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Question Number 96383 by bobhans last updated on 01/Jun/20

$$\mathrm{Find}\mathrm{domain}\mathrm{\&}\mathrm{range}\mathrm{of}\mathrm{function}$$ $$\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{{\mathrm{x}}^2-5\mathrm{x}+6}$$

Answered by john santu last updated on 01/Jun/20

$$\mathrm{The}\mathrm{domain}\mathrm{consists}\mathrm{of}\mathrm{all}\mathrm{real}$$ $$\mathrm{number}\mathrm{except}2\mathrm{and}3.\mathrm{To}\mathrm{determine}$$ $$\mathrm{the}\mathrm{range},\mathrm{set}\mathrm{y}=\frac{1}{x^2-5x+6}$$ $$\mathrm{and}\mathrm{solve}\mathrm{for}x.\Rightarrow x^2-5x+(6-\frac{1}{y})=0$$ $$\mathrm{This}\mathrm{has}\mathrm{solution}\mathrm{when}\mathrm{and}\mathrm{only}$$ $$\mathrm{when}{b}^2-4ac=25-24+\frac{4}{y}⩾0$$ $$\mathrm{This}\mathrm{holds}\mathrm{if}\mathrm{only}\mathrm{if}\frac{4}{\mathrm{y}}⩾-1$$ $$\mathrm{this}\mathrm{holds}\mathrm{when}\mathrm{y}>0,\mathrm{and}\mathrm{if}$$ $$\mathrm{y}<0,\mathrm{when}\mathrm{y}⩽-4.\mathrm{Hence}\mathrm{the}$$ $$\mathrm{range}\mathrm{is}(-\mathrm{\infty },-4]\cup (0,\mathrm{\infty })$$

Commented bybemath last updated on 01/Jun/20

thank you for the explanation

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