How do you find the domain of $g (x) = ln (2 - x - x^{2})$ $g(x) = ln(2 - x - x ^2)$ ?

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Answer 1 · 2018-03-18T21:32:37

$- 2 < x < 1$ $-2< x <1$

The domain of the function are the values of $x$ $x$ that give 1 value for $y$ $y$ .

For this function, $g (x)$ $g(x)$ is valid when $2 - x - x^{2} > 0$ $2-x-x^2>0$

$- (x + 2) (x - 1) > 0$ $-(x+2)(x-1)>0$

$x + 2 > 0 or - x + 1 > 0$ $x+2>0 or -x+1>0$

$x > - 2 or x < 1$ $x> -2 or x<1$

$- 2 < x < 1$ $-2< x <1$

How do you find the domain of g(x)=ln(2−x−x2)g(x) = ln(2 - x - x ^2)?