$f(x) = (x^2+2ax+a^2-1)^{1/4}$. I want to find the range of $a$ such that the union of domain and range = set of all real numbers.

My attempt:

The range is $[0,\infty)$ The domain: $x^2+2ax+a^2-1\geq 0$ $\Rightarrow (x+1-a)(x+1+a) \geq 0$ If $a>0, x\leq -1-a or x\geq a-1$ If $a<0, x\leq a-1 or x\geq -a-1 $

After this point I was unable to go further.

  • $\begingroup$ btw, $(x+1-a)(x-1+a)=(x+1)^2-a^2$ $\endgroup$ – Hagen von Eitzen Jun 9 '17 at 6:46

As the range is $[0,\infty)$, we need that $f(x)$ is defined for all negative $x$, that is, we need $(x+a)^2-1\ge0$ for all negative $x$. But then we also need this inequality in the limit as $x\to 0$. In other words: Necessarily $a^2\ge 1$. If $a$ is positive, then $f(-a)$ is not defined; hence necessarily $a<0$. So far we have found $$a\le -1,$$ and that is indeed sufficient: If $a\le -1$ then for all negative $x$, $x+a<-1$ and $(x+a)^2-1>0$.


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