Axiom Learning·AP PrecalculusChapter 2: Polynomial Functions
多项式函数 · Extrema · Zeros · Symmetry · End Behavior · Graphing · Inequalities · Modeling
2.1 · Polynomial Functions — Definition & Properties 多项式函数的定义与性质
Definition 定义
A polynomial function $p$ can be written in the form:
- $n$ = degree (次数) — a non-negative integer
- $a_n$ = leading coefficient (首项系数), $a_n \neq 0$
- $a_0$ = constant term (常数项)
- $a_n x^n$ = leading term (首项)
• 缺失的项意味着其系数为 0。
• 多项式的定义域是全体实数 $(-\infty, +\infty)$，图像连续（无间断、无孔洞、无渐近线）。
• 常数函数（如 $f(x)=5$）视为次数为 0 的多项式。
- $f(x) = x^{-1} = \frac{1}{x}$ — negative exponent 负指数 ❌
- $g(x) = x^{1/2} = \sqrt{x}$ — fractional exponent 分数指数 ❌
- $h(x) = 2^x$ — variable in exponent 指数是变量 ❌
- $p(x) = 3x^2 - 5x + 7$ ✅ $q(x) = x^4 - \tfrac{1}{2}x^3 + 10$ ✅
Which of the following are polynomial functions? For each that is, state the degree and leading coefficient.
(a) $f(x) = 5x + \sqrt{x}$ (b) $g(x) = 3x^{-1}$ (c) $h(x) = 7x^4 - \tfrac{1}{2}x^3 + 10$
Degree = $4$; Leading coefficient = $7$; Constant term = $10$.
Relative Extrema 相对极值
A relative (local) maximum is a point where the function switches from increasing to decreasing.
A relative (local) minimum is a point where the function switches from decreasing to increasing.
If the domain is restricted (e.g. $[a,b]$), an included endpoint can also be a local maximum or minimum.
相对极小值（local min）：函数在该点从递减变为递增的"谷底"。
定义域受限时，被包含的端点也可以是局部极大/极小值（见下方过山车例题）。
注意：相对极值不一定是整个函数的最大/最小值（那叫绝对极值，见下方"全局极值"）。
Extreme Value Theorem 极值定理
If $f$ is a polynomial on a closed interval $[a,b]$, then $f$ has both an absolute maximum and an absolute minimum on $[a,b]$.
闭区间上的多项式函数必有绝对最大值和绝对最小值。
Local Extrema Theorem 局部极值定理
A polynomial of degree $n$ has at most $n-1$ relative maxima/minima.
$n$ 次多项式最多有 $n-1$ 个相对极值点。例如：3次多项式最多有2个；4次最多有3个。
Point of Inflection Theorem 拐点定理
- A polynomial of degree $n \ge 2$ has at most $n-2$ points of inflection.
$n\ge2$ 次多项式最多有 $n-2$ 个拐点（凹凸性改变的点）。 - A polynomial of odd degree $n \ge 3$ has at least 1 point of inflection.
次数 $\ge 3$ 的奇数次多项式至少有 1 个拐点。（一次函数是直线，没有拐点。）
用变化率的语言（AP 原文定义）：拐点出现在变化率由增变减、或由减变增的输入值处。所以在拐点处，变化率达到局部最大或局部最小，图像在这里"最陡"或"最平缓"。
State the number of possible relative maxima/minima and points of inflection for
$f(x) = -x^4 - 3x^3 - 5x^2 + 1$.
最多 3 个极值点，最多 2 个拐点。
For each polynomial, state: (i) degree, (ii) max number of relative extrema, (iii) max number of inflection points.
(a) $p(x) = x^5 - 3x^3 + 2$ (b) $q(x) = -2x^3 + 4x$ (c) $r(x) = x^6 - x^4 + 1$
(a) Degree 5 (odd): at most $5-1=\mathbf{4}$ extrema; at most $5-2=\mathbf{3}$ inflections. (At least 1 inflection, since odd degree.)
(b) Degree 3 (odd): at most $\mathbf{2}$ extrema; at most $\mathbf{1}$ inflection. (At least 1, since odd degree.)
(c) Degree 6 (even): at most $\mathbf{5}$ extrema; at most $\mathbf{4}$ inflections.
Global Extrema & Two Guarantees 全局极值与两条必然结论
Of all the local maxima, the greatest is the global (absolute) maximum. Of all the local minima, the least is the global (absolute) minimum.
定义域受限时，一定要把端点也算进候选名单。
Guarantee 1 · Between Two Zeros 两零点之间必有极值
Between every two distinct real zeros of a nonconstant polynomial, there is at least one local maximum or local minimum.
图像从 x 轴出发，要回到 x 轴，中间必须"掉头"一次。所以 $k$ 个不同实零点 ⇒ 至少 $k-1$ 个极值点。
Guarantee 2 · Even Degree 偶次多项式
A polynomial of even degree always has either a global maximum or a global minimum.
两端同向：都朝上（$a_n>0$）⇒ 有全局最小值；都朝下（$a_n<0$）⇒ 有全局最大值。奇次多项式（定义域为全体实数）两端反向，既没有全局最大值也没有全局最小值。
The height of a roller coaster track is modeled by $h(x) = -0.001x^3 + 0.12x^2 - 3.6x + 50$ meters, where $x$ is the horizontal distance in meters, for $0 \le x \le 80$. The graph is shown; the marked points lie on grid points.
(a) On what intervals is $h$ increasing? Decreasing?
(b) Identify every local maximum and local minimum of $h$, including endpoints.
(c) Find the global maximum and global minimum of $h$ on $[0, 80]$.
(d) Estimate the point of inflection. Where is the track rising most steeply? Explain using rates of change.
(e) If $h$ were defined for all real numbers, would it have a global maximum? Why?
Local minima: $h(20) = 18$ and $h(80) = 18$ (right endpoint: $h$ decreases into it).
Global maximum $= 50$ m (at $x = 0$ and $x = 60$). Global minimum $= 18$ m (at $x = 20$ and $x = 80$).
