$f(x)=x^9+3x^3+3x-3$, there is only on $c$ to $f(c)=2c$

Question

Let $f(x)=x^9+3x^3+3x-3$. I want to show that there is only one $c\in(0,1)$ such that $f(c)=2c$. How can i prove this?

Answer 1

Hint:

Define $g(x)=f(x)-2x=x^9+3x^3+x-3$ then

$f(c)=2c$ is equivalent to $c$ is a root of $g(x)$.

Show $g(0)=-3<0,g(1)>0$, the root exists by Intermediate value theorem

Show $g$ is strictly increasing on $(0,1)$ by 1st derivative test.

Answer 2

Let $h(x)=f(x)-2x$ then

$$h'(x)=9x^8+9x^2+1>0,\quad \forall x\in(0,1)$$ then $h$ is strictly increasing on this interval and $$h(0)h(1)\le0$$ so we deduce the desired result using the intermediate value theorem.