The domain of f(x) = sqrt{log_2left(frac{10x | vedclass

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Question

(A) For the function $f(x)$ to be defined,the expression inside the square root must be non-negative and the argument of the logarithm must be positiv

Vedclass · Accepted Answer

$\left[ -6, -2 \right) \cup \left[ 1, 2 \right)$

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(A) For the function $f(x)$ to be defined,the expression inside the square root must be non-negative and the argument of the logarithm must be positive.$1$. Argument of logarithm must be positive:$\frac{10x - 4}{4 - x^2} > 0 \Rightarrow \frac{2(5x - 2)}{(2 - x)(2 + x)} > 0 \Rightarrow \frac{5x - 2}{(x - 2)(x + 2)} Using the wavy curve method,the solution for this inequality is $x \in \left( -\infty, -2 \right) \cup \left( 0.4, 2 \right)$.$2$. Expression inside square root must be non-negative:$\log_2\left(\frac{10x - 4}{4 - x^2}\right) - 1 \geq 0 \Rightarrow \log_2\left(\frac{10x - 4}{4 - x^2}\right) \geq 1$$\Rightarrow \frac{10x - 4}{4 - x^2} \geq 2^1 \Rightarrow \frac{10x - 4}{4 - x^2} - 2 \geq 0$$\Rightarrow \frac{10x - 4 - 2(4 - x^2)}{4 - x^2} \geq 0 \Rightarrow \frac{10x - 4 - 8 + 2x^2}{4 - x^2} \geq 0$$\Rightarrow \frac{2x^2 + 10x - 12}{4 - x^2} \geq 0 \Rightarrow \frac{2(x^2 + 5x - 6)}{-(x^2 - 4)} \geq 0$$\Rightarrow \frac{(x + 6)(x - 1)}{(x - 2)(x + 2)} \leq 0$Using the wavy curve method for $\frac{(x + 6)(x - 1)}{(x - 2)(x + 2)} \leq 0$,the intervals are $\left[ -6, -2 \right) \cup \left[ 1, 2 \right)$.Since this interval is a subset of the domain condition from step $1$,the final domain is $\left[ -6, -2 \right) \cup \left[ 1, 2 \right)$.

The domain of $f(x) = \sqrt{\log_2\left(\frac{10x - 4}{4 - x^2}\right) - 1}$ is

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