On $(20, 40)$ the rate of change of $h$ is positive and increasing; on $(40, 60)$ it is positive but decreasing. So the rate of change is greatest at $x = 40$: the track rises most steeply there.
(a) A polynomial $p$ has exactly three distinct real zeros: $x=-3$, $x=1$, and $x=4$. What is the least number of local extrema $p$ must have? Explain.
(b) Does $q(x) = 2x^4 - 7x^3 + x - 5$ have a global maximum? A global minimum?
(c) Does $r(x) = -x^6 + 4x^2 + 1$ have a global maximum? A global minimum?
(d) $s(x) = x^3 - 12x$ has a local maximum at $(-2, 16)$ and a local minimum at $(2, -16)$. Find the global maximum and minimum of $s$ on the interval $[-3, 5]$.
(a) At least 2: one between $-3$ and $1$, and one between $1$ and $4$ (there must be a local max or min between every two distinct real zeros).
(b) Even degree, $a_n = 2 > 0$: both ends go up. It has a global minimum but no global maximum.
(c) Even degree, $a_n = -1 < 0$: both ends go down. It has a global maximum but no global minimum.
(d) Candidates: endpoints $s(-3) = -27+36 = 9$ and $s(5) = 125-60 = 65$; local extrema $16$ and $-16$.
Global maximum $= 65$ at $x = 5$ (an endpoint!). Global minimum $= -16$ at $x = 2$.
Degree from Finite Differences 用差分判断多项式次数
For equally spaced input values, the degree of a polynomial is the least $n$ for which the $n$th differences are constant.
$x$ 等间距时，一阶差分恒定 → 一次；二阶恒定 → 二次；三阶恒定 → 三次……第一次出现"常数"的那一阶，就是多项式的次数。
Bonus: with step $h$, the constant $n$th difference equals $n!\,a_n\,h^n$ (for $n=2$ this is the $2ah^2$ from Chapter 1).
补充：间隔为 $h$ 时，恒定的 $n$ 阶差分等于 $n!\,a_n h^n$，可以反求首项系数 $a_n$。
The table gives values of a polynomial function $p$. Find the degree of $p$ and its leading coefficient.
| $x$ | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| $p(x)$ | 1 | 0 | 5 | 22 | 57 | 116 |
| 1st differences | $-1$ | $5$ | $17$ | $35$ | $59$ |
| 2nd differences | $6$ | $12$ | $18$ | $24$ | |
| 3rd differences | $6$ | $6$ | $6$ |
For each table, decide whether a polynomial could fit the data exactly. If so, give its degree.
Table A:
| $x$ | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| $f(x)$ | 0 | 0 | 14 | 78 | 252 | 620 |
Table B:
| $x$ | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| $g(x)$ | 1 | 3 | 9 | 27 | 81 | 243 |
Table A: 1st: $0, 14, 64, 174, 368$; 2nd: $14, 50, 110, 194$; 3rd: $36, 60, 84$; 4th: $24, 24$ → constant.
Degree 4; $4!\,a_4 = 24 \Rightarrow a_4 = 1$. (In fact $f(x) = x^4 - x$.)
Table B: 1st: $2, 6, 18, 54, 162$; 2nd: $4, 12, 36, 108$; 3rd: $8, 24, 72$; … Each row is 3 times the one before, so the differences never become constant. No polynomial fits exactly. (This is $g(x) = 3^x$, an exponential function; see Chapter 5.)
2.2 · Real and Complex Zeros 实零点与复零点
Key Properties of Zeros 零点的关键性质
A polynomial of degree $n$ has exactly $n$ complex zeros, counting multiplicity.
一个 $n$ 次多项式在复数范围内恰好有 $n$ 个零点（重根重复计算）。
If $p(a)=0$, then $a$ is a zero (root) of $p(x)$.
若 $p(a)=0$，则 $a$ 是 $p(x)$ 的一个零点（zero）或根（root）。
If $a$ is a zero, then $(x-a)$ is a factor of $p(x)$. If $a$ is also a real number, the graph has an x-intercept at $(a,0)$.
若 $a$ 是零点，则 $(x-a)$ 是 $p(x)$ 的因子；若 $a$ 还是实数，图像在 $(a,0)$ 有 x-intercept。非实数（复数）零点不产生 x 截距。
If $a+bi$ ($b\neq0$) is a zero of a real-coefficient polynomial, then its conjugate $a-bi$ is also a zero.
如果 $a+bi$（$b\neq0$）是实系数多项式的零点，那么 $a-bi$ 也一定是零点。
Real roots can appear alone; complex roots always come in conjugate pairs.
A degree-$n$ polynomial has at most $n-1$ turning points (local maxima/minima).
$n$ 次多项式最多有 $n-1$ 个 turning points（极值点）。
When Do Complex Roots Appear? 什么时候出现复数根？
当方程没有足够的实数根时，剩下的根就会出现在复数范围里。
When a polynomial doesn't have enough real zeros, the remaining zeros appear as complex numbers.
- 若 $\Delta > 0$ → 两个不同实根
- 若 $\Delta = 0$ → 一个重实根
- 若 $\Delta < 0$ → 两个复数根（共轭对）
| Example 例子 | $\Delta = b^2-4ac$ | Zeros 零点 | Type 类型 |
|---|---|---|---|
| $x^2 - 5x + 6 = 0$ | $25-24=1>0$ | $x=2,\;3$ | 两个不同实根 |
| $x^2 + 2x + 1 = 0$ | $4-4=0$ | $x=-1$ (mult. 2) | 重实根 |
| $x^2 + 1 = 0$ | $0-4=-4<0$ | $x=\pm i$ | 复数根（共轭对） |
| $x^2 + 4x + 13 = 0$ | $16-52=-36<0$ | $x=-2\pm 3i$ | 复数根（共轭对） |
Zeros: $x = 1,\,-1,\,i,\,-i$
四次多项式 → 4个零点；2个实根，2个虚根（共轭对 $\pm i$）。
对于三次多项式 $x^3 - 1 = 0$：
一个实根 $x=1$，另外两个是复数根：$x = -\dfrac{1}{2}\pm\dfrac{\sqrt{3}}{2}i$（共轭对）。
对于实系数多项式，复数根总是成共轭对出现，所以奇数次多项式至少有一个实根。
- $2+3i$ 是根 → $2-3i$ 一定也是根
- $-5+i$ 是根 → $-5-i$ 一定也是根
- $4i$ 是根（即 $0+4i$）→ $-4i$ 一定也是根
复数 $Z=a+bi$ 与其共轭 $\bar{Z}=a-bi$ 关于实数轴（x轴）对称。
A complex number and its conjugate are reflections across the real axis.
