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4.1 Maximum and Minimum Values goo.gl/8FNtRj 1

Chapter 4

Applications of Differentiation

4.1

Maximum and Minimum Values, page 276

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Definition 1 (page 276). Let c be a number in the domain D of a function f . Then f (c) is the

(1) absolute maximum value (絕對極大值_{) of f on D if f (c) ≥ f (x) for all x in D.}

(2) absolute minimum value (絕對極小值_{) of f on D if f (c) ≤ f (x) for all x in D.}





x x

y y

Figure 1: Absolute maximum value and absolute minimum value of f .



Definition 2 (page 276). The number f (c) is a

(1) local maximum value (局部極大值_{) of f on D if f (c) ≥ f (x) when x is near c.}

(2) local minimum value (局部極小值_{) of f on D if f (c) ≤ f (x) when x is near c.}

We say that something is true near c, we mean that it is true on “some open interval containing c.”

x x

y y

Figure 2: Local maximum value and local minimum value of f .

Example 3. State the absolute (and local) maximum (and minimum) values of the function y = f (x). efbompPs1hs x y y = f (x) 1 1

Figure 3: Find absolute (and local) maximum (and minimum) values of the function.

Solution.

(a) Absolute maximum: (b) Local maximum:

(c) Absolute minimum: (d) Local minimum:

Theorem 4(The Extreme Value Theorem,極值定理, page 278). If f is continuous on a closed interval [a, b], then f attains an absolute maximum value f (c) and an absolute minimum value f (d) at some numbers c and d in [a, b].





Example 5. Give examples that if f is not continuous, or f is continuous on (a, b), the Extreme Value Theorem does not hold. Plot a continuous function that it attains maximum values and minimum values at more than one number.

x

x x

y y

y

4.1 Maximum and Minimum Values goo.gl/8FNtRj 3

Theorem 6 (Fermat’s Theorem費馬定理, page 279). If f has a local maximum or minimum at c, and if f′(c) exists, then f′(c) = 0.

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Proof. Here we prove the local maximum case. Since f (c) ≥ f (x) if x is sufficiently close to c, this implies that if h is sufficiently close to 0, with h being positive or negative, then f (c) ≥ f (c + h), or equivalently, f (c + h) − f (c) ≤ 0.

If h > 0, we have f(c+h)−f(c)_{h ≤ 0. Since f}′_{(c) exists, we get}

f′(c) = lim

h→0

f (c + h) − f (c)

h =

If h < 0, we have f(c+h)−f(c)_{h ≥ 0. Since f}′_{(c) exists, we get}

f′(c) = lim h_→0 f (c + h) − f (c) h = Hence f′_{(c) = 0.} TIa3VmnB5ek

Definition 7 (page 280). A critical number (臨界點) of a function f is a number c in the domain of f such that f′_{(c) = 0 or f}′_{(c) does not exist.}



Example 8 (page 280). Find the critical numbers of f (x) = x35(4 − x) = 4x 3 5 − x 8 5. Solution. We compute f′(x) =

Therefore the critical numbers are .

Theorem 9 (page 280). If f has a local maximum or minimum at c, then c is a critical number of f .



The Closed Interval Method (page 280). To find the absolute maximum and minimum values of a piecewise continuous function on a closed interval [a, b]:

(1) Find the values of f at the critical numbers of f in (a, b).

(2) Find the values of f at the endpoints of the interval, that is, f (a) and f (b).

(3) The largest and smallest of the values from (1) and (2) are absolute maximum value and absolute minimum value, respectively.

4.2

The Mean Value Theorem, page 287

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Question 1. A highway from Taipei to Kaohsiung is 330 km and the speed limit is 110 km/h. Man A drove the car on the high way from Taipei at 9 : 00AM to Kaohsiung at 11 : 59 AM. Did he exceed the speed limit?

Theorem 2(Rolle’s Theorem, page 287). Let f be a function that satisfies the following three hypotheses:

(1) f is continuous on the closed interval [a, b]. (2) f is differentiable on the open interval (a, b). (3) f (a) = f (b).

Then there is a number c in (a, b) such that f′(c) = 0.

x y

Figure 1: Rolle’s Theorem.

Proof. There are three cases.

