Derivative and inequality - 2

Expnding $(ax^2+x+a)e^{-x}$ we get

$xe^{-x} + a(x^2+1)e^{-x}$ . But since $a \leq 0$ and $(x^2+1)e^{-x}$ is always positive, the second term is negative. This means that $f(x) \leq xe^{-x}$ and equality happens when $a=0$ , which is the worst case scenario.

The best approximation we can possibly make is one where their derivatives at $x=0$ match. Indeed, since they have the same value at $x=0$ , if the derivative of $g(x)=b \ln{(x+1)}$ is smaller, then its guaranteed that there exsists some x where $g(x) < f(x)$ .

Let's check if matching derivatives yelds a solution:

The derivative of $f(x)$ at $0$ is $e^0-0e^0 = 1$ ; the derivative of $g(x)$ is $\frac{b}{0+1} = b$ . Setting b = 1, yelds their derivatives:

$f'(x)=(1-x)e^{-x}$ and $g'(x)=\frac{1}{1+x}$ . We can see that $f'(x) \leq g'(x)$ . Indeed, if it weren't, we'd have:

$(1-x)e^{-x} > \frac{1}{1+x}$

$(1-x^2) > e^x$ . Which is clearly false for $x \geq 1$ .

Therefore, their derivatives and values match initially, then for all $x\geq 0$ , $f'(x)\leq g'(x)$ , and thus $f(x) \leq g(x)$ .