Find all zeros of $f(x) = x^3 - 3x^2 - 4x + 12$.
$= x^2(x - 3) - 4(x - 3)$
$= (x^2 - 4)(x - 3)$
So $f(x) = (x-2)(x+2)(x-3)$
$x + 2 = 0 \Rightarrow x = -2$
$x - 3 = 0 \Rightarrow x = 3$
三个实零点，与次数3相符。
Find all zeros of $f(x) = x^3 - 1$. Verify the total count matches the degree.
$x^3 - 1 = (x-1)(x^2+x+1)$
$\Delta = b^2 - 4ac = 1 - 4 = -3 < 0$ → complex roots!
$x = \dfrac{-1 \pm \sqrt{-3}}{2} = \dfrac{-1 \pm i\sqrt{3}}{2}$
$\sqrt{-3} = \sqrt{3}\cdot\sqrt{-1} = i\sqrt{3}$
These are conjugate pairs ✓ (共轭对，关于实轴对称)
Total zeros: $1$ real $+$ $2$ complex $= 3 =$ degree ✓
• Real: $x=1$
• Complex: $x=-\dfrac{1}{2}\pm\dfrac{\sqrt{3}}{2}i$ (conjugate pair 共轭对)
Total: 3 zeros = degree 3 ✓
Multiplicity 重数
When a linear factor $(x-c)$ is repeated $k$ times in the fully factored form, the zero $x=c$ has multiplicity $k$.
Example: $f(x) = (x-4)^2$ → zero at $x=4$ with multiplicity 2. 例：$f(x)=(x-4)^2$ 在 $x=4$ 处的零点重数为 2。
图像穿过 x 轴（符号改变）。
图像切触 x 轴（不穿越，符号不变）。
Sum of all multiplicities = degree of polynomial.
所有零点重数之和 = 多项式次数。
Given $f(x) = (1-x)^3(4+x)^2$.
(a) Find the degree. (b) Find all zeros and their multiplicities.
(c) Describe the graph's behavior at each x-intercept.
Total degree $= 3 + 2 = \mathbf{5}$.
$(4+x)^2 = 0 \Rightarrow x = -4$, multiplicity 2 (even).
Check: $3 + 2 = 5 =$ degree ✓
$x = -4$: multiplicity 2 (even) → graph touches and bounces off the x-axis.
Find a polynomial $f$ for which $x = 2$ is a zero of multiplicity 2 and $x = -5$ is also a zero (multiplicity 1).
Zero $x=-5$ with mult. 1 → factor $(x+5)$
$f(x) = a(x-2)^2(x+5)$ for any $a\neq 0$.
$= x^3+5x^2-4x^2-20x+4x+20 = x^3+x^2-16x+20$
For $g(x) = x^2(x-3)^3(x+1)$:
(a) State the degree. (b) List all zeros and multiplicities. (c) Describe graph behavior at each intercept. (d) Find the y-intercept.
(a) Degree $= 2 + 3 + 1 = \mathbf{6}$.
(b) $x=0$ (mult. 2); $x=3$ (mult. 3); $x=-1$ (mult. 1).
Check: $2+3+1=6=$ degree ✓
(c) $x=0$: even mult. → touches 切触. $x=3$: odd mult. → crosses 穿过. $x=-1$: odd mult. → crosses 穿过.
(d) $g(0) = 0$ → y-intercept: $(0, 0)$.
Even and Odd Functions 奇函数与偶函数
Even function 偶函数
$f(-x) = f(x)$ for every $x$ in the domain.
Graph is symmetric over the $y$-axis (the line $x = 0$).
图像关于 $y$ 轴对称。
Odd function 奇函数
$f(-x) = -f(x)$ for every $x$ in the domain.
Graph is symmetric about the origin $(0, 0)$ (rotate 180°).
图像关于原点中心对称。
- $p(x) = a_n x^n$ with $n$ even is an even function; with $n$ odd it is an odd function. 单项式：偶次幂为偶函数，奇次幂为奇函数。
- Only even powers (the constant counts as $x^0$) → even. Only odd powers → odd. A mix → neither. 只含偶次项 → 偶；只含奇次项 → 奇；混合 → 非奇非偶。
- Always confirm by computing $f(-x)$; that is the justification AP expects. 答题时要写出 $f(-x)$ 的计算作为理由。
"Even function" and "even degree" are different ideas. $x^4 + x$ has even degree but is neither even nor odd.
"偶函数"和"偶次多项式"不是一回事：$x^4+x$ 是偶次多项式，但非奇非偶。
Determine whether each function is even, odd, or neither. Justify using $f(-x)$.
(a) $f(x) = 3x^4 - 5x^2 + 2$ (b) $g(x) = x^5 - 2x^3 + x$ (c) $h(x) = x^3 + 1$
(a) $f$ is an odd function and $f(3) = -7$. Find $f(-3)$.
(b) The graph of an even function $g$ passes through $(-2, 5)$. Name another point on the graph.
(c) If $p$ is an odd polynomial function, what is $p(0)$? Explain.
(d) Is $k(x) = (x-1)^2$ even, odd, or neither?
(e) Is $m(x) = x^4 + x$ even, odd, or neither? What is its degree?
(a) $f(-3) = -f(3) = 7$.
(b) $(2, 5)$, by symmetry over the $y$-axis.
(c) $p(0) = -p(0)$, so $2p(0) = 0$ and $p(0) = 0$. The graph passes through the origin.
