If $f(x) = (x^2+2\alpha x + \alpha^2-1)^{\frac{1}{4}}$ has its domain and range so that union is $\mathbb{R}$, what does $\alpha$ satisfy? My question asks:
If $f(x) = (x^2+2\alpha x + \alpha^2-1)^{\frac{1}{4}}$ has its domain and range such that their union is set of real numbers, what does $\alpha$ satisfy? (Answer is stated to be $\alpha \leq -1 $)
My approach was as follows, I can clearly see that the range has to be positive so that the domain must comprise of all negative real numbers, so that the union of range and domain consists of the full set of real numbers. For domain, the part under the radical sign must be $\geq 0$
$$x^2+2\alpha x + \alpha^2-1 \geq 0$$
$$\implies (x+\alpha-1)(x+\alpha +1)\geq 0$$
But now I'm stuck, how do I get the range for $\alpha$, and what condition do I apply such that the domain is negative? How do I proceed from here?
 A: Opting for a different route than the one suggested by @Gregory in the comments.
Going from what you have
$$g(x) = (x-(-\alpha +1))(x-(-\alpha-1)) \geq 0$$
So, if you plot $1-\alpha$ and $-1-\alpha$ on a number line; notice that the function $g$ is positive everywhere to the right of $(1-\alpha)$ and also positive to the left of $(-1-\alpha)$ and thus satisfies the inequality when
$$x \in D = (-\infty, -1-\alpha] \cup [1-\alpha, \infty)$$
Moreover, notice that the vertex coordinates of the upward open parabola representing $g$ is simply $(-\alpha, -1)$ (thereby making its range $[-1,\infty)$). Therefore, the range of $f$, regardless of the value of $\alpha$ is $R = [0,\infty)$ (because of the fourth root on the outside).
Now, we know $$D \cup R = \mathbb{R} \implies (-\infty, -1-\alpha] \cup [1-\alpha, \infty) \cup [0,\infty) = \mathbb{R}$$
Notice that when $\alpha \leq -1$, we have the first interval in the above equality becomes $(-\infty, c]$ where $c$ is non-negative, the second interval becomes redundant because of the third interval. Therefore, the inequality is held true.  So $\alpha \leq -1$.
Edit: If the last bit of the above argument is a little messy, here's neater a way by contradiction. If $\alpha > 1$, then $-\alpha$ can never be in $D \cup R$, can you see why?
Hint: $-1-\alpha + 1 = -\alpha$, and $-1-\alpha + 2 = 1- \alpha$.