(I) f (x) = k, a constant. We have f′(x) = 0, so the number c can be taken to be any number in (a, b).

(II) f (x) > f (a) for some x in (a, b). By the , f has a maximum somewhere in [a, b]. Since f (a) = f (b), it must attain this maximum value at a number c in the open interval (a, b). Then f has a at c, and f is differentiable at c. By , we know f′(c) = 0.

(III) f (x) < f (a) for some x in (a, b). By the , f has a minimum value in [a, b], and since f (a) = f (b), it attains this local minimum value at a number c ∈ (a, b). By , f′_{(c) = 0.}





4.2 The Mean Value Theorem goo.gl/ZUU6e5 5

Example 3. Give examples that each condition in Rolle’s Theorem is required.

x6lRgE7n8o4 Solution. x x x y y y

Figure 2: Study Rolle’s Theorem.

Example 4 (page 287). Prove that x3_{+ x − 1 = 0 has exactly one real root.} Solution.

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Theorem 5 (The Mean Value Theorem, 平均值定理, page 288). Let f be a function that satisfies the following hypotheses:

(1) f is continuous on the closed interval [a, b]. (2) f is differentiable on the open interval (a, b). Then there is a number c in (a, b) such that

f′(c) = f (b) − f (a)

b − a or equivalently, f (b) − f (a) = f

′_{(c)(b − a).}

x y

Proof of Mean Value Theorem. Define a new function

h(x) =

We will verify that h(x) satisfies the three hypotheses of Rolle’s Theorem.

(1) The function h is continuous on [a, b]: It is the sum of f and a first-degree polynomial, both of which are continuous.

(2) The function h is differentiable on (a, b): Both f and the first-degree polynomial are differentiable. In fact, we have

h′(x) =

(3) h(a) = h(b) = 0:

h(a) =

h(b) =

By _{, there is a number c ∈ (a, b) such that h}′(c) = 0. Therefore,





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Theorem 6. If f′_{(x) = 0 for all x in an interval (a, b), then f is constant on (a, b).}

Proof. Let x1 and x2 be any two numbers in (a, b) with x1 < x2. Since f is differentiable

on (a, b), it must be differentiable on (x1, x2) and continuous on [x1, x2]. By applying the

to f on the interval [x1, x2], we get a number c such that x1 < c < x2

and

Therefore f has the same value at any two numbers x1 and x2 in (a, b). So f (x) is constant

on (a, b).

4.2 The Mean Value Theorem goo.gl/ZUU6e5 7

Corollary 7. If f′(x) = g′_{(x) for all x in an interval (a, b), then f − g is constant on (a, b);} that is, f (x) = g(x) + c where c is a constant.

Proof. Let F (x)



|x| 與 g(x) = 1,它在 x ∈ (−1, 1) 當中並不滿足推論當中的條件。

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Example 8. Show that tanx₂ − tany₂ 

≥ |x−y|₂ for any x, y ∈ (−π, π).

Solution. _{If x = y, the inequality holds. If x 6= y, without loss of generality, we assume} −π < x < y < π. Consider the function f (t) = tan t

2, then

• f (t) is .

• f (t) is .

By the _{, there is a number c ∈ (x, y) such that f (x) − f (y) = f}′_{(c)(x −} y), which implies |f (x) − f (y)| = |f′(c)||x − y|. Since f′(t) = _{, we have |f}′_{(c)| =}

. So |f (x) − f (y)| ≥ 12|x − y|, which means

  tan x 2 − tan y 2    ≥ |x − y| 2 . biygd5TQf38

Theorem 9 (Cauchy’s Mean Value Theorem, (柯西均值定理) Appendix F, A45). Suppose that the functions f and g are continuous on [a, b] and differentiable on (a, b), and g′_{(x) 6= 0} for all x in (a, b). Then there is a number c ∈ (a, b) such that

f′(c) g′_(c) =

f (b) − f (a) g(b) − g(a).

Proof. The key point is to find a new function F (x) and apply the Mean Value Theorem.

4.4

Indeterminate Forms and l’Hospital’s Rule, page

304

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In this section, we want to introduce a new method to deal with the limit such as

lim x_→∞ ln x x − 1, or xlim_→∞ x2 ex. Definition 1 (page 304–305).

