Track / Stage 4 / 305 of 340 #565 of 1964

Problem 565

AMC 12 late, AIME early

Algebra Difficulty 4.9 Find the answer

Let $f(x)$ be continuous on $\mathbf{R}$ , $f(0)=1$ , and $f(x+y) \geqslant f(x) f(y)$ , find $f(x)$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

12. $f(x+y)-f(x) \geqslant f(x)[f(y)-1]$ , that is,
$f(x+y)-f(x) \geqslant f(x)[f(y)-f(0)],$
(1) Dividing both sides by $y$ ,
when $y>0$ , we get
$\frac{1}{(x+y)-x}[f(x+y)-f(x)] \geqslant f(x) \frac{f(y)-f(0)}{y-0}$

Let $y \rightarrow 0$ . We get $f^{\prime}(x) \geqslant f(x) \cdot f^{\prime}(0)$
head $y>0)$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.