The domain of f(x) = frac{log_{(x+1)}(x-2)}{x | vedclass

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Question

(B) For the function $f(x) = \frac{\log_{(x+1)}(x-2)}{x^2 - 2x - 3}$ to be defined:$1$. The argument of the logarithm must be positive: $x - 2 > 0 \Ri

Vedclass · Accepted Answer

$(2, \infty) - \{3\}$

Vedclass · Answer

(B) For the function $f(x) = \frac{\log_{(x+1)}(x-2)}{x^2 - 2x - 3}$ to be defined:$1$. The argument of the logarithm must be positive: $x - 2 > 0 \Rightarrow x > 2$.$2$. The base of the logarithm must be positive and not equal to $1$: $x + 1 > 0 \Rightarrow x > -1$ and $x + 1 
eq 1 \Rightarrow x 
eq 0$.$3$. The denominator must not be zero: $x^2 - 2x - 3 
eq 0$.Factoring the denominator: $(x - 3)(x + 1) 
eq 0$,which implies $x 
eq 3$ and $x 
eq -1$.Combining all conditions: $x > 2$ and $x 
eq 3$.Thus,the domain is $(2, \infty) - \{3\}$.

The domain of $f(x) = \frac{\log_{(x+1)}(x-2)}{x^2 - (2x + 3)}$ for $x \in R$ is

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