(1) If we have a limit of the form lim

x_→a f(x)

g(x), where both f (x) → 0 and g(x) → 0 as x → a,

it is called an indeterminate form of type 0₀ (零分之零的不定型). (2) If we have a limit of the form lim

x_→a f(x)

g(x), where both f (x) → ∞ (or −∞) and g(x) → ∞

(or −∞) as x → a, it is called an indeterminate form of type ∞_∞ (無限大分之無限大的不 定型).

L’ Hospital’s Rule(page 305). Suppose f and g are differentiable and g′_{(x) 6= 0 on an open}

interval I that contains a (except possibly at a). Suppose that a limit has an indeterminate form of type 0₀ or ∞_∞. Then

lim x→a f (x) g(x) = limx→a f′(x) g′_(x)

if the limit on the right side exists (or is ∞ or −∞).







aBKa8sOdzCM (a) Find lim

t→0+

t_−ln(1+t) t2 .

(b) Use (a) to find lim

t→0+ √

t_−ln(1+t) t .

4.4 Indeterminate Forms and l’Hospital’s Rule goo.gl/QwjMrw 9

Indeterminate Products, page 308

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Definition 3 (page 305). If we have a limit of the form lim

x_→af (x)g(x), where limx_→af (x) = 0

and lim

x_→ag(x) = ∞ (or −∞) as x → a, it is called an indeterminate form of type 0 · ∞. (零乘

以無限大的不定型)

We can deal with it by writing the product f g as a quotient:

f g = f

1/g or f g = g 1/f,

and this converts the given limit into an indeterminate form of type 0₀ or ∞_∞. Example 4 (page 308). Evaluate lim

x→0+x ln x. Solution.



Indeterminate Differences, page 309

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Definition 5 (page 305). If lim

x_→af (x) = ∞ and limx_→ag(x) = ∞, then the limit

lim

x→a(f (x) − g(x))

is called an indeterminate form of type ∞ − ∞ (無限大減無限大的不定型).

We can try to convert the difference into a quotient (for instance, by using a common denominator (通分), or rationalization (有理化), or factoring out a common factor (提公因式)) so that we have an indeterminate form of type 0₀ or ∞_∞.

Example 6. Find the limit lim

x→0 1

x2 −_sin12_x
. Solution.

Indeterminate Powers, page 310

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Definition 7 (page 310). Several indeterminate forms arise from the limit

lim

x_→a(f (x)) g(x)

(1) lim

x→af (x) = 0 and limx→ag(x) = 0: type 0 0_.

(2) lim

x_→af (x) = ∞ and limx_→ag(x) = 0: type ∞ 0_.

(3) lim

x_→af (x) = 1 and limx_→ag(x) = ±∞: type 1 ∞_.

Each of these three cases can be treated either by taking the natural logarithm: let y = (f (x))g(x)_{, then ln y = g(x) ln f (x) or by writing the function as an exponential: (f (x))}g(x)₌

eg(x) ln f (x). In either method we are led to the indeterminate product g(x) ln f (x), which is of type 0 · ∞.

Example 8 (page 310). Find lim

x→0+x

x_.

Solution.

4.4 Indeterminate Forms and l’Hospital’s Rule goo.gl/QwjMrw 11

幾個使用

_{l’Hospital Rule}

的經驗

• 避免鬼打牆;像是 _{sin x, cos x (as x → ∞)}或是 sin1_x, cos_x1,_x1,_{ln x}1 _{(as x → 0)}。

lim x→0 x2_sin1 x sin x (0 0),L = lim x→0 2x sin1_x+ x2_cos1 x(− 1 x2) cos x = limx→0 2x sin1_{x− cosx}1 cos x . . .? lim x→0+x ln x = lim_x→0+ x 1 ln x (0 0),L ′ = lim x→0+ 1 −(ln x)1 2 ·_x1 = lim x→0+−x(ln x) 2_{. . . 鬼打牆} • 分子分母適時地整理、 重新分配, 或是變數變換, 有助於計算。 lim x→0+ 1 x2_e1x = lim x→0+ e−x1 x2 (0 0),L ′ = lim x→0+ e−x1 · 1 x2 2x = limx→0+ e−1x 2x3 . . .鬼打牆 lim x→0+ 1 x2_e1x = lim x→0+ 1 x2 e1x (∞ ∞),L = lim x→0+ −2x13 e1x · −1 x2
= lim x→0+ 2 xex1 . . . 情況變好, 再用一次 lim x→0+ 1 x2_e1x lim x→0 tan x − x x − sin x lim x→0+ tan 3x √ 1 − cos 2x • l’Hospital Rule 某種程度是“大絕”, 但並非萬能。 lim x_→∞ (sin x)ex (x + sin x)e2x