(d) $k(x) = x^2 - 2x + 1$ has both even and odd powers; $k(-x) = x^2 + 2x + 1$ is neither $k(x)$ nor $-k(x)$ → neither. (Its graph is symmetric about $x = 1$, not the $y$-axis.)
(e) $m(-x) = x^4 - x$ → neither, even though its degree (4) is even.
2.3 · End Behavior 端点行为
Limit Notation 极限记号
$\displaystyle\lim_{x \to c} f(x) = L$ means: as $x$ gets closer and closer to $c$, $f(x)$ gets closer and closer to $L$.
For end behavior, we use: $\displaystyle\lim_{x\to+\infty}f(x)$ and $\displaystyle\lim_{x\to-\infty}f(x)$.
端点行为告诉我们函数在图像最左端（$x\to-\infty$）和最右端（$x\to+\infty$）的走向。
End Behavior Rules 端点行为规律
End behavior is determined only by the leading term $a_n x^n$. 端点行为只由首项决定。
$\lim_{x\to+\infty}f=+\infty$
两端均向上 · like $x^2$
$\lim_{x\to+\infty}f=-\infty$
两端均向下 · like $-x^2$
$\lim_{x\to+\infty}f=+\infty$
左下右上 · like $x^3$
$\lim_{x\to+\infty}f=-\infty$
左上右下 · like $-x^3$
Why the Leading Term Wins 为什么只看首项
As $|x|$ grows, the leading term becomes so large that the lower-degree terms hardly matter (CED 1.6.A.3). Compare $f(x) = -4x^3 - 17x^2 - 14x + 6$ with its leading term $-4x^3$:
| $x$ | $f(x)$ | $-4x^3$ | $\dfrac{f(x)}{-4x^3}$ |
|---|---|---|---|
| 10 | $-5{,}834$ | $-4{,}000$ | 1.459 |
| 100 | $-4{,}171{,}394$ | $-4{,}000{,}000$ | 1.043 |
| 1000 | $-4{,}017{,}013{,}994$ | $-4{,}000{,}000{,}000$ | 1.004 |
State the left and right end behavior of $f(x) = -4x^3 - 17x^2 - 14x + 6$ using limit notation.
$\displaystyle\lim_{x\to-\infty}f(x)=+\infty$ $\displaystyle\lim_{x\to+\infty}f(x)=-\infty$
$\displaystyle\lim_{x\to+\infty}f(x)=-\infty$ (right end falls)
奇次负系数：左上右下。
State the end behavior using limit notation for each function:
(a) $f(x) = 3x^4 - 7x^2 + 1$ (b) $g(x) = -x^5 + 4x^3$ (c) $h(x) = -2x^6 + x^4 - 3$
(a) Degree 4 (even), $a_n=3>0$ → ↑↑: $\lim_{x\to\pm\infty}f(x)=+\infty$
(b) Degree 5 (odd), $a_n=-1<0$ → ↑↓: $\lim_{x\to-\infty}g=+\infty$, $\lim_{x\to+\infty}g=-\infty$
(c) Degree 6 (even), $a_n=-2<0$ → ↓↓: $\lim_{x\to\pm\infty}h(x)=-\infty$
2.4 · Graphing Polynomial Functions 绘制多项式函数图像
7-Step Strategy 七步绘图策略
- State the degree — determines max x-intercepts.
确定次数，最多可有多少个 x 轴截距。 - State max relative extrema & inflection points — at most $n-1$ and $n-2$.
极值点最多 $n-1$ 个，拐点最多 $n-2$ 个。 - Test for end behavior — check leading term sign and degree.
由首项系数和次数判断端点行为（四种情形）。 - Find x-intercepts (zeros) — factor completely, set $= 0$.
完全因式分解，令 $p(x)=0$ 求所有零点。 - State multiplicity of each zero — cross (odd) or touch (even).
判断每个零点的重数：奇数重数穿过，偶数重数切触。 - Find y-intercept — compute $p(0)$.
令 $x=0$ 求 y 轴截距。 - Sketch the graph using all the above.
综合以上信息绘制草图。
Sketch the graph of $f(x) = -2x^3 + 2x^2 + 8x - 8$.
$\lim_{x\to-\infty}f=+\infty$; $\lim_{x\to+\infty}f=-\infty$
$=-2[x^2(x-1)-4(x-1)]=-2(x^2-4)(x-1)$
$=-2(x-2)(x+2)(x-1)$
Zeros: $x=-2,\;1,\;2$
End behavior: ↑ (left) → ↓ (right). 左上右下。
For $p(x) = x^4 - 5x^2 + 4$:
(a) Factor completely. (b) Find all zeros and multiplicities. (c) State end behavior. (d) Find y-intercept. (e) Describe the sketch.
(a) $p(x)=x^4-5x^2+4=(x^2-1)(x^2-4)=(x-1)(x+1)(x-2)(x+2)$
(b) Zeros: $x=\pm1, \pm2$ — all multiplicity 1 (odd) → all cross.
(c) Degree 4 (even), $a_n=1>0$ → ↑↑: $\lim_{x\to\pm\infty}p=+\infty$. 两端均上升。
(d) $p(0)=0-0+4=\mathbf{4}$. y-intercept: $(0,4)$.
(e) Comes from top-left, crosses at $x=-2$, dips below (between $-2$ and $-1$), crosses at $x=-1$, rises above, passes through $(0,4)$, comes down, crosses at $x=1$, dips below, crosses at $x=2$, then rises to top-right.
Standard Form vs. Factored Form 标准式与因式分解式
| Form 形式 | Example | Reveals directly 直接看出 |
|---|---|---|
| Factored 因式分解式 | $-2(x-2)(x+2)(x-1)$ | Real zeros, $x$-intercepts, multiplicities (cross/touch), sign chart for inequalities 零点、x 截距、重数、不等式符号表 |
| Standard 标准式 | $-2x^3 + 2x^2 + 8x - 8$ | Degree, leading term → end behavior; constant term → $y$-intercept 次数、首项（端点行为）、常数项（y 截距） |
$g(x) = 2(x+1)(x-3)^2 = 2x^3 - 10x^2 + 6x + 18$. State which form you use, then find:
(a) the zeros of $g$ and the behavior of the graph at each; (b) the end behavior of $g$; (c) the $y$-intercept; (d) the solution of $g(x) \ge 0$.
