Science:Math Exam Resources/Courses/MATH100/December 2015/Question 04 (d)

MATH100 December 2015

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Question 04 (d)
Find the domain of $f(x)={\frac {x^{2}-1}{\sqrt {x^{2}-x-6}}}$ .

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
What is the domain of a function which can be written in the form ${\frac {F(x)}{G(x)}}$ ?

Hint 2
What's the domain of a function which has the form ${\sqrt {H(x)}}$ ?

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Solution

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Recall that the domain of the function of the form ${\frac {F(x)}{G(x)}}$ is $\{x\in \mathbb {R} :G(x)\neq 0\}$ . Also, the domain of the function of the form ${\sqrt {H(x)}}$ is $\{x\in \mathbb {R} :H(x)\geq 0\}$ .

This implies that the domain of $f(x)={\frac {x^{2}-1}{\sqrt {x^{2}-x-6}}}$ is

$\{x\in \mathbb {R} :{\sqrt {x^{2}-x-6}}\neq 0\}\ {\text{and}}\ \{x\in \mathbb {R} :x^{2}-x-6\geq 0\}=\{x\in \mathbb {R} :x^{2}-x-6=(x-3)(x+2)>0\}.$ .

We use the following sign chart to determine where $x^{2}-x-6=(x-3)(x+2)>0$ :

	$\displaystyle (-\infty ,-2)$	$\displaystyle (-2,3)$	$\displaystyle (3,\infty )$
$x-3$	$\displaystyle -$	$\displaystyle -$	$\displaystyle +$
$x+2$	$\displaystyle -$	$\displaystyle +$	$\displaystyle +$
$x^{2}-x-6$	$\displaystyle +$	$\displaystyle -$	$\displaystyle +$