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find the domain of f(x)^-1if f(x)=7x+3
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\[f(x)^{-1}=\frac{1}{7x+3}\] find the value of x for which denominator is zero 7x+3=0 7x=-3 x=-3/7 at x=-3/7 th value of f(x)^-1 is undefined, so domain is all real numbers -{-7/3}
\[f(x)^{-1}=\frac{1}{7x+3}\] \[7x+3\neq0, 7x \neq -3, x \neq \frac{-3}{7}\] \[(-\infty, \frac{-3}{7})U(\frac{-3}{7}, \infty)\]
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