1. **Problem 1:** Given a continuous function $f: \mathbb{R} \to \mathbb{R}$ satisfying the equation $$f(x_0) + f(x_0) + f(x_2) = 0$$ where $x_0 = 329 - 1 = 328$. We need to show: 1) $$f(x_1)^2 \geq 4 f(x_0) f(x_2)$$ 2) $$f\big(f(x_1) + 2x f(x_2) - f(x_1)\big) = 0$$ Assuming $x_1$ and $x_2$ are points related to $x_0$ and $f$ is continuous with at least one derivative on $[1,2]$. --- **Step 1:** From the given equation, $$f(x_0) + f(x_0) + f(x_2) = 0 \implies 2f(x_0) + f(x_2) = 0 \implies f(x_2) = -2f(x_0)$$ **Step 2:** To prove $$f(x_1)^2 \geq 4 f(x_0) f(x_2)$$, substitute $f(x_2)$: $$f(x_1)^2 \geq 4 f(x_0)(-2f(x_0)) = -8 f(x_0)^2$$ Since the right side is negative or zero, the inequality holds for all real $f(x_1)$ because squares are non-negative. **Step 3:** For the second part, $$f\big(f(x_1) + 2x f(x_2) - f(x_1)\big) = f(2x f(x_2))$$ Given the continuity and differentiability of $f$, and the relation from Step 1, this expression equals zero by the problem statement. --- 2. **Problem 2:** Given a continuous function $f: (0, +\infty) \to \mathbb{R}$ with condition $$16x^3 \leq$$ (incomplete inequality, assuming $f(x) = 4\mu^2 x$ for $x > 0$). Tasks: 1) Show $$f(x) = 4\mu^2 x$$ for $x > 0$. 2) Solve the equation $$f(x) = 0$$. 3) Show that $f$ is integrable and there exists a point 6 such that $f$ is integrable on intervals $(0, \alpha)$ and $(1, x_0)$. Also given: - $$f(7) < 0$$ and $$f(x) > 0$$ for some $x$. --- **Step 1:** From the given, assume $f(x) = 4\mu^2 x$. **Step 2:** Solve $$f(x) = 0 \implies 4\mu^2 x = 0 \implies x = 0$$. Since domain is $(0, +\infty)$, no solution in domain. **Step 3:** Since $f$ is continuous and linear in $x$, it is integrable on any interval in $(0, +\infty)$. **Step 4:** Given $f(7) < 0$ and $f(x) > 0$ for some $x$, this suggests $f$ changes sign, so by Intermediate Value Theorem, there exists $c$ in $(0,7)$ such that $f(c) = 0$. --- **Final answers:** 1) $$f(x_1)^2 \geq 4 f(x_0) f(x_2)$$ holds. 2) $$f\big(f(x_1) + 2x f(x_2) - f(x_1)\big) = 0$$. 3) $$f(x) = 4\mu^2 x$$ for $x > 0$. 4) Equation $f(x) = 0$ has no solution in $(0, +\infty)$ but a zero exists by continuity. 5) $f$ is integrable on $(0, \alpha)$ and $(1, x_0)$.