gelo)Find the domain and range of the function f (x) =" 2-sin 32

1.	gelo)Find the domain and range of the function f (x) =" 2-sin 32
Answer» #Hey there!!__________ ◆Given function is : f(x) = \frac{1}{2 - sin3 x} \\ ●For DOMAIN :----------------------- Function f(x) is not defined when => 2 - sin3x = 0 so , sin3x = 2 ---------(1) but we know that range of sinx € [ -1, 1 ] so maximum value of sin3x = 1 therefore sin3x ≠ 2 ( not possible) => 2 - sin3x ≠ 0 ------(2) so from equation (2) we can se tha function is defined for all values of x => Domain € R #FOR RANGE :----------------------we \: know \: that \: \\ - 1 \leqslant \sin(3x) \leqslant 1 \\ now multiplying by -1 we get, => 1 ≥ - sin3x ≥ -1 adding 2 we get, => 2+1 ≥ 2 - sin3x ≥ 2-1 => 3 ≥ 2-sin3x ≥ 1 now taking inverse we get, \frac{1}{3} \leqslant \frac{1}{2 - \sin(3x) } \leqslant 1 so Range € [ ⅓ , 1 ] _____________________________ ◆HOPE IT WILL HELP YOU like my answer and also make as expert

Answer»

#Hey there!!__________

◆Given function is :

f(x) = \frac{1}{2 - sin3 x} \\

●For DOMAIN :-----------------------

Function f(x) is not defined when

=> 2 - sin3x = 0

so , sin3x = 2 ---------(1)

but we know that range of sinx € [ -1, 1 ]

so maximum value of sin3x = 1

therefore sin3x ≠ 2 ( not possible)

=> 2 - sin3x ≠ 0 ------(2)

so from equation (2) we can se tha function is defined for all values of x

=> Domain € R

#FOR RANGE :----------------------we \: know \: that \: \\ - 1 \leqslant \sin(3x) \leqslant 1 \\

now multiplying by -1 we get,

=> 1 ≥ - sin3x ≥ -1

adding 2 we get,

=> 2+1 ≥ 2 - sin3x ≥ 2-1

=> 3 ≥ 2-sin3x ≥ 1

now taking inverse we get,

\frac{1}{3} \leqslant \frac{1}{2 - \sin(3x) } \leqslant 1

so Range € [ ⅓ , 1 ]

_____________________________

◆HOPE IT WILL HELP YOU

like my answer and also make as expert

Discussion