2
$\begingroup$

This is the function:

$$f(x) = \sqrt{x^2-2x+5}$$

Edit: normally what I would do is this: Since it's a square root function, the thing inside the root has to be $\ge 0$. So, $(x^2 - 2x+5)\ge 0$. Then I would factor the stuff in the brackets so that I get () (__). But since this has complex roots, I don't know what to do

Edit: Thanks for the help with the domain! For the range, I found the inverse of the function and did this:

$$x = \sqrt{(y-1)^2 +4}$$ $$x^2 -4 = (y-1)^2 $$ $$\sqrt{x^2 -4}+1 = y$$

And then proceeded to find the domain of the inverse:

$$x^2 -4 \ge 0$$ $$x^2 \ge 4$$ $$x \ge +-2$$

Is this correct? How do I know if it is correct without graphing it out?

$\endgroup$
10
  • 1
    $\begingroup$ $x^2 - 2 x + 5 = (x-1)^2 + 4$ $\endgroup$
    – Will Jagy
    Commented Aug 19, 2014 at 22:36
  • 1
    $\begingroup$ What are your thoughts about it? Show us some previous work you have done on this problem so we can help you better. $\endgroup$
    – Dmoreno
    Commented Aug 19, 2014 at 22:40
  • $\begingroup$ (to continue Dmoreno's comment) For example, you could reassure us that you know what the teminologies "domain" and "range" mean. $\endgroup$ Commented Aug 19, 2014 at 22:41
  • $\begingroup$ Okay, so normally what I would do is this: Since it's a square root function, the thing inside the root has to be >=0. So, (x^2 - 2x+5) >= 0. Then I would factor the stuff in the brackets so that I get (_____) (_______). But since this has complex roots, I don't know what to do $\endgroup$
    – Alantin14
    Commented Aug 19, 2014 at 22:43
  • $\begingroup$ That's precisely the sort of thing we're looking for. Why not edit the original question to include this information? $\endgroup$ Commented Aug 19, 2014 at 22:46

2 Answers 2

1
$\begingroup$

Hint:$$x^2-2x+5=(x-1)^2+4\geq0$$ for all$x\in(-\infty,+\infty)=\mathbb R$

$\endgroup$
7
  • $\begingroup$ Fine now. Just sayin $\endgroup$ Commented Aug 19, 2014 at 22:45
  • $\begingroup$ factoring is wrong. the roots are actually $1 \pm 2i$ $\endgroup$
    – Will Jagy
    Commented Aug 19, 2014 at 22:46
  • $\begingroup$ I edit my answer $\endgroup$
    – Adi Dani
    Commented Aug 19, 2014 at 22:47
  • $\begingroup$ But how did you go from that step to finding out that the domain is all real numbers? I understand that you completed the square, but I don't get how you jumped from that to the domain... $\endgroup$
    – Alantin14
    Commented Aug 19, 2014 at 22:56
  • $\begingroup$ What the hint shows is that the expression inside the square root is positive for any $x$, so you can take the square root for any $x$. So the domain is...? $\endgroup$
    – user169852
    Commented Aug 19, 2014 at 22:58
0
$\begingroup$

I found the inverse of the function ... And then proceeded to find the domain of the inverse

I understand the logic of doing this when finding the range, since range is kind of the domain of inverse function... provided there is an inverse function. Not every function has an inverse: if several values of $x$ correspond to the same $y$, we cannot invert this relationship.

Since the range concerns the values of dependent variable ($y$ in your case), it helps to look at a formula where $y$ is isolated -- which it was from the beginning, $y = \sqrt{x^2-2x+5}$. We want to know what kind of values this square root can produce. That depends on what goes under the square root, of course. Under the square root we have $x^2-4x+5 = (x-1)^2+4$. So, this is something that takes on all values from $4$ upward, and none below $4$. So you consider what sort of square root comes out of this, and arrive at the answer for the range.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .