Question:

Step 1: Find the roots of $f(x) = 0$. We need to solve $x^3 - 3x + 1 = 0$. Using the discriminant formula for cubic equations $\Delta = -4a^3 - 27b^2$ where the cubic is in the form $x^3 + ax + b = 0$: $$\Delta = -4(-3)^3 - 27(1)^2 = 108 - 27 = 81 > 0$$ Since $\Delta > 0$, the equation has three distinct real roots. Let us denote them as $\alpha_1 < \alpha_2 < \alpha_3$. Step 2: Reformulate $f(f(x)) = 0$ in terms of the roots found. The equation $f(f(x)) = 0$ means that $f(x)$ must equal one of the roots of $f(x) = 0$. Therefore: $$f(f(x)) = 0 \iff f(x) \in \{\alpha_1, \alpha_2, \alpha_3\}$$ This means we need to solve three separate equations: $f(x) = \alpha_1$, $f(x) = \alpha_2$, and $f(x) = \alpha_3$. Step 3: Analyze the behavior of $f(x)$ using calculus. To understand how many solutions each equation $f(x) = \alpha_i$ has, we find the critical points: $$f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$$ Critical points occur at $x = -1$ and $x = 1$. Step 4: Determine local extrema. Evaluating $f(x)$ at the critical points: - At $x = -1$: $f(-1) = (-1)^3 - 3(-1) + 1 = -1 + 3 + 1 = 3$ (local maximum) - At $x = 1$: $f(1) = (1)^3 - 3(1) + 1 = 1 - 3 + 1 = -1$ (local minimum) Therefore, $f(x)$ has a local maximum value of $3$ and a local minimum value of $-1$. Step 5: Determine the number of solutions to $f(x) = k$ for different values of $k$. Based on the local extrema, the horizontal line $y = k$ intersects the curve $y = f(x)$ as follows: - If $-1 < k < 3$: the line intersects the curve at **3 points** (3 real roots) - If $k = -1$ or $k = 3$: the line intersects the curve at **2 points** (2 real roots) - If $k < -1$ or $k > 3$: the line intersects the curve at **1 point** (1 real root) Step 6: Locate the three roots $\alpha_1, \alpha_2, \alpha_3$ relative to the interval $[-1, 3]$. We evaluate $f(x)$ at key points: - $f(0) = 1 > 0$ and $f(1) = -1 < 0$, so $\alpha_2 \in (0,1)$ - $f(1) = -1 < 0$ and $f(2) = 8 - 6 + 1 = 3 > 0$, so $\alpha_3 \in (1,2)$ - By the intermediate value theorem and the shape of the cubic, $\alpha_1 < -1$ Therefore: - $\alpha_1 < -1$ - $-1 < \alpha_2 < 3$ - $-1 < \alpha_3 < 3$ Step 7: Count the total number of real roots of $f(f(x)) = 0$. For each equation $f(x) = \alpha_i$: - $f(x) = \alpha_1$ where $\alpha_1 < -1$: This gives **1 real root** - $f(x) = \alpha_2$ where $-1 < \alpha_2 < 3$: This gives **3 real roots** - $f(x) = \alpha_3$ where $-1 < \alpha_3 < 3$: This gives **3 real roots** The minimum total number of real roots is: $$1 + 3 + 3 = 7$$ **Final Answer: The minimum number of real roots of $f(f(x)) = 0$ is 7.** This corresponds to **Option 4: 7**.