Question

Find the domain of definition of the following function:
$f\left(x\right)=\sqrt{{\log }_{0.5}(-x^2+x+6)}+\frac{1}{x^2+2x}$

Answer 1

Since, ${\log }_{0.5}(-x^2+x+6)\ge 0$

As base is less than $1$

$\Rightarrow 0<-x^2+x+6\le 1$

Also for $\frac{1}{x^2+2x}$

$x^2+2x\ne 0\Rightarrow x(x+2)\ne 0\Rightarrow x\ne 0,-\mathrm{2....}\left(i\right)$

Now,

$-x^2+x+6>0x^2-x-6<0(x-3)(x+2)<0x\in (-2,3)$ and $-x^2+x+6\le 1-x^2+x+5\le 0x^2-x-5\le 0x\in [\frac{1-\sqrt{21}}{2},\frac{1+\sqrt{21}}{2}]...\left(2\right)$

Taking intersection from (i) and (ii) , we get

$x\in [\frac{1-\sqrt{21}}{2},\frac{1+\sqrt{21}}{2}]-\{0,2\}$