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Question Number 13598 by Tinkutara last updated on 21/May/17

If g(x) = x^2 + x − 2 and (1/2) gof(x) = 2x^2 − 5x + 2, then prove that f(x) = 2x − 3.

$$\mathrm{If}\:{g}\left({x}\right)\:=\:{x}^{\mathrm{2}} \:+\:{x}\:−\:\mathrm{2}\:\mathrm{and} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:{gof}\left({x}\right)\:=\:\mathrm{2}{x}^{\mathrm{2}} \:−\:\mathrm{5}{x}\:+\:\mathrm{2},\:\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{that}\:{f}\left({x}\right)\:=\:\mathrm{2}{x}\:−\:\mathrm{3}. \\ $$

Commented by mrW1 last updated on 21/May/17

(1/2)gof=(1/2)(f^2 +f−2)=2x^2 −5x+2 f^2 +f−2(2x^2 −5x+3)=0 f=((−1±(√(1+4×2(2x^2 −5x+3))))/2) f=((−1±(√(16x^2 −40x+25)))/2) f(x)=((−1±(4x−5))/2)=((4x−6)/2)=2x−3 or ((−4x+4)/2)=2(1−x) that means there are 2 solutions: f(x)=2x−3 or f(x)=2(1−x)

$$\frac{\mathrm{1}}{\mathrm{2}}{gof}=\frac{\mathrm{1}}{\mathrm{2}}\left({f}^{\mathrm{2}} +{f}−\mathrm{2}\right)=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{2} \\ $$$${f}^{\mathrm{2}} +{f}−\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{3}\right)=\mathrm{0} \\ $$$${f}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}×\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{3}\right)}}{\mathrm{2}} \\ $$$${f}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{16}{x}^{\mathrm{2}} −\mathrm{40}{x}+\mathrm{25}}}{\mathrm{2}} \\ $$$${f}\left({x}\right)=\frac{−\mathrm{1}\pm\left(\mathrm{4}{x}−\mathrm{5}\right)}{\mathrm{2}}=\frac{\mathrm{4}{x}−\mathrm{6}}{\mathrm{2}}=\mathrm{2}{x}−\mathrm{3}\:{or}\:\frac{−\mathrm{4}{x}+\mathrm{4}}{\mathrm{2}}=\mathrm{2}\left(\mathrm{1}−{x}\right) \\ $$$${that}\:{means}\:{there}\:{are}\:\mathrm{2}\:{solutions}: \\ $$$${f}\left({x}\right)=\mathrm{2}{x}−\mathrm{3} \\ $$$${or}\:{f}\left({x}\right)=\mathrm{2}\left(\mathrm{1}−{x}\right) \\ $$

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