If $f\circ g$ is $"1-1"$ then show that $g$ is $"1-1"$ and more

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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...

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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

$$h(-2)<0\;,\;\;h(0)>0\;,\;\;h(1)<0\;,\;\;h(2)>0\;\ldots$$

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Suppose $x\neq x'$. If $g(x)$ were equal $g(x')$ then $f[g(x)]=f[g(x')]$, a contradiction. So $g(x)\neq g(x')$. This deals with I.

For II, use I: $g[f(x)+x^3-x]=g[f(x)+2x-1]$ implies $$ f(x)+x^3-x=f(x)+2x-1\iff x^3-3x+1=0 $$ Now, let $h(x)=x^3-3x+1$. Check: $$ h(-2)<0,\quad h(-1)>0,\quad h(0)>0,\quad h(1)<0,\quad h(2)>0. $$ So $h$ has 3 zeros: one in $(-2,-1)$, one in $(0,1)$, and the last one in $(1,2)$.