Math, asked by nikhilsam465, 1 year ago

Find the range of the function f (x) = 1 ÷ 2+sin 3x

Answers

Answered by Santosh1729
0

 we \: know \: that \:  \:  \  - 1\leqslant sin(x)   \leqslant 1  \: for \: all \: values \: of \: x\\  \:  \:   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: =  >  - 1 \leqslant  \sin(3x)  \leqslant 1 \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \:  \:  \:  =  >  \:  - 1 + 2 \leqslant  \ 2+ sin(3x)   \leqslant 1 + 2 \\  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:=  >  \:  \: 1 \leqslant  2 + sin(3x)  \leqslant 3 \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  =  >  \: 1 \div 3 \leqslant 1 \div (2 +  \sin(3x) ) \leqslant 1.

I hope you will get it.

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