Question 17
Given $$f(x) = \frac{2025x}{x+1}$$ where $$x \neq -1$$. Then the value of $$x$$ for which $$f(f(x)) = (2025)^2$$ is ------.
Correct Answer: -1/2025
Solution
Substituting $$f(x)$$ into itself gives $$f(f(x)) = \frac{2025^2 x}{2026x+1}$$. Setting this equal to $$2025^2$$ gives $$x = 2026x+1$$, so $$-2025x = 1$$ and $$x = -\frac{1}{2025}$$.
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