1. **Problem statement:** Given the graph of the derivative $y = f'(x)$ of a quartic function $y = f(x)$, analyze the following statements: a) $f(x)$ is decreasing on $(-\infty, -2)$. b) $f(x)$ has three critical points. c) The minimum value of $f(x)$ on $[-2, 2]$ is $f(0)$. d) Given $f(0) > 0$, the equation $f(x) = 0$ has at most 3 distinct solutions. 2. **Recall:** - $f'(x) < 0$ means $f(x)$ is decreasing. - Critical points of $f(x)$ occur where $f'(x) = 0$. - Local extrema of $f(x)$ correspond to zeros of $f'(x)$ where the sign of $f'(x)$ changes. - The number of roots of $f(x)$ relates to the behavior of $f'(x)$ and values of $f(x)$. 3. **Analyze (a):** - On $(-\infty, -2)$, the graph of $f'(x)$ is below the $x$-axis (negative). - Therefore, $f'(x) < 0$ on $(-\infty, -2)$. - Hence, $f(x)$ is decreasing on $(-\infty, -2)$. 4. **Analyze (b):** - The graph of $f'(x)$ crosses the $x$-axis at $x = -2, 0, 2$. - These are the critical points of $f(x)$. - There are exactly 3 critical points. 5. **Analyze (c):** - On $[-2, 2]$, $f'(x)$ changes sign at $x = -2, 0, 2$. - $f'(x)$ is positive on $(-2, 0)$ and negative on $(0, 2)$, so $f(x)$ increases then decreases. - Thus, $f(0)$ is a local maximum, not minimum. - Check endpoints: $f(-2)$ and $f(2)$ could be smaller. - So the minimum on $[-2, 2]$ is not necessarily $f(0)$. 6. **Analyze (d):** - Since $f(0) > 0$ and $f(x)$ is continuous, - $f(x)$ has 3 critical points, so $f'(x)$ has 3 zeros. - The quartic $f(x)$ can have up to 4 real roots. - But with $f(0) > 0$ and shape of $f'(x)$, $f(x)$ crosses zero at most 3 times. **Final answers:** - (a) True - (b) True - (c) False - (d) True