To preface this, I have had calculus, but not analysis. I am working on a discrete dynamical system which gives rise to the function $f(x) = (-x^2-x+1)(e^{-x^2})$.
I need to show that $|f(x)|<1$ for all values $x > 0$. I know that this is true from plotting the function, but I have no idea how to begin approaching this formally (probably because I haven't had any training in analysis). How do I approach this proof?
Since $e^{x^2}$ is positive, we can multiply through on both sides to see that $$ |f(x)|\leq 1\iff |1-x-x^2|\leq e^{x^2}. $$ For $x\geq 0$ (the domain you are interested in), obviously $1-x-x^2\leq 1$. Since $1\leq e^{x^2}$ for all $x$, $$ 1-x-x^2\leq 1\leq e^{x^2}, $$ proving half of the inequality. For the other half, we must show that $$ x^2+x-1\leq e^{x^2}, $$ which can by handled by using the inequality $e^y\geq y+1$ valid for all $y\geq 0$.
Specifically, for $x\leq 1$ we can take $y=x^2$ to conclude that $$ x^2+x-1\leq x^2+1\leq e^{x^2}, $$ whereas for $x\geq 1$ we can take $y=x^2-1$ to conclude that $$ x^2+x-1\leq 2x^2\leq ex^2\leq e\cdot e^{x^2-1}=e^{x^2}, $$ as desired.