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Find the maximal domain and range of:

$$f(x) = \ln(x^2 - 7)$$

I reasoned that the domain can be found by finding the values of $x$ such that $x^2 - 7 > 0$ so that the logarithm is defined. So:

$$x^2 > 7\\ x > \pm \sqrt{7}$$

Shouldn't I have arrived at $x > \sqrt{7}$ and $x < -\sqrt{7}$ somehow? From inspection I can see this, but I couldn't seem to understand why I didn't arrive at that algebraically.

Also, by inspection, I would guess that the range is $\{y \in \mathbb{R}\mid y > 0\}$. Is it standard to just derive this from a graph? or should I discuss the limiting behaviour as $x \rightarrow \pm~ \infty$ and as $x \rightarrow \pm\sqrt{7}$ from $+$ and $-$ respectively?

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3 Answers 3

up vote 3 down vote accepted

For the domain, your reasoning is correct, but $$\begin{align*}x^2 > 7 \implies |x| > \sqrt{7} & \implies -x > \sqrt{7} \; \text{or }\; x > \sqrt{7}\\ & \implies x < -\sqrt{7} \; \text{or } \; x > \sqrt{7},\end{align*}$$ as you were expecting. You just skipped a step when you wrote $x < \pm \sqrt{7}$, forgetting to reverse the inequality when you moved the negative across.

For the range, this is all the values that $\ln(x^2 -7)$ can take, so you could say that it goes to $-\infty$ as $x \to \sqrt{7}$ from the right and as $x \to -\sqrt{7}$ from the left. It also goes to $\infty$ as $x \to \pm \infty$. So the range is all of $\mathbb{R}$.

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You can break $x^2-7 > 0 \Leftrightarrow (x-\sqrt 7)(x+\sqrt 7) >0$ in two cases :

1) $(x-\sqrt 7)>0 ~\text{and}~(x+\sqrt 7) >0$

2) $(x-\sqrt 7)<0 ~\text{and}~(x+\sqrt 7) <0$

So , domain should be :

$x \in (-\infty,-\sqrt 7) \cup (\sqrt 7 ,+\infty)$

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$x^2-7>0$ means that $x \in (-\infty,-\sqrt 7) \cup (\sqrt 7 ,+\infty)$

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Please use $\LaTeX$ for the posts you make. –  user21436 Mar 24 '12 at 7:49
    
So, what does $x^2 - 7 > 0$ mean? –  stariz77 Mar 24 '12 at 7:52
    
it means that in logarithm argument can't be negative or 0 –  dato datuashvili Mar 24 '12 at 7:56

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