(a) Factored form: $x = -1$ (mult. 1, crosses) and $x = 3$ (mult. 2, touches / tangent).
(b) Standard form, leading term $2x^3$: $\displaystyle\lim_{x\to-\infty}g(x) = -\infty$, $\displaystyle\lim_{x\to+\infty}g(x) = +\infty$.
(c) Standard form, constant term: $(0, 18)$.
(d) Factored form: $(x-3)^2 \ge 0$ always, so the sign of $g$ is the sign of $2(x+1)$. $g(x) \ge 0$ when $x \ge -1$ → $[-1, +\infty)$.
2.5 · Binomial Theorem 二项式定理
For any positive integer $n$:
Pascal's Triangle 帕斯卡三角
| $n$ | Coefficients 系数 | ||||||
| 0 | 1 | ||||||
| 1 | 1 1 | ||||||
| 2 | 1 2 1 | ||||||
| 3 | 1 3 3 1 | ||||||
| 4 | 1 4 6 4 1 | ||||||
| 5 | 1 5 10 10 5 1 | ||||||
例：$(a+b)^3 = 1\cdot a^3 + 3\cdot a^2b + 3\cdot ab^2 + 1\cdot b^3$（系数来自第3行 1 3 3 1）。
Write the full expansion of $(3-x)^4$.
$= 1\cdot81\cdot1 + 4\cdot27\cdot(-x) + 6\cdot9\cdot x^2 + 4\cdot3\cdot(-x^3) + 1\cdot1\cdot x^4$
Expand each expression fully:
(a) $(x+2)^3$ (b) $(2x-1)^4$ (c) Find the coefficient of $x^2$ in $(x+3)^5$.
(a) Row 3 coefficients: 1, 3, 3, 1.
$(x+2)^3 = x^3 + 3(x^2)(2) + 3(x)(4) + 8 = x^3 + 6x^2 + 12x + 8$
(b) $a=2x$, $b=-1$, $n=4$. Coefficients: 1, 4, 6, 4, 1.
$(2x)^4 + 4(2x)^3(-1) + 6(2x)^2(1) + 4(2x)(-1)^3 + (-1)^4$
$= 16x^4 - 32x^3 + 24x^2 - 8x + 1$
(c) For $(x+3)^5$, the $x^2$ term: $\binom{5}{2}x^2 \cdot 3^3 = 10 \cdot x^2 \cdot 27 = 270x^2$. Coefficient: $\mathbf{270}$.
2.6 · Polynomial Inequalities 多项式不等式
$f(x) > 0$ when the graph is above the x-axis. 当图像在 x 轴上方时，$f(x)>0$。
$f(x) < 0$ when the graph is below the x-axis. 当图像在 x 轴下方时，$f(x)<0$。
- Rewrite in the form $f(x) > 0$, $\ge 0$, $< 0$, or $\le 0$.
将不等式整理成 $f(x)$ 与 0 比较的形式。 - Find all roots of $f(x)$ (set $f(x) = 0$).
令 $f(x) = 0$，求出所有根（零点）。 - Plot roots on a number line — they divide it into intervals.
在数轴上标出零点，将数轴分成若干区间。 - Test a point in each interval → determine sign of $f(x)$.
在每个区间内取测试点，代入确定 $f(x)$ 的正负。 - Write the solution based on whether you need $f > 0$ or $f < 0$.
根据题目要求（$>0$ 还是 $<0$）写出解集。
• Strict inequalities ($>$ or $<$): use open intervals $(\;)$ — zeros are excluded.
严格不等式：零点不包含在解集内，用圆括号。
• Non-strict ($\ge$ or $\le$): use closed intervals $[\;]$ at zero points — zeros are included.
非严格不等式：零点包含在解集内，用方括号。
Solve: $2x^3 - 15 < x^2$. Give the boundary value to three decimal places.
Solve: $(x-1)(x+2)(x-3) \le 0$
| Interval | $(-\infty,-2)$ | $(-2,1)$ | $(1,3)$ | $(3,+\infty)$ |
|---|---|---|---|---|
| Test point | $x=-3$ | $x=0$ | $x=2$ | $x=4$ |
| $(x-1)$ | $-$ | $-$ | $+$ | $+$ |
| $(x+2)$ | $-$ | $+$ | $+$ | $+$ |
| $(x-3)$ | $-$ | $-$ | $-$ | $+$ |
| $f(x)$ | $-$ ✓ | $+$ | $-$ ✓ | $+$ |
$(-\infty,-2)$: $(-)\cdot(-)\cdot(-) = -$ ✓ $(-2,1)$: $(-)\cdot(+)\cdot(-) = +$ $(1,3)$: $(+)\cdot(+)\cdot(-) = -$ ✓
非严格不等式（$\le$），零点包含在解集中。
Solve: $(x+15)(x-2)^6(x-10) \le 0$
$x = 2$ (mult. 6, even → touches but does NOT change sign);
$x = 10$ (mult. 1, odd → crosses).
| Interval | $(-\infty,-15)$ | $(-15,2)$ | $(2,10)$ | $(10,+\infty)$ |
|---|---|---|---|---|
| Test pt | $-20$ | $0$ | $5$ | $15$ |
| $(x+15)$ | $-$ | $+$ | $+$ | $+$ |
| $(x-2)^6$ | $+$ | $+$ | $+$ | $+$ |
| $(x-10)$ | $-$ | $-$ | $-$ | $+$ |
| $f(x)$ | $+$ | $-$ ✓ | $-$ ✓ | $+$ |
注意：$x=2$ 处函数值为 0（是解），且两侧均为负，所以 $[-15,10]$ 是完整解集。
Solve each inequality. Write the solution in interval notation.
