$f(x) = x^{4} - 6x^{2}$. Full curve analysis: stationary points, inflections, concavity, sketch.$f(x) = x^{4} - 6x^{2}$。完整曲线分析：稳定点、拐点、凹凸性、草图。
Answers:答案： (b) max $(0, 0)$, min $(\pm\sqrt{3}, -9)$极大 $(0, 0)$、极小 $(\pm\sqrt{3}, -9)$ · (c) inflections $(\pm 1, -5)$拐点 $(\pm 1, -5)$
(a) Derivatives M1·A1
$f'(x) = 4x^{3} - 12x = 4x(x^{2} - 3)$ and $f''(x) = 12x^{2} - 12 = 12(x^{2} - 1)$.
(b) Stationary points M1·A1·A1·A1·A1
Solve $f'(x) = 0$: $4x(x^{2} - 3) = 0 \Rightarrow x = 0,\; \pm \sqrt{3}$.
Apply second derivative test using $f''(x) = 12(x^{2} - 1)$:
- $x = 0$: $f''(0) = -12 < 0 \Rightarrow$ local max. $f(0) = 0$, so $(0, 0)$.
- $x = \sqrt{3}$: $f''(\sqrt{3}) = 12 \cdot 2 = 24 > 0 \Rightarrow$ local min. $f(\sqrt{3}) = 9 - 18 = -9$, so $(\sqrt{3}, -9)$.
- $x = -\sqrt{3}$: $f''(-\sqrt{3}) = 24 > 0 \Rightarrow$ local min. $f(-\sqrt{3}) = -9$, so $(-\sqrt{3}, -9)$.
(c) Inflection points M1·A1·A1·R1·A1
Solve $f''(x) = 0$: $12(x^{2} - 1) = 0 \Rightarrow x = \pm 1$. Candidates: $(\pm 1, f(\pm 1)) = (\pm 1, -5)$.
Sign chart for $f''$:
- $x < -1$: $x^{2} - 1 > 0 \Rightarrow f'' > 0$ (concave up).
- $-1 < x < 1$: $x^{2} - 1 < 0 \Rightarrow f'' < 0$ (concave down).
- $x > 1$: $x^{2} - 1 > 0 \Rightarrow f'' > 0$ (concave up).
$f''$ changes sign at both $x = -1$ and $x = 1$, so both are genuine inflection points. Coordinates: $(-1, -5)$ and $(1, -5)$.
(d) Concavity intervals and sketch A1·A1·A1
Concave up on $(-\infty, -1) \cup (1, \infty)$; concave down on $(-1, 1)$.
$x$-intercepts: $f(x) = x^{2}(x^{2} - 6) = 0 \Rightarrow x = 0$ (double root) or $x = \pm\sqrt{6} \approx \pm 2.449$. At $x = 0$ the curve has a local maximum at height $0$: near the origin $f(x) \approx -6x^{2} \le 0$ with equality only at $x = 0$, so the curve sits on the $x$-axis at the origin and dips below it on either side before climbing through $(\pm\sqrt{6}, 0)$ to $+\infty$. Geometrically, the $x$-axis is tangent to the curve at the origin (a double root) — not a transverse crossing.
End behaviour: $f(x) \to +\infty$ as $x \to \pm \infty$ (even quartic, positive leading coefficient). Even symmetry: $f(-x) = f(x)$, so the graph is symmetric about the $y$-axis. Sketch shows a "W"-shape with peak at $(0, 0)$, troughs at $(\pm\sqrt{3}, -9)$, inflections at $(\pm 1, -5)$, and outer $x$-intercepts at $\pm\sqrt{6}$.
Sign-chart tabulation is the marker-friendly way. On Paper 3 with three or more critical points, write an explicit table:
\[
\begin{array}{c|ccccc}
x & (-\infty, -1) & -1 & (-1, 1) & 1 & (1, \infty) \\ \hline
f''(x) & + & 0 & - & 0 & + \\
\text{concavity} & \cup & \text{inflect} & \cap & \text{inflect} & \cup
\end{array}
\]
The table earns the R1 (the reasoning mark for sign change) for both inflections simultaneously, and it eliminates the most common HL error: forgetting that the same $f''$ sign on both sides of a candidate means it is not an inflection. For $f(x) = x^{4}$ at $0$: $f''(x) = 12x^{2} \ge 0$ everywhere, $f''$ does not change sign through $0$, so $(0, 0)$ is not an inflection despite $f''(0) = 0$. Tabulate, then conclude.