• 記得其他求極限的方法,像是 Squeeze Theorem, definition of derivative, 也很好用。(之後還

會介紹用積分方法求極限,下學期會介紹使用泰勒展式法求極限。) lim x→0 sin x x = xlim_→∞x sin 1 x = xlim_→∞ sin x x = xlim→0x sin 1 x = lim x_→∞  1 + 1 x x = lim x→0(1 + x) 1 x = lim x→0 ln(1 + x) x = lim x_→0  1 + 1 x x = lim x→∞(1 + x) 1 x = lim x→∞ ln(1 + x) x =

Appendix

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Proof of l’Hospital’s Rule (Appendix A46)

We are assuming that lim

x_→af (x) = 0 and limx_→ag(x) = 0. Let limx_→a f′ (x) g′ (x) = L. Define F (x) = ( f (x) if x 6= a 0 if x = 1 , G(x) = ( g(x) if x 6= a 0 if x = 1 .

Then both F and G are continuous on I since f and g are continuous on_{{x ∈ I|x 6= a} and}

lim

x_→aF (x) = limx_→af (x) = 0 = F (a), xlim_→aG(x) = limx_→ag(x) = 0 = G(a).

Furthermore, F and G are differentiable on (a, x) (or (x, a)) since F′ = f′ and G′ = g′. Since G′ _{6= 0, by the Cauchy’s Mean Value Theorem, there is a number y such that a < y < x (or} x < y < a) and F′(y) G′_(y) = F (x) − F (a) G(x) − G(a) = F (x) G(x). Hence lim x→a+ f (x) g(x) = limx→a+ F (x) G(x) = limy→a+ F′(y)

G′_(y) = limy→a+ f′(y) g′_(y) = L,  and lim x→a− f (x) g(x) = limx→a− F (x) G(x) = limy→a− F′(y)

G′_(y) = limy→a− f′(y) g′_(y) = L.  Therefore, lim x→a f (x) g(x) = L. This proves l’Hospital’s Rule for the case where a is finite.

If a is infinite, we let t = _x1_{. Then t → 0}+ _{as x → ∞, so we have}

lim x_→∞ f (x) g(x) = limt→0+ f (1_t) g(1_t) L = lim t→0+ f′₍1 t) · − 1 t2
g′₍1 t) · − 1 t2
= lim t→0+ f′(1_t) g′₍1 t) = lim x_→∞ f′(x) g′_(x).

4.4 Indeterminate Forms and l’Hospital’s Rule goo.gl/QwjMrw 13

L’Hospital Rule

與極限法則的合併使用

回想極限的四則運算法則與羅必達法則:

Limit Laws (page 99). Suppose that lim

x_→ap(x) and limx_→aq(x) exist. Then

lim x_→a p(x) q(x) = lim x→ap(x) lim x→aq(x) if lim x_→aq(x) 6= 0.

L’ Hospital’s Rule (page 302). Suppose f and g are differentiable and g′_{(x) 6= 0 on an open} interval I that contains a (except possibly at a). Suppose that

lim

x→af (x) = 0 and xlim→ag(x) = 0

or that

lim

x→af (x) = ±∞ and xlim→ag(x) = ±∞

(In other words, we have an indeterminate form of type 0₀ or ∞_∞.) Then

lim x→a f (x) g(x) = limx→a f′(x) g′_(x)

if the limit on the right side exists (or is ∞ or −∞).

