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Compute the domain of the function $$f(x)=\frac{1}{\lfloor x^2-17x+18\rfloor}.$$

Guest Apr 19, 2022

Guest · Answer 1 · 2022-04-19T06:02:31

f(x) = [x^2 - 17x + 18] ^-1

f'(x) = [2x - 17]^-1

f'(x) = 1/(2x - 17)

MaxWong · Answer 2 · 2022-04-19T06:21:12

For the domain, $\text{denominator} \neq 0$. Then we have $\lfloor x^2 - 17x + 18\rfloor \neq 0$.

By properties of floor function, we have $x^2 - 17x + 18 < 0\text{ or }x^2 - 17x + 18 \geq 1$.

Now you can solve the compound inequality to get the domain of f(x).