The domain of the definition of the function | vedclass

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(C) The function is $f(x) = \frac{1}{4 - x^2} + \log(x^3 - x)$.For the function to be defined,two conditions must be satisfied simultaneously:$1$. The

Vedclass · Accepted Answer

$\left( -1, 0 \right) \cup \left( 1, 2 \right) \cup \left( 2, \infty \right)$

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(C) The function is $f(x) = \frac{1}{4 - x^2} + \log(x^3 - x)$.For the function to be defined,two conditions must be satisfied simultaneously:$1$. The denominator of the rational part must not be zero: $4 - x^2 
eq 0 \implies x^2 
eq 4 \implies x 
eq \pm 2$.$2$. The argument of the logarithm must be strictly positive: $x^3 - x > 0$.Factoring the inequality: $x(x^2 - 1) > 0 \implies x(x - 1)(x + 1) > 0$.Using the sign scheme (Wavy Curve Method),the solution for $x(x - 1)(x + 1) > 0$ is $x \in (-1, 0) \cup (1, \infty)$.Now,we must exclude the points where the denominator is zero ($x = 2$ and $x = -2$).Since $x = -2$ is not in the interval $(-1, 0) \cup (1, \infty)$,we only need to exclude $x = 2$.Thus,the domain is $x \in (-1, 0) \cup (1, 2) \cup (2, \infty)$.

The domain of the definition of the function $f(x) = \frac{1}{4 - x^2} + \log(x^3 - x)$ is

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