這兩個定理可以搭配起來靈活運用, 比方說:

Theorem 9. Suppose that p(x), q(x) satisfy the assumptions of Limit Laws with lim

x_→ap(x) 6= 0,

and suppose f (x), g(x) satisfy the assumptions of L’Hospital Rule. Then

lim x→a p(x)f (x) q(x)g(x) = limx→a p(x) q(x) · limx→a f′(x) g′_(x). 上述定理要強調的是說: 雖然分子 p(x)f (x) 整體看取極限是 0, 分母 q(x)g(x) 整體看取極限是 0, 於是 p(x)f (x) q(x)g(x) 是 00 的不定型, 但是當分子與分母各自只有一部分是不定型 (只有 f g 是 00), 而 p, q 具有非零的極限,那麼就可以把p, q 的極限抽出來,考慮剩下的不定型之極限, 再相乘。 理由很簡單, 因為 lim x_→a f(x) g(x) = lim_x→a f′ (x) g′ (x) = L,所以下式的第一個等式合法 (乘法法則) lim x→a p(x)f (x) q(x)g(x) = limx→a p(x) q(x)· limx→a f (x) g(x) = limx→a p(x) q(x) · limx→a f′_(x) g′_(x). 定理一就是大大化簡計算的一個方法, 因為有很多人一看到 0 0 就不加思索的地上下微分(典型 的“看見黑影就開槍”), 雖然還是可以算出答案, 但是過程中可能會添上很多計算上的麻煩。

4.3

How Derivatives Affect the Shape of a Graph,

page 293

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Increasing/Decreasing Test (page 293).

(a) If f′(x) > 0 on an interval, then f is increasing on that interval. (b) If f′_{(x) < 0 on an interval, then f is decreasing on that interval.}

Proof.

(a) Let x1 < x2. By the , there is c ∈ (x1, x2) such that

.

(b) Let x1 < x2. By the , there is c ∈ (x1, x2) such that

.

Example 1. Find where the function f (x) = 3x4_{− 4x}3_{− 12x}2+ 5 is increasing and where it is decreasing.

Solution. We compute f′_{(x) =}

Solutions of f′_{(x) = 0 are . Hence}

f (x) is increasing on ; f (x) is decreasing on .

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The First Derivative Test(page 294). Suppose that c is a critical number of a continuous function f .

(a) If f′ changes from positive to negative at c, then f has a local maximum at c. (b) If f′ changes from negative to positive at c, then f has a local minimum at c.

(c) If f′ _{does not change sign at c (for example, if f}′ _{is positive on both side of c or negative}

on both sides), then f has no local maximum or minimum at c.

x x x y y y

4.3 How Derivatives Affect the Shape of a Graph goo.gl/11C7Ku 15

Example 2. Find the local minimum and maximum values of the function f (x) = 3x4 ₋ 4x3_{− 12x}2+ 5 in Example 1.

Solution.

x ₋₁ 0 2

f 0 5 ₋₂₇

f′

Hence f has local maximum ; f has local minimum .

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Definition 3 (page 296). If the graph f lies above all of it tangents on an interval I, then it is called concave upward (凹口朝上) on I. If the graph f lies below all of it tangents on an interval I, then it is called concave downward (凹口朝下) on I.



x x

y y

Figure 2: Concave upward and concave downward.

Concavity Test (page 296).

(a) If f′′_{(x) > 0 for all x in I, then the graph of f is concave upward on I.}

(b) If f′′_{(x) < 0 for all x in I, then the graph of f is concave downward on I.}

Proof of (a). Since f′′_{(x) > 0 in I, we know that f}′_{(x) is increasing in I. Given x0 ∈ I, the}

tangent line equation to the graph of f (x) at (x0, f (x0)) is

y − f (x0) = f′(x0)(x − x0) ⇒ y = f′(x0)(x − x0) + f (x0).

We will show that f (x) ≥ f′(x0)(x − x0) + f (x0) for all x ∈ I.

Consider the function

F (x) = f (x) − f′(x0)(x − x0) − f (x0) for x ∈ I.

First, we know that F (x0) = 0. Next, we compute F′(x) = f′(x) − f′(x0), which implies

F′(x0) = f′(x0) − f′(x0) = 0. Since F′(x) < 0 for x < x0 and F′(x) > 0 for x > x0, we know

that F (x0) is a local (and hence absolute) minimum at x = x0 in I. That means F (x) ≥ 0

Definition 4 (page 297). A point P on a curve y = f (x) is called an inflection point (反曲 點) if f is continuous there and the curve changes from concave upward to concave downward or from concave downward to concave upward at P .

