Sufficiency without second derivative test

Question

Consider a function $\mathcal C^1 \ni f : \mathbb R_+ \to \mathbb R$. The following properties are given: \begin{align} &f(0) = 0,\\ &f(\infty) = -\infty,\\ &f'(\bar x) = 0,\\ &|\{x : f'(x) = 0\}| = 1. \end{align} Claim: There exists a unique global maximizer $x^*$ defined as \begin{align} x^* = \begin{cases} 0 \quad &\text{if } f(\bar x) < 0,\\ \bar x &\text{if } f(\bar x) \geq 0. \end{cases} \end{align} I'm pretty sure the statement is correct, but would not know how to prove it formally.

Answer 1

Let $q$ be such that $f(x) < -1$ if $x > q$. As continuous function on compact $[0, q]$, $f$ achieves global maximum in at least one point $x_0$ (and as this maximum is non-negative, it's also global maximum on $\mathbb R_+$). If for some $x_1 \neq x_0$ we have $f(x_0) = f(x_1)$, assuming wlog $x_0 < x_1$ we have $f'(x_1) = 0$ and also $f'(x) = 0$ for some $x \in (x_0, x_1)$ - so $f' = 0$ in at least two points. Thus $f$ has unique global maximizer.

If this maximizer isn't $0$, then $f'$ in it is $0$. So global maximizer is either $x$ such that $f'(x) = 0$ or $0$ - in whichever $f$ is greater. This is exactly your last expression.