(a) $(x-2)(x+1)(x-5) > 0$
(b) $x^2(x-4) \le 0$
(c) $x^3 - 4x \ge 0$
(a) Zeros: $x=-1,2,5$ (all cross). Sign chart:
$(-\infty,-1)$: $(-)(-)(-) = -$ $(-1,2)$: $(-)(+)(-) = +$ ✓ $(2,5)$: $(+)(+)(-) = -$ $(5,\infty)$: $(+)(+)(+) = +$ ✓
Solution ($>0$, strict): $(-1,2)\cup(5,+\infty)$
(b) $x^2(x-4)\le0$. Zeros: $x=0$ (mult. 2, touches, no sign change); $x=4$ (mult. 1, crosses).
$x^2\ge0$ always. So sign of $f$ = sign of $(x-4)$: negative when $x<4$, zero at $x=4$.
But we need $\le0$: all $x<4$ give $f\le0$, plus $x=4$. Also $x=0$ gives $f=0$.
Solution: $(-\infty,4]$
(c) $x^3-4x=x(x-2)(x+2)\ge0$. Zeros: $x=-2,0,2$ (all cross).
Sign: $(-\infty,-2)$: $-$ $(-2,0)$: $+$ ✓ $(0,2)$: $-$ $(2,\infty)$: $+$ ✓
Solution: $[-2,0]\cup[2,+\infty)$
2.7 · Polynomial Models in Context 多项式情境建模
- Variables & units: what are the input and output? 输入、输出分别是什么，单位是什么？
- Restrict the domain: which inputs make sense physically? 情境允许哪些输入？
- Choose the form: factored for zeros, standard for end behavior. 选对形式。
- Use technology for extrema and intersections. 用计算器求极值和交点。
- Answer in context with units and correct rounding (3 decimal places). 用情境语言、带单位、保留三位小数作答。
- Enter the function: Y=, type it into $Y_1$ (and a horizontal line into $Y_2$ if needed).
- Set a window that matches the domain: WINDOW, then GRAPH.
- 2nd TRACE (CALC) → 2: zero · 3: minimum · 4: maximum · 5: intersect.
- For zero / min / max: move to the Left Bound → ENTER, Right Bound → ENTER, Guess → ENTER. For intersect: ENTER three times near the crossing.
更多操作见同目录下的《TI-84 Plus CE 常用功能手册》。
A 20 cm by 30 cm sheet of cardboard has a square of side $x$ cm cut from each corner. The sides are folded up to make an open-top box with volume $$V(x) = x(20-2x)(30-2x) = 4x^3 - 100x^2 + 600x.$$
(a) What is a reasonable domain for $V$ in this context?
(b) Find the zeros of $V$ and explain what they mean in context.
(c) Use a calculator to find the value of $x$ that gives the greatest volume, and the greatest volume.
(d) For what values of $x$ is the volume at least $800$ cm³?
(e) The standard form shows $\displaystyle\lim_{x\to\infty}V(x) = \infty$. Does this mean the box can be made as large as we like?
$V(x) \ge 800$ for $1.876 \le x \le 6.360$ (approximately).
A company's weekly profit is modeled by $P(x) = -0.5x^3 + 6x^2 - 10$ thousand dollars, where $x$ is the number of units sold (in hundreds), $0 \le x \le 12$.
(a) Find $P(0)$ and explain its meaning.
(b) Find the global maximum and global minimum of $P$ on $[0, 12]$. Where do they occur?
(c) For what values of $x$ does the company make a profit?
(a) $P(0) = -10$: with no units sold, the company loses \$10,000 per week (fixed costs).
(b) Calculator maximum at $x = 8$: $P(8) = -256 + 384 - 10 = 118$. Endpoints: $P(0) = -10$ and $P(12) = -864 + 864 - 10 = -10$.
Global maximum \$118,000 at $x = 8$ (800 units). Global minimum $-$\$10,000 at both endpoints $x = 0$ and $x = 12$.
(c) Zeros by calculator: $x \approx 1.372$ and $x \approx 11.858$. $P(x) > 0$ for $1.372 < x < 11.858$, i.e. about 138 to 1185 units per week.
Summary 总结
| Concept 概念 | Key Fact / Formula | 中文要点 |
|---|---|---|
| Polynomial | $a_nx^n+\cdots+a_0$, all non-negative integer exponents | 所有指数为非负整数，定义域 $\mathbb{R}$ |
| Relative extrema | At most $n-1$ for degree $n$ | $n$ 次多项式最多 $n-1$ 个极值点 |
| Inflection points | At most $n-2$; odd degree $n\ge3$ has $\ge 1$; where the rate of change switches between increasing and decreasing | 最多 $n-2$ 个；次数 ≥3 的奇次至少 1 个；变化率由增变减（或反之）处 |
| Total zeros | Exactly $n$ (complex, counting mult.) | $n$ 次多项式恰好有 $n$ 个复数零点（含重数） |
| Odd mult. zero | Graph crosses x-axis | 奇数重数 → 穿过 x 轴 |
| Even mult. zero | Graph touches x-axis | 偶数重数 → 切触 x 轴 |
| End behavior | Determined by leading term $a_nx^n$ | 只由首项决定；偶次同向，奇次反向 |
| Binomial Theorem | $(a+b)^n=\sum_{k=0}^n\binom{n}{k}a^{n-k}b^k$ | 展开公式；系数从帕斯卡三角读取 |
| Polynomial inequality | Find zeros → sign chart → select intervals | 求零点 → 符号表 → 选解集；注意严格与非严格 |
| Global extrema | Greatest local max / least local min; include endpoints of a restricted domain | 全局极值 = 局部极值中的最大/最小；受限定义域要比较端点 |
| Guarantees | Between two distinct real zeros → a local extremum; even degree → a global max or min | 两零点间必有极值；偶次必有全局最大或最小值 |
| Finite differences | Degree = least $n$ with constant $n$th differences ($=n!\,a_nh^n$) | 第几阶差分首次恒定，就是几次 |
| Even / odd | Even: $f(-x)=f(x)$, $y$-axis symmetry; odd: $f(-x)=-f(x)$, origin symmetry | 偶函数关于 y 轴对称；奇函数关于原点对称 |
| Forms | Factored → zeros, inequalities; standard → end behavior, $y$-intercept | 问零点用因式式；问端点行为用标准式 |
| Context & technology | Restrict domain; calculator zero / max / min / intersect; answer with units | 情境定义域、计算器求极值和交点、三位小数、带单位 |
核心口诀：
- 多项式：次数 $n$ → 最多 $n$ 个实零点，$n-1$ 个极值，$n-2$ 个拐点。
- 端点行为：偶次同向，奇次反向；系数正右升，系数负右降。
- 零点重数：奇数穿，偶数切。
- 不等式解法：整理为 $f\lessgtr0$ → 求零点 → 符号表 → 写解集。
- 全局极值：比较所有局部极值和端点；两零点之间必有极值；偶次多项式必有全局最大或最小值。
- 拐点 = 变化率由增变减（或由减变增）的点。
- 差分定次数：第 $n$ 阶差分首次恒定 → $n$ 次。
- 奇偶：$f(-x)=f(x)$ 偶（关于 y 轴对称），$f(-x)=-f(x)$ 奇（关于原点对称）。
- 情境题：先限定义域，再用计算器求零点/极值/交点，最后带单位回答。
Chapter 2 Review 章末习题
The function $f$ is given by $f(x)=4x^3-7x^5+2x-1$. Which of the following describes the end behavior of $f$?