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x x

y y

Figure 3: Inflection points.

Example 5. Find the concave upward and downward intervals, and inflection points of the function f (x) = 3x4_{− 4x}3_{− 12x}2+ 5 in Example 1. Sketch the graph of f .

Solution. We compute f′′(x) = . So x ₋₁ x1 0 x2 2 f 0 f (x1) 5 f (x2) −27 f′ ₋ 0 + 0 ₋ 0 + f′′

The points of inflections are . f is concave upward on . f is concave downward on .

x y

4.3 How Derivatives Affect the Shape of a Graph goo.gl/11C7Ku 17

The Second Derivative Test (page 297). Suppose f′′ is continuous near c.

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(a) If f′(c) = 0 and f′′(c) > 0, then f has a local minimum at c. (b) If f′(c) = 0 and f′′(c) < 0, then f has a local maximum at c. Example 6. Show that f (x) = sin x_x is decreasing on (0, π₂).

Solution.



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Example 7. Classify all cubic functions f (x) = ax3+ bx2+ cx + d. Solution.

4.5

Summary of Curve Sketching, page 315

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Definition 1 (page 320). If

lim

x_→∞(f (x) − (mx + b)) = 0,

where m 6= 0, then the line y = mx + b is called a slant asymptote (斜漸近線). Proposition 2. The graph of f (x) has a slant asymptote if and only if

lim x→∞ f (x) x = m 6= 0 and xlim→∞(f (x) − mx) = b. Proof. When x > 0, f (x) x = f (x) − (mx + b) x + m + b x ⇒ limx_→∞ f (x) x = 0 + m + 0 = m. lim x→∞(f (x) − mx) = limx→∞(f (x) − (mx + b) + b) = lim x→∞(f (x) − (mx + b)) + limx→∞b = 0 + b = b. Conversely, we have lim x_→∞(f (x) − (mx + b)) = limx_→∞((f (x) − mx) − b) = lim x→∞(f (x) − mx) − limx→∞b = b − b = 0.





Guidelines for sketching a curve

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定 Domain: the set of x for which f (x) is defined.

交 Intercepts: y-intercept f (0), x-intercepts: let y = 0 and solve for x.

對 Symmetry: even function, odd function, periodic function.

漸 Asymptotes: horizontal asymptotes, vertical asymptotes, slant asymptotes.

一 Intervals of increase or decrease: use the Increasing/Decreasing test.

極 Local maximum and minimum values: find the critical numbers of f (f′(c) = 0 or f′(c) does not exist.)

二 Concavity and points of inflection: compute f′′(x) and use the Concavity Test.

4.5 Summary of Curve Sketching goo.gl/gJhoUh 19

Example 1 (page 317). Sketch the curve y = f (x) = _x2x2₋₁2 .

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Solution.

A. The domain is . B. The x- and y-intercept are both .

C. Since , the function f is . D. Since

lim

x_→±∞

2x2

x2_{− 1} = ,

the line is a . The denominator is 0 when . we compute the following limits:

lim x→1+ 2x2 x2_{− 1} = _xlim_→1− 2x2 x2_{− 1} = lim x_→−1+ 2x2 x2_{− 1} = _x→−1lim− 2x2 x2_{− 1} =

Therefore the lines and are vertical asymptotes.

E. Direct computation gives

y′ = .

Since f′(x) > 0 when and f′(x) < 0 when , f is increasing on and decreasing on .

F. The only critical number is . Since f′ changes from positive to negative at 0, f (0) = 0 is a by the First Derivative Test.

G. Direct computation gives

f′′(x) = .

We know f′′_{(x) > 0 on and f}′′_{(x) < 0 on . Thus the curve is concave}

upward on the interval and concave downward on . It has no point of inflection since .

Example 2 (page 317). Sketch the curve y = f (x) = √x2 x+1.

3W1Yt_jtLQg Solution.

A. The domain is .

B. The x- and y-intercept are both . C. Symmetry: None. D. Since lim x_→∞ x2 √ x + 1= , there is no horizontal asymptote. Since

lim

x→−1+ x2 √

x + 1 = , the line is a vertical asymptotes.

