Let the functions $f,g$ with domain:$\mathbb R$. The function of the composition $f\circ g$ is $"1-1"$.
I) Show that $g$ is $"1-1".$
II) Show that the equation $g(f(x)+x^3-x)=g(f(x)+2x-1)$ has got exactly 2 positive and one positive roots.
Personal work:
I) All I can recall regarding the composition is this: The domain of $f\circ g$ is:
$x\in D_g$ and $g(x)\in D_f$.
II) $$g(f(x)+x^3-x)=g(f(x)+2x-1) \iff{g:"1-1"} f(x)+x^3-x=f(x)+2x-1 \iff x^3-3x+1=0$$ and I can't seem to find any obvious roots...
Ideas for you (fill in details):
$$(I)\;\;g(x)=g(y)\implies (f\circ g)(x)=(f\circ g)(y)\implies x=y$$
and now
$$(II)\;\;g(f(x)+x^3-x)=g(f(x)+2x-1)\stackrel{\text{by}\;(I)}\implies f(x)+x^3-x=f(x)+2x-1\implies$$
$$h(x):=x^3-3x+1=0$$
But
$$h(-2)<0\;,\;\;h(0)>0\;,\;\;h(1)<0\;,\;\;h(2)>0\;\ldots$$