How do I determine x for the following problems

Question

I have to put the ranges where the following exists: $$\sqrt[6-x^2]x$$ I put the condition that $$6-x^2>1$$ and $x\ge0$ so I have the following interval: $$[0,\sqrt5]$$ which contains 3 elements but the answer in my book is 7 so what am I missing here? 2) the interval for x in order for the following expression to be true : $$\log_x (x+1)+\log_{x^3}(x^3+1)=2\log_{x^2} (x^2+1)$$

Answer 1

The function exists for $x>0$.

The limit for x tending to zero from the right is zero.

The limit for x tending to $\sqrt{6}$ from the left is infinite.

The limit for x tending to $\sqrt{6}$ from the right is null.

The limit for x tending to infinity is 1.

That is, it has a horizontal asymptotic, $y =1$.

The first derivative allows us to say that:

a) In the path that goes from 0 to 1, the function is increasing, in $x=1$ has an inflection, and then grows to infinity for $x=\sqrt{6}$;

b) In the path from $\sqrt{6}$ to an additional inflection point, the function is increasing, after the concavity is facing down to $x=$infinity, reaches the value 1. (horizontal asymptot).