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Question Number 165376 by MikeH last updated on 31/Jan/22

Verify wether f is invertible   f (x) = (1+2x)^3

$$\mathrm{Verify}\:\mathrm{wether}\:{f}\:\mathrm{is}\:\mathrm{invertible}\: \\ $$$${f}\:\left({x}\right)\:=\:\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{3}} \\ $$

Answered by Rasheed.Sindhi last updated on 31/Jan/22

f (x) = (1+2x)^3   y=(1+2x)^3   1+2x=(y)^(1/3)    x=(((y)^(1/3)  −1)/2)  f^(−1) (x)=(((x)^(1/3)  −1)/2)

$${f}\:\left({x}\right)\:=\:\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{3}} \\ $$$${y}=\left(\mathrm{1}+\mathrm{2}{x}\right)^{\mathrm{3}} \\ $$$$\mathrm{1}+\mathrm{2}{x}=\sqrt[{\mathrm{3}}]{{y}}\: \\ $$$${x}=\frac{\sqrt[{\mathrm{3}}]{{y}}\:−\mathrm{1}}{\mathrm{2}} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=\frac{\sqrt[{\mathrm{3}}]{{x}}\:−\mathrm{1}}{\mathrm{2}} \\ $$

Commented by aleks041103 last updated on 01/Feb/22

If x∈R, g(x)=x^3  is bijective.  For all x, l(x)=1+2x is bijective.  Therefore f=g○l is a composition  of 2 bijective functions⇒f is bijective.  ⇒f is invertible, but only on R.

$${If}\:{x}\in\mathbb{R},\:{g}\left({x}\right)={x}^{\mathrm{3}} \:{is}\:{bijective}. \\ $$$${For}\:{all}\:{x},\:{l}\left({x}\right)=\mathrm{1}+\mathrm{2}{x}\:{is}\:{bijective}. \\ $$$${Therefore}\:{f}={g}\circ{l}\:{is}\:{a}\:{composition} \\ $$$${of}\:\mathrm{2}\:{bijective}\:{functions}\Rightarrow{f}\:{is}\:{bijective}. \\ $$$$\Rightarrow{f}\:{is}\:{invertible},\:{but}\:{only}\:{on}\:\mathbb{R}. \\ $$

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