# How do I find the range of $\sqrt{x^2-4}$?

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?

• 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 $[0,\infty)$. – Evan William Chandra May 16 at 1:06
• 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. – Vasya May 16 at 2:02
• @Vasya squaring will result in extraneous solutions. – Unique May 16 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 $$[0,\infty)$$.