1. **Problem statement:** Given the function $y = f(x) = ax^3 + bx^2 + cx + d$ with the derivative sign table and function behavior: - $f'(x)$ changes sign: positive on $(-\infty, -1)$, zero at $x=-1$, negative on $(-1, 2)$, zero at $x=2$, positive on $(2, +\infty)$. - $f(x)$ increases from $-\infty$ to 1 at $x=-1$, then decreases to $-2$ at $x=2$, then increases to $+\infty$. We analyze the statements: a) The function has a local minimum at $x=2$. b) The function is decreasing on $(0,1)$. c) On $(-\infty, 2)$, the function's maximum is 1 and minimum is $-2$. Also, the function $y = \frac{2024}{f'(x) + 1}$ has 4 asymptotes. 2. **Key formulas and rules:** - Critical points occur where $f'(x) = 0$. - If $f'(x)$ changes from negative to positive at a point, $f$ has a local minimum there. - If $f'(x)$ changes from positive to negative, $f$ has a local maximum. - A function is decreasing where $f'(x) < 0$. - The maximum and minimum values on an interval can be found at critical points or endpoints. - Vertical asymptotes of $y = \frac{2024}{f'(x) + 1}$ occur where $f'(x) + 1 = 0$, i.e., $f'(x) = -1$. 3. **Step-by-step analysis:** **a) Local minimum at $x=2$?** - From the table, $f'(x)$ changes from negative (on $( -1, 2)$) to zero at $x=2$, then positive on $(2, +\infty)$. - This sign change from negative to positive means $f$ has a local minimum at $x=2$. - The statement is **true**. **b) Is $f$ decreasing on $(0,1)$?** - On $( -1, 2)$, $f'(x) < 0$, so $f$ is decreasing on the entire interval $( -1, 2)$. - Since $(0,1) \subset (-1,2)$, $f$ is decreasing on $(0,1)$. - The statement is **true**. **c) Maximum and minimum on $(-\infty, 2)$?** - On $(-\infty, -1)$, $f'(x) > 0$, so $f$ is increasing. - At $x=-1$, $f'(x)=0$, $f(x)=1$ (local max). - On $(-1, 2)$, $f'(x) < 0$, so $f$ is decreasing. - At $x=2$, $f'(x)=0$, $f(x)=-2$ (local min). - So on $(-\infty, 2)$, the maximum is $1$ at $x=-1$ and minimum is $-2$ at $x=2$. - The statement is **true**. **d) Number of asymptotes of $y = \frac{2024}{f'(x) + 1}$?** - Vertical asymptotes occur where denominator is zero: $f'(x) + 1 = 0 \Rightarrow f'(x) = -1$. - Since $f'(x)$ is a quadratic (derivative of cubic), it can equal $-1$ at up to 2 points. - However, the problem states there are 4 asymptotes, implying $f'(x) + 1$ has 4 zeros or the function has other asymptotes (horizontal or oblique). - Without explicit $f'(x)$, we cannot confirm 4 vertical asymptotes. 4. **Summary:** - a) True - b) True - c) True - d) Cannot confirm with given data **Final answer:** - a), b), c) are correct.