Solutions

a) By definition, there exists $N\in\mathbb{N}$ such that for all $x$,

$$\begin{align*} |x|>N & \implies \left|\frac{\phi(x)}{x^n}\right|<1 \\ & \implies |\phi(x)|<|x^n|. \end{align*}$$

Pick positive $x_1$ such that $x_1>N$. It follows that

$$x_1^n=|x_1^n|>-\phi(x_1).$$

Likewise, we can pick negative $x_2$ such that $x_2<-N$, where it follows that

$$-x_2^n=|x_2^n|>\phi(x_2).$$

Hence, since $\phi(x)+x^n$ is a continuous function by virtue of being the sum of continuous functions, there must exist $y\in[x_2,x_1]$ such that

$$\phi(y)+y^n=0.$$

b) By definition, there exists $N>0$ such that for all $x$,

$$\begin{align*} |x|>N &\implies \left|\frac{\phi(x)}{x^n}\right|<\frac14\\ &\implies -\frac14<\frac{\phi(x)}{x^n}<\frac14\\ &\implies \frac34<1+\frac{\phi(x)}{x^n}<\frac54. \end{align*}$$

In addition, we can find $b>N$ such that

$$\frac34 b^n>\phi(0).$$

For $|x|>b$, it follows that

$$x^n+\phi(x)=x^n\left(1+\frac{\phi(x)}{x^n}\right)>\frac34 x^n>\frac34 b^n>\phi(0).$$

Since $x^n+\phi(x)$ is continuous, there exists $y\in[-b,b]$ such that $f(y)\le f(x)$ for all $x\in[a,b]$. But $f(y)\le f(0)$. Hence, $f(x)\ge f(y)$ for all $x$.