How does one prove that this function is injective? As the title suggests, I am trying to prove that the following function is injective but at the moment I am stuck.
$$f(x)=-x^5-16x -1 $$
$$-x_1^5 - 16x_1 -1 = -x_2^5 - 16x_2 -1$$
$$x_1^5 + 16x_1= x_2^5 + 16x_2$$
What should be the next step?
Could you also tell me if the following identities are correct?
$$sin\frac{1}{1+x_1^2}=sin\frac{1}{1+x_2^2}$$
$$\frac{1}{1+x_1^2}=\frac{1}{1+x_2^2}$$

I suppose that your function is defined on $\mathbb{R}$. Assume that $f(x)=f(y)$. This means that $-x^{5}-16x-1=-y^{5}-16y-1$, so $x^5+16x=y^5+16y$. From this we get $$(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4+16)=0.$$
Then, $x=y$ or $x^4+x^3y+x^2y^2+xy^3+y^4+16=0$. If $x=y$ we're done. Otherwise, $x^4+x^3y+x^2y^2+xy^3+y^4+16=0$. But if $x,y>0$ thus the LHS is positive, and therefore cannot be equal to $0$. So at least one of $x$ and $y$ must be negative. Suppose that it's $y$ and write $y=-z$ with $z>0$. We have $$x^4-x^3z+x^2z^2-xz^3+z^4+16=0.$$
But $x^4-x^3z+x^2z^2-xz^3+z^4+16=(x-z)^2(x^2+xz+z^2)+x^2z^2+16>0$. So it must be $x=y$, and hence $f$ is injective.
