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Answer 1 · 2016-12-11T22:08:29

$- 1 \leq x < 1 and 1 < x < 2$ $-1 <= x<1 and 1 < x<2$

1) $x + 1 \geq 0 (\sqrt{x + 1})$ $x+1>=0 color(red)((sqrt(x+1)))$

2) $2 - x > 0 log (2 - x)$ $2-x>0 color(red)(log(2-x))$

3) $log (2 - x) \neq 0 \frac{1}{log (2 - x)}$ $log(2-x)!=0 color(red)(1/log(2-x))$

$x \geq - 1 and x < 2 and 2 - x \neq 1$ $x >= -1 and x<2 and 2-x!=1$

$x \geq - 1 and x < 2 and x \neq 1$ $x>= -1 and x<2 and x!=1$

that's

$- 1 \leq x < 1 and 1 < x < 2$ $-1 <= x<1 and 1 < x<2$