How do you find the domain of $f (x) = a r c cot (\frac{x - 1}{{(1 - x)}^{0.5}}) + {log}_{\frac{1}{3}} (2 x + 21)$ $f(x) = arc cot( (x−1) / ((1−x)^0.5) ) + log_(1/3) (2x + 21)$ ?

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How do you find the domain of $f (x) = a r c cot (\frac{x - 1}{{(1 - x)}^{0.5}}) + {log}_{\frac{1}{3}} (2 x + 21)$ $f(x) = arc cot( (x−1) / ((1−x)^0.5) ) + log_(1/3) (2x + 21)$ ?

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Answer 1 · 2016-07-11T15:33:22

Domain: $(- 10 \frac{1}{2}, 1)$ $(-10 1/2, 1)$

Explanation:

The given equation is
$f (x) = {cot}^{- 1} (\frac{x - 1}{{(1 - x)}^{0.5}}) + {log}_{\frac{1}{3}} (2 x + 21)$ $f(x)=cot^-1((x-1)/(1-x)^0.5)+log_(1/3) (2x+21)$

For the term ${cot}^{- 1} (\frac{x - 1}{{(1 - x)}^{0.5}})$ $cot^-1((x-1)/(1-x)^0.5)$

Domain: $(- \infty, 1)$ $(-oo, 1)$

For the term ${log}_{\frac{1}{3}} (2 x + 21)$ $log_(1/3) (2x+21)$

Domain: $(- 10 \frac{1}{2}, + \infty)$ $(-10 1/2, +oo)$

Therefore the domain for the entire equation is the intersection which is

Domain: $(- 10 \frac{1}{2}, 1)$ $(-10 1/2, 1)$

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How do you find the domain of $f (x) = a r c cot (\frac{x - 1}{{(1 - x)}^{0.5}}) + {log}_{\frac{1}{3}} (2 x + 21)$ $f(x) = arc cot( (x−1) / ((1−x)^0.5) ) + log_(1/3) (2x + 21)$ ?

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Explanation:

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How do you find the domain of $f (x) = a r c cot (\frac{x - 1}{{(1 - x)}^{0.5}}) + {log}_{\frac{1}{3}} (2 x + 21)$ $f(x) = arc cot( (x−1) / ((1−x)^0.5) ) + log_(1/3) (2x + 21)$ ?