The domain of the definition of the function | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

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

Function	Range
$A. f(x) = \|x\|$	$I. [0, \infty)$
$B. f(x) = x^2$	$II. \mathbb{R}$
$C. f(x) = x^3$	$III. [0, \infty)$
$D. f(x) = \text{sgn}(x)$	$IV. \{-1, 0, 1\}$

The domain of the definition of the function $f(x) = \frac{1}{4-x^2} + \log_{10}(x^3-x)$ is

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