The domain of the real-valued function f(x) = | vedclass

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(C) For the function $f(x) = \frac{3}{4-x^2} + \log_{10}(x^3-x)$ to be defined:$1$. The denominator must not be zero: $4-x^2 
eq 0 \implies x^2 
eq

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) For the function $f(x) = \frac{3}{4-x^2} + \log_{10}(x^3-x)$ to be defined:$1$. The denominator 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 positive: $x^3-x > 0$.Factoring the expression: $x(x^2-1) > 0 \implies x(x-1)(x+1) > 0$.Using the wavy curve method (sign scheme) for the critical points $x = -1, 0, 1$:The inequality $x(x-1)(x+1) > 0$ holds for $x \in (-1, 0) \cup (1, \infty)$.Combining the conditions: $x \in (-1, 0) \cup (1, \infty)$ and $x 
eq \pm 2$.Since $x 
eq 2$ and $x 
eq -2$,we exclude $2$ from the interval $(1, \infty)$.Thus,the domain is $x \in (-1, 0) \cup (1, 2) \cup (2, \infty)$.

The domain of the real-valued function $f(x) = \frac{3}{4-x^2} + \log_{10}(x^3-x)$ is

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