(a) 各阶导 M1·A1
$f'(x) = 4x^{3} - 12x = 4x(x^{2} - 3)$，$f''(x) = 12x^{2} - 12 = 12(x^{2} - 1)$。
(b) 稳定点 M1·A1·A1·A1·A1
解 $f'(x) = 0$：$4x(x^{2} - 3) = 0 \Rightarrow x = 0,\; \pm \sqrt{3}$。
用 $f''(x) = 12(x^{2} - 1)$ 做二阶导数判别：
- $x = 0$：$f''(0) = -12 < 0 \Rightarrow$ 极大。$f(0) = 0$，即 $(0, 0)$。
- $x = \sqrt{3}$：$f''(\sqrt{3}) = 24 > 0 \Rightarrow$ 极小。$f(\sqrt{3}) = 9 - 18 = -9$，即 $(\sqrt{3}, -9)$。
- $x = -\sqrt{3}$：$f''(-\sqrt{3}) = 24 > 0 \Rightarrow$ 极小。$f(-\sqrt{3}) = -9$，即 $(-\sqrt{3}, -9)$。
(c) 拐点 M1·A1·A1·R1·A1
解 $f''(x) = 0$：$12(x^{2} - 1) = 0 \Rightarrow x = \pm 1$。候选点 $(\pm 1, f(\pm 1)) = (\pm 1, -5)$。
$f''$ 符号表：
- $x < -1$：$x^{2} - 1 > 0 \Rightarrow f'' > 0$（上凹）。
- $-1 < x < 1$：$x^{2} - 1 < 0 \Rightarrow f'' < 0$（下凹）。
- $x > 1$：$x^{2} - 1 > 0 \Rightarrow f'' > 0$（上凹）。
$f''$ 在 $x = \pm 1$ 处均变号，故两点都是真正的拐点。坐标 $(-1, -5)$ 与 $(1, -5)$。
(d) 凹凸区间与草图 A1·A1·A1
上凹于 $(-\infty, -1) \cup (1, \infty)$；下凹于 $(-1, 1)$。
$x$ 截距：$f(x) = x^{2}(x^{2} - 6) = 0 \Rightarrow x = 0$（重根）或 $x = \pm \sqrt{6} \approx \pm 2.449$。$x = 0$ 处曲线与 $x$ 轴相切（$x$ 轴是该处切线，因 $(0, 0)$ 是局部极大且高度为 $0$，邻近点 $f(x) < 0$）。
端行为：$x \to \pm \infty$ 时 $f \to +\infty$（偶次四次、首项系数正）。偶对称：$f(-x) = f(x)$，图像关于 $y$ 轴对称。草图为"W"形：峰 $(0, 0)$，谷 $(\pm\sqrt{3}, -9)$，拐点 $(\pm 1, -5)$，外侧 $x$ 截距 $\pm\sqrt{6}$。
列符号表是评卷友好的做法。Paper 3 临界点 $\ge 3$ 时，明确写出表：
\[
\begin{array}{c|ccccc}
x & (-\infty, -1) & -1 & (-1, 1) & 1 & (1, \infty) \\ \hline
f''(x) & + & 0 & - & 0 & + \\
\text{shape} & \cup & \mathrm{IP} & \cap & \mathrm{IP} & \cup
\end{array}
\]
表格一次性赢得两个拐点的 R1（变号推理分），同时杜绝 HL 最常见错误：忘记候选点两侧 $f''$ 同号意味着不是拐点。例 $f(x) = x^{4}$ 在 $0$：$f''(x) = 12x^{2} \ge 0$ 恒成立，$f''$ 未变号，故 $(0, 0)$ 不是拐点，尽管 $f''(0) = 0$。列表，再下结论。