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

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Question

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

Vedclass · Accepted Answer

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

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

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

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