答案：(C) 首项（最高次项）是 $-7x^5$，不是写在最前面的 $4x^3$。奇次、首项系数为负：左端趋向 $+\infty$，右端趋向 $-\infty$。
The polynomial function $p$ is given by $p(x)=(x-2)(x^2+4x-12)$. Which of the following is true?
答案：(B) $x^2+4x-12=(x-2)(x+6)$，所以 $p(x)=(x-2)^2(x+6)$。
不同的实零点只有 $2$ 和 $-6$ 两个；$x=2$ 重数为 2（偶）→ 图像在这里切触 $x$ 轴；$x=-6$ 重数为 1（奇）→ 穿过。
A cubic polynomial function $p$ with real coefficients has a non-real zero $x=2-i$. The graph of $y=p(x)$ has an $x$-intercept at $(-3,\,0)$. Which of the following could be an expression for $p(x)$?
答案：(A) 实系数多项式的复数零点成对出现，所以 $2+i$ 也是零点。
$\big(x-(2-i)\big)\big(x-(2+i)\big)=(x-2)^2-i^2=x^2-4x+5$；$x$ 截距 $(-3,0)$ 给出因式 $(x+3)$。所以 $p(x)=(x+3)(x^2-4x+5)$。
The polynomial function $p$ is given by $p(x)=(x+2)(x-1)^2(x-4)$. What are all values of $x$ for which $p(x)\le 0$?
答案：(B) 零点 $-2$（奇）、$1$（偶）、$4$（奇）。首项系数为正、4 次：$x>4$ 时为正；过 $x=4$ 变号 → $(1,4)$ 为负；$x=1$ 是偶重根不变号 → $(-2,1)$ 仍为负；过 $x=-2$ 变号 → $x<-2$ 为正。
所以 $p(x)\le0$ 的解是 $[-2,4]$（$x=1$ 处 $p=0$，也包含在内）。(A) 把 $x=1$ 当成变号点；(C) 漏了等号；(D) 把 $x=1$ 错误地排除了。
The binomial theorem can be used to expand $(2x-1)^6$. What is the coefficient of the $x^2$ term in the expansion?
答案：(C) 含 $x^2$ 的项：$\dbinom{6}{4}(2x)^2(-1)^4=15\cdot4x^2\cdot1=60x^2$。
(A) 忘了 $(2x)^2$ 里的 $2^2$；(B) 符号错（$(-1)^4=1$）；(D) $\dbinom{6}{2}\cdot2^4=240$ 是 $x^4$ 项的系数。
The function $f$ is given by $f(x)=x^4-3x^3-2x^2+5x+1$. What is the absolute maximum value of $f$ on the closed interval $-1.2\le x\le 3.2$?
答案：(B) 用计算器画图、求极值：相对极大值 $f(0.621)\approx2.764$；相对极小值 $f(-0.821)\approx-2.339$，$f(2.450)\approx-6.843$。端点：$f(-1.2)\approx-0.622$，$f(3.2)\approx3.074$。
把相对极值和两个端点放在一起比较，最大的是 $f(3.2)\approx3.074$。闭区间上的最大值可能出现在端点，不一定是相对极大值。
An open-top box is made from a $16$-inch by $10$-inch rectangular sheet of cardboard by cutting a square with side length $x$ inches from each corner and folding up the sides. The volume of the box is $V(x)=x(16-2x)(10-2x)$ cubic inches. What is the maximum possible volume of the box?
答案：(C) 情境定义域 $0<x<5$（短边 $10-2x>0$）。在这个区间上用计算器求最大值：$x=2$ 时 $V(2)=2\cdot12\cdot6=144$ 立方英寸。
(A) $V(1)=112$，(B) $V(2.5)=137.5$ 都不是最大值；(D) $160$ 是纸板面积的数值，不是体积。
The polynomial function $f$ is an odd function with domain $-5\le x\le5$. On $0\le x\le5$, the only relative extrema of $f$ are a relative maximum of $6$ at $x=2$ and a relative minimum of $-1$ at $x=4$, and $f(5)=7$. Which of the following is true about $f$ on its domain?
答案：(B) 奇函数关于原点对称：$f(-x)=-f(x)$。所以 $f(-2)=-6$（相对极小），$f(-4)=1$（相对极大），$f(-5)=-7$，$f(0)=0$。
最大值候选：$6,\ 1,\ 7$（端点 $x=5$）→ 绝对最大值 $7$；最小值候选：$-6,\ -1,\ -7$（端点 $x=-5$）→ 绝对最小值 $-7$。
The graph of the polynomial function $p$ is shown. The points $A$ and $D$ are the $x$-intercepts at $x=-3$ and $x=2$; the graph also crosses the $x$-axis at $x=-1$. $B$ and $C$ are the locations of relative extrema, and $E$ is a point of inflection.
