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Word for word:

Consider the functions $f(x)=\ln x$ and $g(x)=x^2−1$, find the domain and range of $(f\circ g)(x)$


I think this is asking to find the domain and range of $\ln(x)^{2}-1$ and the domain of $f(x)=ln(x)$ I believe are

$D: (0, \infty)$ $R:(-\infty,\infty)$

and $x^2 -1$ is all real numbers on the domain and range.

So the final function $(f\circ g)(x)$ would just inherit the domain and range of $\ln(x)$ and be

$D: (0, \infty)$ $R:(-\infty,\infty)$

EDIT

Okay, I made a dumb mistake. The function $(f\circ g)(x)$ is $\ln(x^2-1)$. But I still think my final answer is correct. Right?

Ok according the answers so far I believe the Domain is $(-\infty, -1) \cup (1, \infty)$?

I'm still getting used to that notation.

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  • $\begingroup$ Is that function defined for $x=0$? $\endgroup$ – Thomas Andrews Aug 13 '14 at 1:37
  • $\begingroup$ no. you can't take the log of a number 0 or less $\endgroup$ – Gᴇᴏᴍᴇᴛᴇʀ Aug 13 '14 at 1:38
  • $\begingroup$ Is it defined for $x=1$? Because your domain contains $1$. $\endgroup$ – Thomas Andrews Aug 13 '14 at 1:39
  • $\begingroup$ no. do you want me to change my answer now? I didn't think I should change it now I started waiting for others to answer $\endgroup$ – Gᴇᴏᴍᴇᴛᴇʀ Aug 13 '14 at 1:42
  • $\begingroup$ It is all right to change your posted answer or thoughts on same at later times as you did, by marking the revised portion as an "edit" or "update". People here make and correct mistakes all the time (as anywhere else). What does not make people happy is changing the question while they are attempting to respond to it... $\endgroup$ – colormegone Aug 13 '14 at 6:58
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$$f\circ g (x) = \ln(x^2-1)$$

As you reasoned, the domain of $\ln$ is $(0, \infty)$.

So what you need to find is the domain of $x$ such that $x^2-1 \in (0,+\infty)$

Then find the range of $f\circ g$ mapped by that domain.

So the function is undefined between -1 and 1. Could you include the notation for that?

Thus the domain of $f\circ g$ is: $(-\infty, -1)\cup (+1,+\infty)$

Which may also be denoted as: $\mathbb{R}\backslash[-1,+1]$

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  • $\begingroup$ So the function is undefined between -1 and 1. Could you include the notation for that? $\endgroup$ – Gᴇᴏᴍᴇᴛᴇʀ Aug 13 '14 at 1:58
  • $\begingroup$ @Zack It's just a union of two domains, or an exclusion from the whole. $\endgroup$ – Graham Kemp Aug 13 '14 at 2:21
  • $\begingroup$ but I have to find the domain and range. Is what I wrote in the edit correct? $\endgroup$ – Gᴇᴏᴍᴇᴛᴇʀ Aug 13 '14 at 2:28
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$$(f \circ g)(x)=(f(g(x))=f(x^2-1)=\ln(x^2-1)$$

What happens if:

  • $x > 1$?
  • $x < 1$?
  • $-1 \le x \le 1$?
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No, it's the domain and range of ln (x^2-1)

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  • $\begingroup$ oohhh I'll edit that. Just putting down what I had done so far $\endgroup$ – Gᴇᴏᴍᴇᴛᴇʀ Aug 13 '14 at 1:30
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The systematic way I show students for resolving the domain of a composite function is to start by finding the domains of the individual functions (as you did):

$$ f(x) \ = \ \ln \ x \ : \quad \text{domain --} \ \ \mathbb{R}^+ \ \ \text{or} \ \ x \ > \ 0 \ \ , $$

$$ g(x) \ = \ x^2 \ - \ 1 \ : \quad \text{domain --} \ \ \mathbb{R} \ \ . $$

Since we have $ \ f(x) \ $ composed upon $ \ g(x) \ $ , we start with the domain of $ \ g(x) \ $ , and then exclude any values of $ \ x \ $ for which $ \ f( \ g(x) \ ) \ $ is undefined. So we remove from the set of all real numbers those values of $ \ x \ $ for which $ \ \ln \ g(x) \ $ is undefined, which is the set

$$ g(x) \ \le \ 0 \ \ \Rightarrow \ \ x^2 \ - \ 1 \ \le \ 0 \ \ \Rightarrow \ \ x^2 \ \le \ 1 \ \ \Rightarrow \ \ |x| \ \le \ 1 \ \ ; $$

in terms of the function domains, we are removing the complement of the set for which $ \ g(x) \ > \ 0 \ $ .

Removing this interval from $ \ \mathbb{R} \ $ leaves us with the disjoint intervals $ \ x \ < \ -1 \ $ and $ \ x \ > \ +1 \ $ , written in inequality notation. The answer Graham Kemp and your edit show is then the domain in interval notation.

Finding the range of a function is not as easy a task, unless the function has a reasonably simple behavior. Here, though, we just have $ \ (f \circ g) (x) \ = \ \ln \ (x^2 \ - \ 1) \ $ . In the domain we've determined, $ \ x^2 \ - \ 1 \ $ may be any positive number, so all real values for a logarithm can be produced by the composite function. So the range is $ \ \mathbb{R} \ $ .

[Just as an exercise -- one which exam-writers like to ask (I oughta know...) -- we will do the same for $ \ (g \circ f) (x) \ $ . We now start with $ \ x \ > \ 0 \ $ (the domain of $ \ f \ $ ) and remove any values for which $ \ [ \ f(x) \ ]^2 \ - \ 1 \ $ is undefined. Since that polynomial can always be computed, there is no further restriction of the domain; hence, the domain of $ \ (g \circ f) (x) \ $ is $ \ x \ > \ 0 \ $ . The range of $ \ (g \circ f) (x) \ = \ (\ln \ x)^2 \ - \ 1 \ $ is found from the fact that, for the domain we've determined, the logarithm of $ \ x \ $ can be any real number, so the minimum of the polynomial is $ \ -1 \ $ . The range of this composite function is therefore $ \ y \ \ge \ -1 \ $ . ]

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