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Question: The domain of definition of \(f(x) = \sqrt{\log_{0.4}\left( \frac{x - 1}{x + 5} \right)}\text{ x }\f...

The domain of definition of f(x)=log⁡0.4(x−1x+5) x 1x2−36f(x) = \sqrt{\log_{0.4}\left( \frac{x - 1}{x + 5} \right)}\text{ x }\frac{1}{x^{2} - 36} is

A

{x:x<0,x≠−6}\left\{ x:x < 0,x \neq - 6 \right\}

B

{x:x>0,x≠1,x≠6}\left\{ x:x > 0,x \neq 1,x \neq 6 \right\}

C

{x:x>1,x≠6}\left\{ x:x > 1,x \neq 6 \right\}

D

{x:x≥1,x≠6}\left\{ x:x \geq 1,x \neq 6 \right\}

Answer

{x:x>1,x≠6}\left\{ x:x > 1,x \neq 6 \right\}

Explanation

Solution

For log⁡0.4(x−1x+5)\sqrt{\log_{0.4}\left( \frac{x - 1}{x + 5} \right)}to be defined. We must have

0<x−1x+5<1,0 < \frac{x - 1}{x + 5} < 1, which is true if x>1x > 1. Morever, 1x2−36\frac{1}{x^{2} - 36} is defined for x≠6,−6x \neq 6, - 6. Hence the domain of ffis {x:x>1,x≠6}\left\{ x:x > 1,x \neq 6 \right\}