Question

Find the domain of
$f(x) = \frac{1}{ log(2 - x) } + \sqrt{x + 1}$
​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

x ∈ [ - 1 , 2 ) - { 1 }

Step-by-step explanation:

Given :

$\displaystyle{f(x)=\dfrac{1}{\log(2-x)}+\sqrt{x+1}}$

We have to find Domain here

$\displaystyle{f(x)=\dfrac{1}{\log(2-x)}+\sqrt{x+1}}\\\\\displaystyle \text{Condition first}\\\\\displaystyle{\sqrt{x+1} \ \geq0}\\\\\displaystyle{x+1\ \geq0}\\\\\displaystyle{x \ \geq-1}\\\\\displaystyle \text{Condition second}\\\\\displaystyle{2-x \ > 0}\\\\\displaystyle{x \ >2}\\\\\displaystyle \text{Condition third}\\\\\displaystyle{2-x \ \neq 1}\\\\\displaystyle{x \ \neq 1}\\\\\displaystyle \text{Now Domain of f(x) }$

From all above three conditions we can write Domain of f ( x )  as :

$\displaystyle{f(x)=\dfrac{1}{\log(2-x)}+\sqrt{x+1}}$

where , x ∈ [ - 1 , 2 ) - { 1 }

Thus we get Domain of f ( x ) .