Spivak's Calculus Chapter $7$ $15$b) Prove $f$ is bounded below Q15) Suppose that $\phi$ is continuous and $\lim_{x \rightarrow \infty}(\frac{\phi(x)}{x^n})=0=\lim_{x \rightarrow -\infty}(\frac{\phi(x)}{x^n})$
b) Prove that if $n$ is even, then there is a number $y$ such that $y^n + \phi(y) \leq x^n + \phi(x)$ for all $x$.
I have been wrestling with the question all day but can't get the result. In fact not only did I not get the result, I'm becoming convinced it's false. For example, say $\phi(x) = \frac{1}{x}$. Then $\phi$ is continuous and satisfies the limits above, but $x^n + \frac{1}{x}$ doesn't have a lower bound because $\lim_{x \rightarrow 0^-}(x^n + \frac{1}{x}) = -\infty$. Unless $\frac{1}{x}$ is not continuous at $0$, ok, but when we say a function is continuous, don't we mean specifically it's continuous on the domain on which it's defined? I'm pretty sure I'm wrong somewhere. How do we solve this question?

Hint : The function $$x \mapsto x^n + \phi(x)$$
is continuous and tends to $+\infty$ when $x$ tends to $\pm \infty$. It is sufficient to ensure that it has a minimum.
