If $\phi$ is continuous and $\lim_{x \to \infty} \frac{\phi(x)}{x^n}=0=\lim_{x \to -\infty} \frac{\phi(x)}{x^n}$ then(a) Prove that if $n$ is odd then there is a number $x$ such that $x^n+\phi(x)=0$
Proof: Since $\phi$ is continuous it must have an upper bound $N$ and a lower bound $M$. Since every number has an odd root it follows...$$x_{N}^n+N=0$$$$x_{M}^n+M=0$$for some $x_{N}$ and some $x_{M}$. Consider the interval $[x_{N},x_{M}]$ because $x^n$ is continuous it must take on every value between $x_{N}$ and $x_{M}$. Therefore there must be an $x$ such that $x^n+\phi(x)=0$ for every $x$ in the domain of $\phi$.
(b) Prove that if $n$ is even there is a number $y$ such that$y^n+\phi(y)\le x^n+\phi(x)$
Lemma: If $f$ is continuous on $[a,b]$, then there is some $y$ in $[a,b]$ such that $f(y)\le f(x)$ for all $x$ in $[a,b]$
If we let $f(x)=x^n$, it follows that $(f+\phi)(x)$ is continuous. It then follows from the lemma above that $$y^n+\phi(y)\le x^n+\phi(x)$$is automatically true for some $y$. Q.E.D
Afterthoughts: I am slightly worried because I did not use the fact that, $\lim_{x \to \infty} \frac{\phi(x)}{x^n}=0=\lim_{x \to -\infty} \frac{\phi(x)}{x^n}$, anywhere in my proof. This leads me to believe it's wrong, but I'm struggling to find an error. Any guidance would be appreciated.