}\right)=\mathrm{3}−\mathrm{max}\left({g}\right)=\mathrm{1} \\ $$$$\mathrm{max}\left({f}\right)=\mathrm{3}−\mathrm{min}\left({g}\right)=\mathrm{5} \\ $$$$\: \\ $$$$\therefore\:\mathrm{1}\leqslant{f}\left({x}\right)\leqslant\mathrm{5} \\ $$

Commented by tawa tawa last updated on 11/Nov/17

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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