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Q. 16

Expert-verifiedFound in: Page 225

Book edition
6th

Author(s)
Sullivan

Pages
1200 pages

ISBN
9780321795465

Find the domain of the rational function.

$G\left(x\right)=\frac{6}{\left(x+3\right)\left(4-x\right)}$

The domain of the given function is $\left(-\infty \hspace{0.17em},\hspace{0.17em}-3\right)\cup \left(-3,\hspace{0.17em}4\right)\cup \left(4,\hspace{0.17em}\infty \hspace{0.17em}\right)$

The given function is $G\left(x\right)=\frac{6}{\left(x+3\right)\left(4-x\right)}$

For the domain, the denominator should not be equal to zero.

So,

$\left(x+3\right)\left(4-x\right)\ne 0\phantom{\rule{0ex}{0ex}}x+3\ne 0\phantom{\rule{0ex}{0ex}}x\ne -3\phantom{\rule{0ex}{0ex}}or\phantom{\rule{0ex}{0ex}}4-x\ne 0\phantom{\rule{0ex}{0ex}}x\ne 4$

Thus, the domain of the given function is all real numbers except $-3$ and $4$.

Therefore, the domain is $\hspace{0.17em}\left(-\infty \hspace{0.17em},\hspace{0.17em}-3\right)\cup \left(-3,\hspace{0.17em}4\right)\cup \left(4,\hspace{0.17em}\infty \hspace{0.17em}\right)$

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