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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?

4 Answers 4

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Your domain appears to be in the set of Real Numbers - the domain is x is an element of the Real Numbers less than the negative square root of seven or greater than the postive square root of seven.

If you wish your domain to be in set of Complex Numbers, try it out and see what Complex values of x must be excluded. What is the range?

Try it for x as an element of the Integers. What is the range?

What does "maximal" mean in these cases? Maximal domain is the domain which is the super set of all posible domains - in other words, the maximal domain is the domain which has all possible domains as its subsets. For example, you may choose the domain with only the positive values. This is a subset of the maximal domain.

3

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 :

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

3

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 $\mathbb{R}$.

1

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

  • 0
    it means that in logarithm argument can't be negative or 02012-03-24