Question:

Step 1: Determine the minimum value of $F(t)$. Let $G(t) = 2t^3 - 15t^2 + 36t - 25$. To find the critical points of $G(t)$, we compute its derivative: $G'(t) = 6t^2 - 30t + 36 = 6(t^2 - 5t + 6) = 6(t-2)(t-3)$. Setting $G'(t) = 0$ gives $t=2$ or $t=3$. Now, we evaluate $G(t)$ at these critical points: $G(2) = 2(2)^3 - 15(2)^2 + 36(2) - 25 = 16 - 60 + 72 - 25 = 3$. $G(3) = 2(3)^3 - 15(3)^2 + 36(3) - 25 = 54 - 135 + 108 - 25 = 2$. The local minimum of $G(t)$ is $2$. The second function in $F(t)$ is $H(t) = 2|\sin t|$. The minimum value of $H(t)$ is $2 \times 0 = 0$. Therefore, $F(t) = \min\{G(t), H(t)\}$. Since $G(t)$ has a local minimum of $2$ and $H(t)$ has a global minimum of $0$, the minimum value of $F(t)$ is $0$. Step 2: Analyze the equation $3^x + 3^{f(x)} = \min F(t)$. Substituting the minimum value of $F(t)$ into the equation, we get: $3^x + 3^{f(x)} = 0$. However, $3^x > 0$ for all real $x$, and $3^{f(x)} > 0$ for all real $f(x)$. The sum of two positive numbers cannot be zero. This implies that the minimum value of $F(t)$ must be greater than $0$ for $f(x)$ to be defined. The minimum value of $G(t)$ is $2$. The minimum value of $H(t)$ is $0$. The minimum value of $F(t)$ is $0$. If the problem implies that $3^x + 3^{f(x)}$ must be equal to the *local* minimum of $G(t)$ that is greater than $0$, then we consider the value $2$. Let's assume $\min F(t) = 2$. Then $3^x + 3^{f(x)} = 2$. Step 3: Determine the domain of $f(x)$. For $f(x)$ to be a real-valued function, $3^{f(x)}$ must be a real number. From $3^x + 3^{f(x)} = 2$, we have $3^{f(x)} = 2 - 3^x$. For $3^{f(x)}$ to be defined and positive, we must have $2 - 3^x > 0$. $2 - 3^x > 0 \implies 3^x < 2$. Taking $\log_3$ on both sides: $x < \log_3 2$. Thus, the domain of $f(x)$ is $(-\infty, \log_3 2)$.