The domain of the function f(x) = frac{1}{log | vedclass

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(D) For the function $f(x) = \frac{1}{\log_{10}(1-x)} + \sqrt{x+2}$ to be defined:$1$. The term inside the square root must be non-negative: $x + 2 \g

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$[-2, 0) \cup (0, 1)$

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(D) For the function $f(x) = \frac{1}{\log_{10}(1-x)} + \sqrt{x+2}$ to be defined:$1$. The term inside the square root must be non-negative: $x + 2 \geq 0 \implies x \geq -2$.$2$. The argument of the logarithm must be positive: $1 - x > 0 \implies x $3$. The denominator cannot be zero: $\log_{10}(1-x) 
eq 0 \implies 1 - x 
eq 10^0 \implies 1 - x 
eq 1 \implies x 
eq 0$.Combining these conditions: $x \geq -2$,$x Thus,the domain is $x \in [-2, 0) \cup (0, 1)$.

The domain of the function $f(x) = \frac{1}{\log_{10}(1-x)} + \sqrt{x+2}$ is

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