(a) What is the least possible degree of $p$? Is the leading coefficient positive or negative? Give a reason for each answer.
(b) What can you conclude about the multiplicity of the zero at $x=2$? Explain.
(c) Between which two labeled points is $p$ increasing and its graph concave down?
(d) Suppose $p$ has degree $4$ and $p(0)=-12$. Write an expression for $p(x)$.
(a) 图上有 3 个相对极值（$B$、$C$、$D$）→ 次数至少为 4。两端都趋向 $-\infty$ → 偶次，首项系数为负。
(b) 图像在 $D$ 处只碰到 $x$ 轴、不穿过（两侧都在 $x$ 轴下方）→ $x=2$ 的重数为偶数（至少 2）。在 $D$ 处 $p$ 有相对极大值 $0$。
(c) 在 $E$ 和 $D$ 之间：$p$ 递增，而且 $E$ 是拐点、$D$ 是极大值，图像从上凹变成下凹后一直到 $D$ 都下凹 → "递增且下凹"的区间是 $E$ 到 $D$（$x_E<x<2$）。（$C$ 到 $E$ 之间是递增且上凹。）
(d) 零点 $-3,\ -1$（重数 1），$2$（重数 2），次数 4：$p(x)=a(x+3)(x+1)(x-2)^2$。$p(0)=a\cdot3\cdot1\cdot4=12a=-12\Rightarrow a=-1$，所以 $p(x)=-(x+3)(x+1)(x-2)^2$。
Directions: Solve each part without a calculator. Show the work that leads to your answers.
(a) Find all values of $x$ for which $g(x)=0$, where $g(x)=x^3+x^2-12x$.
(b) The function $h$ is given by $h(x)=2x^2-7x-15$. Find all values of $x$ for which $h(x)\ge 0$.
(c) The polynomial function $m$ has real coefficients and degree $4$. Its zeros include $0$, $4$, and $1-3i$. Write a possible expression for $m(x)$.
(d) Use the binomial theorem to expand $(x+2)^4$.
(a) $g(x)=x(x^2+x-12)=x(x+4)(x-3)=0\Rightarrow x=0,\ -4,\ 3$。
(b) $h(x)=(2x+3)(x-5)$，零点 $-\tfrac32$ 和 $5$；开口向上，所以两根之外为正：$x\le-\tfrac32$ 或 $x\ge5$。
(c) 复数零点成对：$1+3i$ 也是零点。$\big(x-(1-3i)\big)\big(x-(1+3i)\big)=(x-1)^2+9=x^2-2x+10$。所以 $m(x)=x(x-4)(x^2-2x+10)$（乘任意非零常数都可以）。
(d) 系数 $1,4,6,4,1$：$x^4+4x^3(2)+6x^2(2)^2+4x(2)^3+2^4=x^4+8x^3+24x^2+32x+16$。
The number of visitors in a park, in hundreds, $t$ hours after the park opens at 8 A.M. is modeled by $V(t)=-0.1t^3+1.2t^2+0.5t+2$ for $0\le t\le 12$. A calculator is allowed.
(a) According to the model, what is the greatest number of visitors in the park, and at what time does it occur?
(b) Find the average rate of change of $V$ from $t=2$ to $t=6$. Interpret your answer in context, with units.
(c) For what times $t$ does the model predict at least $2500$ visitors in the park?
(d) The park closes at 8 P.M. ($t=12$). Explain why the model should not be used for values of $t$ much greater than $12$.
(a) 用计算器求 $[0,12]$ 上的最大值：$t\approx8.203$ 时 $V\approx31.651$，即大约 3165 人，出现在开门后约 8.2 小时（约下午 4:12）。（还要比较端点：$V(0)=2$，$V(12)=8$，都更小。）
(b) $\dfrac{V(6)-V(2)}{6-2}=\dfrac{26.6-7}{4}=4.9$ hundred visitors per hour：从 10 A.M. 到 2 P.M.，园内人数平均每小时增加约 490 人。
(c) 解 $V(t)\ge25$：用计算器求 $V(t)=25$ 的交点，$t\approx5.629$ 和 $t\approx10.328$。所以 $5.629\le t\le10.328$。
(d) 首项 $-0.1t^3$ 使 $t$ 很大时 $V(t)\to-\infty$；模型在 $t\approx12.527$ 之后给出负的人数，没有意义。而且闭园后没有游客进入，模型只适用于营业时间 $0\le t\le12$。
| $x$ | $-2$ | $-1$ | 0 | 1 | 2 | 3 |
|---|---|---|---|---|---|---|
| $q(x)$ | 0 | 12 | 6 | $-6$ | $-12$ | 0 |
The table gives values of a polynomial function $q$.
(a) Use successive differences to find the degree of $q$ and its leading coefficient.
(b) Explain why $q$ must have a zero between $x=0$ and $x=1$.
(c) It is known that $q(x)=2x^3-3x^2-11x+6$. Find all zeros of $q$ exactly.
(d) Is $q$ an even function, an odd function, or neither? Justify using values from the table.
(a) 一阶差分：$12,\ -6,\ -12,\ -6,\ 12$；二阶差分：$-18,\ -6,\ 6,\ 18$；三阶差分：$12,\ 12,\ 12$，第一次恒定 → 三次多项式。$\Delta_3=3!\,a\,h^3=6a=12\Rightarrow a=2$。
(b) 多项式是连续函数，$q(0)=6>0$，$q(1)=-6<0$，函数值由正变负，所以在 $0$ 和 $1$ 之间必有一个零点。
(c) 表中 $q(-2)=0$，所以 $(x+2)$ 是因式：$2x^3-3x^2-11x+6=(x+2)(2x^2-7x+3)=(x+2)(2x-1)(x-3)$。零点：$x=-2,\ \tfrac12,\ 3$（$\tfrac12$ 正是 (b) 里 $0$ 和 $1$ 之间的那个）。
(d) 都不是。$q(1)=-6$，$q(-1)=12$：既不满足 $q(-1)=q(1)$（偶），也不满足 $q(-1)=-q(1)$（奇）。
