# Find the domain of each function. a) f(x)=\frac{x^2-4}{x^2-5x+6} b) f(x)=\sqrt{9-x^2}

Find the domain of each function.
$a\right)f\left(x\right)=\frac{{x}^{2}-4}{{x}^{2}-5x+6}\phantom{\rule{0ex}{0ex}}b\right)f\left(x\right)=\sqrt{9-{x}^{2}}$
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Kendrick Jacobs
a) Consider the function:
$f\left(x\right)=\frac{{x}^{2}-4}{{x}^{2}-5x+6}$
This function is not defined for those values of x for which
${x}^{2}-5x+6=0$
Solve for x:
${x}^{2}-5x+6=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-2x-3x+6=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-2\right)-3\left(x-2\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-2\right)\left(x-3\right)=0\phantom{\rule{0ex}{0ex}}x=2,3$
Thus the given function is not defined when x=2 and x=3
That is the function is defined for those values of x for which
The domain of the function in set notation is

The domain of the function f(x) in interval notation is
$\left(-\mathrm{\infty },2\right)\cup \left(2,3\right)\cup \left(3,\mathrm{\infty }\right)$

Arectemieryf0
b) Consider the function:
$f\left(x\right)=\sqrt{9-{x}^{2}}$
This function is not defined for those values of x for which

The domain of the function in set notation is
$\left\{x\in \mathbb{R}|-3\le x\le 3\right\}$
The domain of the function f(x) in interval notation is [-3,3]