Question:

Step 1: Find the critical points and local extrema of $f(x)$. First, we find the derivative of $f(x)$ with respect to $x$ and set it to zero to find the critical points. $$ f(x) = x^3 - 3x + 1 $$ $$ f'(x) = 3x^2 - 3 $$ Setting $f'(x) = 0$: $$ 3x^2 - 3 = 0 $$ $$ 3(x^2 - 1) = 0 $$ $$ x^2 - 1 = 0 $$ $$ x = \pm 1 $$ Now, we evaluate $f(x)$ at these critical points to find the local extrema: At $x=1$: $$ f(1) = (1)^3 - 3(1) + 1 = 1 - 3 + 1 = -1 $$ This is a local minimum. At $x=-1$: $$ f(-1) = (-1)^3 - 3(-1) + 1 = -1 + 3 + 1 = 3 $$ This is a local maximum. Step 2: Determine the number and approximate locations of the real roots of $f(t) = 0$. The function $f(x)$ is a cubic polynomial, which is continuous everywhere. From the local extrema: - $f(-1) = 3$ (local maximum) - $f(1) = -1$ (local minimum) Since $f(-1) > 0$ and $f(1) < 0$, by the Intermediate Value Theorem, there must be a root between $-1$ and $1$. Also, as $x \to -\infty$, $f(x) \to -\infty$, and $f(-1) = 3 > 0$, so there is a root for $x < -1$. As $x \to \infty$, $f(x) \to \infty$, and $f(1) = -1 < 0$, so there is a root for $x > 1$. Thus, the equation $f(t) = 0$ has exactly three distinct real roots. Let these roots be $\alpha, \beta, \gamma$ such that $\alpha < -1$, $-1 < \beta < 1$, and $1 < \gamma$. Step 3: Formulate the condition for $f(f(x)) = 0$. The equation $f(f(x)) = 0$ implies that $f(x)$ must be equal to one of the roots of $f(t) = 0$. Therefore, we must have $f(x) = \alpha$, $f(x) = \beta$, or $f(x) = \gamma$. The total number of real roots for $f(f(x))=0$ will be the sum of the number of roots for each of these three equations. Step 4: Analyze the number of real roots for $f(x) = \alpha$, $f(x) = \beta$, and $f(x) = \gamma$. We use the local maximum value $f(-1)=3$ and local minimum value $f(1)=-1$ to determine the number of solutions for $f(x)=k$ for different values of $k$. Case 1: $f(x) = \alpha$ We know that $\alpha < -1$. Since the local minimum of $f(x)$ is $f(1) = -1$, and $\alpha < -1$, the horizontal line $y=\alpha$ intersects the graph of $y=f(x)$ only once. This intersection occurs for $x > 1$, where $f(x)$ decreases from $-1$ to $-\infty$. Thus, $f(x) = \alpha$ has **1 real root**. Case 2: $f(x) = \beta$ We know that $-1 < \beta < 1$. Since the local minimum $f(1) = -1 < \beta$ and the local maximum $f(-1) = 3 > \beta$, the horizontal line $y=\beta$ intersects the graph of $y=f(x)$ three times. Specifically, one root for $x < -1$, one root for $-1 < x < 1$, and one root for $x > 1$. Thus, $f(x) = \beta$ has **3 real roots**. Case 3: $f(x) = \gamma$ We know that $\gamma > 1$. Since the local maximum of $f(x)$ is $f(-1) = 3$, and $\gamma > 1$, the horizontal line $y=\gamma$ intersects the graph of $y=f(x)$ only once. This intersection occurs for $x < -1$, where $f(x)$ increases from $-\infty$ to $3$. Thus, $f(x) = \gamma$ has **1 real root**. Step 5: Calculate the total minimum number of real roots of $f(f(x)) = 0$. The total number of real roots for $f(f(x)) = 0$ is the sum of the roots from each case: Total roots = (roots from $f(x)=\alpha$) + (roots from $f(x)=\beta$) + (roots from $f(x)=\gamma$) Total roots = $1 + 3 + 1 = 5$. The minimum number of real roots of $f(f(x))=0$ is 5. The final answer is $\boxed{5}$.