Show that $f(x) = x^7 + x^5 + x^3 + x$ is bijective I want to show that the real polynomial function $f: \mathbb R \to \mathbb R, f(x) = x^7 + x^5 + x^3 + x$ is bijective. I want to show this without using the inverse or the derivative.
I'm struggling to prove injectivity, because I see no easy way to arrive at $x = y$. What I have so far:
Surjective:
Because the degree is odd, we have $\lim_{x \to +\infty}(f(x)) = +\infty$ and $\lim_{x \to -\infty}(f(x)) = -\infty$. Because a polynom is continuous, we can apply the IVT to the interval $I = (-\infty,+\infty) = \mathbb R$ so that for every $y \in I$ there is a $x$ such that $f(x) = y$.
Injective:
Let $f(x) = f(y)$. Then $x^7 + x^5 + x^3 + x = y^7 + y^5 + y^3 + y$. Then ??? , so $x = y$.

Suppose that $f(a)=f(b).$  Then since 
$$
f(x)=x \left( {x}^{2}+1 \right)  \left( {x}^{4}+1 \right) 
$$
we have 
$$
0=f(a)-f(b)={b}^{7}+{b}^{5}+{b}^{3}+b-({a}^{7}+{a}^{5}+{a}^{3}+a) = \left( b-a \right)  \left(  \left( b-a \right) ^{2}+1 \right) 
 \left(  \left( b-a \right) ^{4}+1 \right).
$$
Its implies that $a=b$, so $f$ is injective
