The domain of definition of the function f(x) | vedclass

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(B) For the function $f(x) = \sqrt{1 + \log_{e}(1 - x)}$ to be defined, the expression inside the square root must be non-negative, and the argument o

Vedclass · Accepted Answer

$-\infty < x \leq \frac{e - 1}{e}$

Vedclass · Answer

(B) For the function $f(x) = \sqrt{1 + \log_{e}(1 - x)}$ to be defined, the expression inside the square root must be non-negative, and the argument of the logarithm must be positive.$1$. Condition for logarithm: $1 - x > 0 \implies x $2$. Condition for square root: $1 + \log_{e}(1 - x) \geq 0$.$\log_{e}(1 - x) \geq -1$.Exponentiating both sides with base $e$:$1 - x \geq e^{-1}$.$1 - x \geq \frac{1}{e}$.$x \leq 1 - \frac{1}{e}$.$x \leq \frac{e - 1}{e}$.Combining both conditions: $x Since $\frac{e - 1}{e} < 1$, the domain is $-\infty < x \leq \frac{e - 1}{e}$.

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

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