Please tell me how to calculate the range of $f(x)=\sqrt{2x-6}$?

The solution in my math note says that the range of $f(x)$ is $\lbrace y\in\mathbb{R}:y\ge 0\rbrace$.

As I followed this link and do the sum, it gives me the range as $-\infty$ to $\infty$. Please help me.

  • $\begingroup$ Your link has been deleted, and it is a good thing. Follow this link instead. Do you know that the range of $=\sqrt{x}$ over $\mathbb{R}$ is $[0,+\infty)$? $\endgroup$
    – Julien
    Apr 20, 2013 at 13:44
  • $\begingroup$ for every $y \in [0,\infty)$ you can find $x=\frac {6+y^2} {2}$ such that $f(x)=y$, Can you? $\endgroup$ Apr 20, 2013 at 14:05

1 Answer 1


$$f(x)=\sqrt{2x-6}$$ $$g(x)=\sqrt x$$

Now, what is similarity in $f,g$?

If we substitute $x=(u+6)/2$ in $f$ we get $g(u)$

(Also square root function is always positive.So we get the +ve numbers only)

But $y^2=x$ have soltion $y=\pm \sqrt x$, but $f$ being a function is defined as $y+\sqrt x$. Yields only the positive part.


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