Question

find range and domain of the function?
f(x)=1/2-sin3x

Answer 1

#Hey there!!
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◆Given function is :

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

●For DOMAIN :
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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 :
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$we \: know \: that \: \\ - 1 \leqslant \sin(3x) \leqslant 1$

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 ]

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◆HOPE IT WILL HELP YOU

Answer 2

Answer:

The range and domain of the function is $\left [ \frac{1}{3},1 \right ]$

Step-by-step explanation:

Given that $f(x)=\frac{1}{2-sin3x}$

As we know the domain of $f(x)$ is $\left ( -\infty, \infty \right )$

So,

$2-sin3x=\frac{1}{y} \\sin3x=2-\frac{1}{y} \\sin3x=\frac{2y-1}{y} \\3x=sin^-1(\frac{2y-1}{y})$

Further solving,

$x=\frac{1}{3} sin^(-1)(\frac{2y-1}{y} )$

Now for x to be real

$-1\leq \frac{2y-1}{y}\leq 1\\ -y\leq 2y-1\leq y$

$2y-1\geq -y$ and $2y-1\leq -y$

$y\geq \frac{1}{3} , y\leq 1$

Hence, The range and domain of the function is $\left [ \frac{1}{3},1 \right ]$