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 $x = y$.
 A: For injectivity: the sum of strictly increasing functions is strictly increasing. And $x^k$ is strictly increasing for odd $k$.
A: For injectivity we may use Descarte's rule of signs. Suppose we want solutions to $f(x)=c$. This is equivalent to showing the following has exactly one root: $$x^7+x^5+x^3+x-c=0$$ Suppose $c$ is positive (this works for $c<0$ but I'll leave that to you. 
Now $f(x)-c=0$ has exactly one sign change, so there is exactly 1 positive root, or one positive solution to $f(x)=c$. To examine negative roots we look at $f(-x)-c=0$, which is 
$$
-x^7-x^5-x^3-x-c=0.
$$
This has no sign changes, hence no roots.
A: Hint: This is an odd function.  
A: 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