E. Direct computation gives

y′= .

We see that f′(x) = 0 when , so the only critical number is . Since f′(x) > 0 when and f′(x) < 0 when , f is increasing on and decreasing on .

F. Since f′(0) = 0 and f′ changes from negative to positive at 0, f (0) = 0 is a by the First Derivative Test.

G. Direct computation gives

f′′(x) = .

Since the numerator is always , we know f′′(x) > 0 for all x in the domain of f , which means f is concave upward on and there is no point of inflection.

4.5 Summary of Curve Sketching goo.gl/gJhoUh 21

Example 3 (page 318). Sketch the curve y = f (x) = xex_.

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Solution.

A. The domain is .

B. The x- and y-intercept are both . C. Symmetry: None.

D. Since

lim

x→∞xe

x_{= ,}

there is no horizontal asymptote. By the l’Hospital Rule, we have

lim x→−∞xe x_{= lim} x→−∞ x e−x = ,

so the is a horizontal asymptote. E. Direct computation gives

y′= .

Since f′_{(x) > 0 when and f}′_{(x) < 0 when , f is increasing on}

and decreasing on .

F. Since f′_{(−1) = 0 and f}′ _{changes from negative to positive at x = −1, f (−1) = −e}−1 is a by the First Derivative Test.

G. Direct computation gives

f′′(x) = .

Since f′′_{(x) > 0 if and f}′′_{(x) < 0 if , f is concave upward on}

and concave downward on . The inflection point is .

Example 4 (page 319). Sketch the curve y = f (x) = _{2+sin x}cos x .

wbkisiSCwIU Solution.

A. The domain is .

B. The x-intercepts are and y-intercept is .

C. Symmetry: f is neither even nor odd. Since f (x + 2π) = f (x) for all x, f is

and has period _{. Thus, the following steps we only consider 0 ≤ x ≤ 2π and then} extend the curve by translation.

D. Asymptotes: None. E. Direct computation gives

y′ = .

Thus f′_{(x) > 0 when . So f is increasing}

on and decreasing on .

F. From part E and First Derivative Test, the local minimum value is and local maximum value is .

G. Direct computation gives

f′′(x) =

Since f′′(x) > 0 if , f is concave upward on and concave downward on . The inflection point is .

4.5 Summary of Curve Sketching goo.gl/gJhoUh 23

Example 5 _{(page 319). Sketch the curve y = f (x) = ln(4 − x}2).

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Solution.

A. The domain is .

B. _{The y-intercept is f (0) = ln 4. To find the x-intercept, we set ln(4 − x}2) = 0, so we have . Therefore the x-intercepts are .

C. _{Since f (−x) = f (x), f is} and the curve is symmetric about the . D. Since lim x_→−2+ln(4 − x 2_{) = , lim} x_→2−ln(4 − x 2_{) = ,}

the lines are vertical asymptotes.

E. Direct computation gives

y′ = .

Since f′_{(x) > 0 when and f}′_{(x) < 0 when , f is increasing on}

and decreasing on .

F. The only critical number is . Since f′ _{changes from positive to negative at 0,}

f (0) = ln 4 is a by the First Derivative Test.

G. Direct computation gives

f′′(x) = .

Since f′′(x) < 0 for all x, the curve is on and has no inflection point.

Example 6 (page 320). Sketch the curve y = f (x) = _x2x₊₁3 .

aoV6RJZFQ2Y Solution.

A. The domain is .

B. The x- and y-intercept are both .

C. Since , the function f is .

D. Since x2_{+ 1 is never 0, there is no vertical asymptote. Since f (x) → ∞ as x → ∞ and}

f (x) → −∞ as x → −∞, there is no horizontal asymptote. Long division gives

f (x) = x

3

x2_{+ 1} = ,

f (x) − x = −_x2x_{+ 1} = .

So the line is a . E. Direct computation gives

y′ = .

Since f′_{(x) > 0 when , f is increasing on .}

F. Although f′_{(0) = 0, f}′ _{does not change sign at 0, so there is or}

.

G. Direct computation gives

f′′(x) = .

Since f′′_{(x) = 0 when , we set up the following chart.}

The points of inflection are .

