The domain of the function f(x) = log_{3 + x} | vedclass

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Question

(C) For the function $f(x) = \log_{3 + x}(x^2 - 1)$ to be defined,the following conditions must be satisfied:$1$. The argument of the logarithm must b

Vedclass · Accepted Answer

$(-3, -2) \cup (-2, -1) \cup (1, \infty)$

Vedclass · Answer

(C) For the function $f(x) = \log_{3 + x}(x^2 - 1)$ to be defined,the following conditions must be satisfied:$1$. The argument of the logarithm must be positive: $x^2 - 1 > 0$,which implies $x^2 > 1$,so $x  1$.$2$. The base of the logarithm must be positive and not equal to $1$: $3 + x > 0$ and $3 + x 
eq 1$.From $3 + x > 0$,we get $x > -3$.From $3 + x 
eq 1$,we get $x 
eq -2$.Combining all these conditions: $x > -3$,$x 
eq -2$,and ($x  1$).Intersection of these sets gives the domain: $D_f = (-3, -2) \cup (-2, -1) \cup (1, \infty)$.

The domain of the function $f(x) = \log_{3 + x}(x^2 - 1)$ is

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