Domain of $\ln\left(\frac{6}{6+x-x^2}-1\right)+\arcsin\left(\frac{x+1}{3}\right)$

Question

blob:https%3A//mail.google.com/ea67134d-45a0-4cc0-9ec7-abf6d5a50852

I believe that my first condition is wrong but I don't understand why. Can somebody please help?

Answer 1

You are correct about the domain of $\arcsin\left(\frac{x+1}{3}\right)$ being $[-4,2]$ but not the domain of $\ln\left(\frac{6}{6+x-x^2}-1\right)$.

After you take the roots make sure to find which outputs of $\left(\frac{6}{6+x-x^2}-1\right)$ between each root is positive or negative. Also evaluate the outputs of the roots themselves.

$x=-2,0,1,3$

If it is negative the values are undefined. For example if we take $\ln(-2)$ it is undefined.

You should end up with $(-2,0)\bigcup(1,3)$

Now compare this to the arcsin domain $[-4,2]$. Remeber when adding both equations if you add an undefined amount to a defined amount it is still undefined because the value is still complex. (To clarify this $\ln(-2)+\arcsin(\frac{4}{3})$ is undefined ).

So when adding both equations you should end up with $$(-2,0)\bigcup(1,2]$$