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Question Number 16457 by Tinkutara last updated on 22/Jun/17

Find the range of f(x) = (3/(2 − x^2 ))

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:{f}\left({x}\right)\:=\:\frac{\mathrm{3}}{\mathrm{2}\:−\:{x}^{\mathrm{2}} } \\ $$

Answered by sandy_suhendra last updated on 22/Jun/17

y = (3/(2−x^2 )) y(2−x^2 )=3 2y−x^2 y=3 x^2 y=2y−3 x^2 = ((2y−3)/y) x = (√((2y−3)/y)) f^(−1) (x) = (√((2x−3)/x)) range of f(x) is domain of f^(−1) (x) ((2x−3)/x) ≥ 0 ⇒ x < 0 or x ≥ 1(1/2) range of f(x) = {y∣ y < 0 or y ≥ 1(1/2)}

$$\mathrm{y}\:=\:\frac{\mathrm{3}}{\mathrm{2}−\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{y}\left(\mathrm{2}−\mathrm{x}^{\mathrm{2}} \right)=\mathrm{3} \\ $$$$\mathrm{2y}−\mathrm{x}^{\mathrm{2}} \mathrm{y}=\mathrm{3} \\ $$$$\mathrm{x}^{\mathrm{2}} \mathrm{y}=\mathrm{2y}−\mathrm{3} \\ $$$$\mathrm{x}^{\mathrm{2}} =\:\frac{\mathrm{2y}−\mathrm{3}}{\mathrm{y}} \\ $$$$\mathrm{x}\:=\:\sqrt{\frac{\mathrm{2y}−\mathrm{3}}{\mathrm{y}}} \\ $$$$\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)\:=\:\sqrt{\frac{\mathrm{2x}−\mathrm{3}}{\mathrm{x}}} \\ $$$$\mathrm{range}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{is}\:\mathrm{domain}\:\mathrm{of}\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right) \\ $$$$\frac{\mathrm{2x}−\mathrm{3}}{\mathrm{x}}\:\geqslant\:\mathrm{0}\:\:\Rightarrow\:\:\mathrm{x}\:<\:\mathrm{0}\:\:\mathrm{or}\:\mathrm{x}\:\geqslant\:\mathrm{1}\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{range}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\left\{\mathrm{y}\mid\:\mathrm{y}\:<\:\mathrm{0}\:\:\mathrm{or}\:\mathrm{y}\:\geqslant\:\mathrm{1}\frac{\mathrm{1}}{\mathrm{2}}\right\} \\ $$

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