question about inverse functions

Question

Functions $f$ and $g$ are defined by

$$ f: x\mapsto 2x+1$$

$$g: x \mapsto \dfrac{2x +1}{x+3}$$

(i) Solve the equation $gf(x) = x $

(iii) Show that the equation $g^{-1} (x) = x$ has no solutions

I need help with these.

Answer 1

$$ f(x)=2x+1,g(x)= \frac{2x +1}{x+3}$$

$$g(f(x)) =\frac{2(2x+1) +1}{2x+1+3}= x$$

$$\frac{4x+3}{2x+4}= x,x\neq-2\Rightarrow2x^2+4x=4x+3,x^2=3/2,x=\pm\sqrt{3/2}$$

$$g(x)=y= \frac{2x +1}{x+3}$$

$$x= \frac{2y +1}{y+3},xy+3x=2y+1,y(x-2)=1-3x,y=\frac{1-3x}{x-2},x\neq2$$

$$g^{-1}(x)=\frac{1-3x}{x-2}=x\Rightarrow x^2+x-1=0,x=\frac{-1\pm\sqrt{5}}{2}$$

Answer 2

For the second part

$g(x)= \dfrac{2x+1}{x+3}$

$g(x)=y$

$y=\dfrac{2x+1}{x+3}$

$xy+3y=2x+1 \implies x= \dfrac{1-3y}{y-2}$

$g^-1(x)=\dfrac{1-3x}{x-2}$

Now equate it to $x$.

$\dfrac{1-3x}{x-2}=x \implies 1-3x=x^2-2x$

$ \implies x^2+x+1=0$ , here $b^2-4ac <0$, you can conclude there exists no solutions in reals.