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Question Number 96383 by bobhans last updated on 01/Jun/20
$$\mathrm{Find}\:\mathrm{domain}\:\&\:\mathrm{range}\:\mathrm{of}\:\mathrm{function}\: \\ $$ $$\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{6}}\: \\ $$
Answered by john santu last updated on 01/Jun/20
![The domain consists of all real number except 2 and 3. To determine the range , set y = (1/(x^2 −5x+6)) and solve for x. ⇒x^2 −5x+(6−(1/y)) =0 This has solution when and only when b^2 −4ac = 25−24+(4/y)≥0 This holds if only if (4/y)≥−1 this holds when y > 0 , and if y<0 , when y ≤ −4. Hence the range is (−∞,−4 ] ∪ (0, ∞ )](Q96384.png)
$$\mathrm{The}\:\mathrm{domain}\:\mathrm{consists}\:\mathrm{of}\:\mathrm{all}\:\mathrm{real}\: \\ $$ $$\mathrm{number}\:\mathrm{except}\:\mathrm{2}\:\mathrm{and}\:\mathrm{3}.\:\mathrm{To}\:\mathrm{determine} \\ $$ $$\mathrm{the}\:\mathrm{range}\:,\:\mathrm{set}\:\mathrm{y}\:=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}}\: \\ $$ $$\mathrm{and}\:\mathrm{solve}\:\mathrm{for}\:{x}.\:\Rightarrow{x}^{\mathrm{2}} −\mathrm{5}{x}+\left(\mathrm{6}−\frac{\mathrm{1}}{{y}}\right)\:=\mathrm{0} \\ $$ $$\mathrm{This}\:\mathrm{has}\:\mathrm{solution}\:\mathrm{when}\:\mathrm{and}\:\mathrm{only} \\ $$ $$\mathrm{when}\:{b}^{\mathrm{2}} −\mathrm{4}{ac}\:=\:\mathrm{25}−\mathrm{24}+\frac{\mathrm{4}}{{y}}\geqslant\mathrm{0} \\ $$ $$\mathrm{This}\:\mathrm{holds}\:\mathrm{if}\:\mathrm{only}\:\mathrm{if}\:\frac{\mathrm{4}}{\mathrm{y}}\geqslant−\mathrm{1}\: \\ $$ $$\mathrm{this}\:\mathrm{holds}\:\mathrm{when}\:\mathrm{y}\:>\:\mathrm{0}\:,\:\mathrm{and}\:\mathrm{if} \\ $$ $$\mathrm{y}<\mathrm{0}\:,\:\mathrm{when}\:\mathrm{y}\:\leqslant\:−\mathrm{4}.\:\mathrm{Hence}\:\mathrm{the}\: \\ $$ $$\mathrm{range}\:\mathrm{is}\:\left(−\infty,−\mathrm{4}\:\right]\:\cup\:\left(\mathrm{0},\:\infty\:\right)\:\: \\ $$
Commented bybemath last updated on 01/Jun/20
thank you for the explanation