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Question Number 32710 by mondodotto@gmail.com last updated on 31/Mar/18

Answered by MJS last updated on 01/Apr/18

$$f\left(x\right)=a(x+1)(x-4)$$$$f\left(0\right)=3$$$$a\times 1\times (-4)=3$$$$a=-\frac{3}{4}$$$$f\left(x\right)=-\frac{3}{4}{x}^2+\frac{9}{4}x+3$$$$$$$$g\left(x\right)=b(x+1)(x-4)$$$$g\left(0\right)=2$$$$b\times 1\times (-4)=2$$$$b=-\frac{1}{2}$$$$g\left(x\right)=-\frac{1}{2}{x}^2+\frac{3}{2}x+2$$$$$$$$\mathrm{now}\mathrm{some}\mathrm{read}f\circ g\left(x\right)=g\left(f\left(x\right)\right)$$$$\mathrm{but}\mathrm{some}\mathrm{read}f\circ g\left(x\right)=f\left(g\left(x\right)\right)$$$$$$$$$$$$g\left(f\left(x\right)\right)=3$$$$-\frac{1}{2}{f}^2\left(x\right)+\frac{3}{2}f\left(x\right)+2=3$$$$f^2\left(x\right)-3f\left(x\right)+2=0$$$$f\left(x\right)=\frac{3}{2}\pm \sqrt{\frac{9}{4}-2}=\frac{3}{2}\pm \frac{1}{2}$$$$f\left(x\right)=1\vee f\left(x\right)=2$$$$$$$$-\frac{3}{4}{x}^2+\frac{9}{4}x+3=1$$$$x^2-3x-\frac{8}{3}=0$$$$x=\frac{3}{2}\pm \sqrt{\frac{9}{4}+\frac{8}{3}}=\frac{3}{2}\pm \frac{\sqrt{177}}{6}$$$$$$$$-\frac{3}{4}{x}^2+\frac{9}{4}x+3=2$$$$x^2-3x-\frac{4}{3}=0$$$$x=\frac{3}{2}\pm \sqrt{\frac{9}{4}+\frac{4}{3}}=\frac{3}{2}\pm \frac{\sqrt{129}}{6}$$$$$$$$x\in \{\frac{3}{2}\pm \frac{\sqrt{177}}{6};\frac{3}{2}\pm \frac{\sqrt{129}}{6}\}$$$$$$$$$$$$f\left(g\left(x\right)\right)=3$$$$-\frac{3}{4}{g}^2\left(x\right)+\frac{9}{4}g\left(x\right)+3=3$$$$g^2\left(x\right)-3g\left(x\right)=0$$$$g\left(x\right)(g\left(x\right)-3)=0$$$$$$$$g\left(x\right)=0\vee g\left(x\right)=3$$$$g\left(x\right)=0\Rightarrow x=-1\vee x=4\left(\mathrm{from}\mathrm{above}\right)$$$$$$$$-\frac{1}{2}{x}^2+\frac{3}{2}x+2=3$$$$x^2-3x+2=0$$$$x=1\vee x=2$$$$$$$$x\in \{-1;1;2;4\}$$

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