Question

Find the domain of
$f(x) = \sqrt{1 - x} - \sin( \frac{2x - 1}{3} )$
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Answer 1

Answer:

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Answer 2

Answer:

$\displaystyle{\implies(-\infty , \ 1]}$

Step-by-step explanation:

Given :  

$\displaystyle{f(x)=\sqrt{1-x}-\sin \left(\dfrac{2x-1}{3}\right)}$

We have to find Domain here

$\displaystyle{f(x)=\sqrt{1-x}-\sin \left(\dfrac{2x-1}{3}\right)}\\\\\displaystyle \text{Condition first since no inside square root value}\\\\\displaystyle \text{should be greater than or equal to zero}\\\\\displaystyle{\sqrt{1-x} \ \geq0}\\\\\displaystyle{1-x\ \geq0}\\\\\displaystyle{1 \ \geq x}\\\\\displaystyle{x\leq 1}\\\\\displaystyle{\implies (- \infty , \ 1]}\\\\\displaystyle \text{Let it is as A}$

$\displaystyle{Now \ for \implies \sin \left(\dfrac{2x-1}{3}\right)}\\\\\displaystyle \text{ x \in\math{R} Let take it B}\\\\\displaystyle \text{Since sine function exist for every value of x}\\\\\displaystyle \text{So final interval is A \cap B }\\\\\displaystyle{\implies (- \infty , \ 1]}$

So , the Domain for f ( x)  is    $\displaystyle{\implies(-\infty , \ 1]}$