4.7 Optimization Problems goo.gl/gNJP2d 25

4.7

Optimization Problems, page 330

Steps in solving optimization problems

1. Understand the problem: What is the unknown? What are the given quantities? What are the given conditions?

2. Draw a diagram: In most problems it is useful to draw a diagram and identify the given and required quantities on the diagram.

3. Introduce notation: Assign a symbol to the quantity that is to be maximized or minimized (call it Q for now). Also select symbols a, b, c, . . . , x, y for other known quantities and label the diagram with these symbols.

4. Express Q in terms of some of the other symbols.

5. If Q has been expressed as a function of more than one variable, use the given informa-tion to find relainforma-tionships among these variables. Then use these equainforma-tions to eliminate all but one of the variables. Thus we get Q = f (x).

6. Use the methods of Section 4.1 and 4.3 to find the absolute maximum or minimum value of f .

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Example 1 (Snell’s Law, 斯乃爾定律, page 268). Let v1 be the velocity of light in air and v2

the velocity of light in water. According to Fermat’s Principle, a ray of light will travel from a point A in the air to a point B in the water by a path ACB that minimizes the time taken. Show that sin θ1 sin θ2 = v1 v2 ,

where θ1 (the angle of incidence) and θ2 (the angle of refraction) are known.

Example 2 (搬家俱, page 268). A steel pipe is being carried down a hallway a m wide. At

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the end of the hall there is a right-angled turn into a narrower hallway b m wide. What is the length of the longest pipe that can be carried horizontally around the corner?

Solution.

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Example 3 (看畫、 教室的風水, page 269). A painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How far from the wall should the observer stand to get the best view? (In other words, where should the observer stand so as to maximize the angle θ subtended at his eye by the painting?)

4.7 Optimization Problems goo.gl/gNJP2d 27

Example 4. A right circular cone is inscribed in a sphere of radius r. Find the largest

hbKrEiTBiqA

possible volume of such a cone. In this case, what is the height and radius of the cone?

Solution.

Example 5 (折紙問題, page 269). The upper right-hand corner of a piece of paper, 30 cm by 20 cm, is folded over to the bottom edge. How would you fold it so as to minimize the length

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of the fold? In other words, how would you choose x to minimize y? Solution.

Example 6.

N-KwPosBmfE (a) Find the point (denote P ) on the line y = x

2 _{that is closest to the point Q(3, 0).}

(b) Show that the line P Q is orthogonal to the tangent line of y = x2 at P . Solution.

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4.9 Antiderivative goo.gl/q3mCPB 29

4.9

Antiderivative, page 350

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Definition 1 (page 350). A function F is called an antiderivative (反導函數) of f on an interval I if F′_{(x) = f (x) for all x in I.}

Theorem 2 (page 351). If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I is

F (x) + C,

where C is an arbitrary constant.

Proof. If F and G are any two antiderivative of f , then F′_{(x) = f (x) = G}′_{(x). Form the}

corollary of the Mean Value Theorem (Section 4.2 Corollary 8), we know G(x) − F (x) = C, where C is a constant. So G(x) = F (x) + C.

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This is a table of antidifferentiation formulas. We use the notation F′(x) = f (x) and G′_{(x) = g(x).}

Function Particular antiderivative Function Particular antiderivative cf (x) cF (x) sec2_{x tan x}

f (x) + g(x) F (x) + G(x) sec x tan x sec x xn_{(n 6= −1)} xn+1 n+1 √_1−x1 2 sin−1x 1 x ln |x| 1 1+x2 tan−1x ex _ex _{cosh x sinh x}

cos x sin x sinh x cosh x sin x _{− cos x}

Example 3. Find the most general antiderivative of the function. (Let F (x) is the antideriva-tive of the function f (x).)

(1) f (x) = e2 F (x) = (2) f (x) = x(2 − x)2 F (x) = (3) f (x) = xπ_{− x}3.14 _{F (x) =} (4) f (x) = 2+x_1+x22 F (x) =

Example 4. Find f (x).

nb2j0fedk9w (1) f′(x) = 2 cos x + sec

2_{x, −}π 2 < x < π 2, f ( π 3) = 4. (2) f′′(x) = 2ex_{+ 3 sin x, f (0) = 0, f (π) = 0.} Solution.