Show that $f(1)$ is the maximum value of $f(x)=\frac{x^2+x}{1+x^2}$ where $x \in [-1, 1]$ without derivatives

Question

Let $f: [-1,1] \to \mathbb{R}$ such that $$f(x)=\frac{x^2+x}{1+x^2}$$ Show that $f(1)$ is the maximum value of $f$.

I thought about maybe breaking it down into simpler steps by finding the maximum of $g(x)=x^2+x$ and $h(x)=1+x^2$, which are $1$ and $\{-1, 1\}$ respectively, but I'm not sure how to show that either. We haven't seen derivatives, but we have seen the Bolzano theorem and the Weierstrass theorem.

Answer 1

You have $f(1)=1$ and\begin{align}1-\frac{x^2+x}{1+x^2}&=\frac{1-x}{1+x^2}\\&\geqslant0,\end{align}since $x\leqslant1$. So, yes, $f$ attains its maximum at $1$.