I tried to find the range of $\sqrt{x^2-4}$ by the method of finding the domain of its inverse function. The inverse function will be $\sqrt{x^2+4}$. It's domain will be $(-\infty,\infty)$ as ${x^2}$ will always be greater than ${-4}$. But the range of the original function is $[0,\infty)$. How is that?

  • 2
    $\begingroup$ First, you have to determine the domain of the function which is $(-\infty,-2]\cup[2,\infty)$. Then, by using the monotonicity of $f(x) = \sqrt{x}$, you will note that the range is $[0,\infty)$. $\endgroup$ May 16 '19 at 1:06
  • $\begingroup$ When you find the inverse, you write $x=\sqrt{y^2-4}$ So what is the domain of the inverse? It's $x \ge 0$. Squaring both parts expands the domain. $\endgroup$
    – Vasya
    May 16 '19 at 2:02
  • $\begingroup$ @Vasya squaring will result in extraneous solutions. $\endgroup$ May 16 '19 at 5:48

The domain of a function $y=f(x)$ is set of all real values of $x$ where the function is real finite and unique. So in this case the domain is $(-\infty, -2] \cup [2, \infty)$. The range is set of all real values of $y$ taken by the function over the domain. So in this case the range is $[0, \infty)$. This is the direct method.

Otherwise, you have to realize the your function $f(x)$ is positive definite so its range is only $[0, \infty)$. In fact, $(-\infty,0]$ is the range of another function $g(x)=-\sqrt{x^2-4}$ which is the other branch of the curve: $x^2-y^2=4$ (hyperbola). So this curve defines two functions: $f(x)$ and $g(x)$. Here the question chooses $f(x)$ hence its range is $[0,\infty)$.


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