1. The problem asks: If $f$ is a decreasing function, what can we say about the function $g = f \circ f$ (the composition of $f$ with itself)? 2. Recall that a function $f$ is decreasing if for any $x_1 < x_2$, we have $f(x_1) \geq f(x_2)$. 3. The composition $g = f \circ f$ means $g(x) = f(f(x))$. 4. Since $f$ is decreasing, applying $f$ once reverses the order: if $x_1 < x_2$, then $f(x_1) \geq f(x_2)$. 5. Now consider $g(x_1) = f(f(x_1))$ and $g(x_2) = f(f(x_2))$. 6. Because $f$ is decreasing, and $f(x_1) \geq f(x_2)$, applying $f$ again reverses the inequality: $$g(x_1) = f(f(x_1)) \leq f(f(x_2)) = g(x_2)$$ 7. This means $g$ is an increasing function. 8. In summary, the composition of a decreasing function with itself is an increasing function. Final answer: If $f$ is decreasing, then $g = f \circ f$ is increasing.