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Chapter 3

Applications of Differentiation (微分的應用)

Hung-Yuan Fan (范洪源)

Department of Mathematics, National Taiwan Normal University, Taiwan

Fall 2018

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本章預定授課範圍

3.1 Extrema on an Interval

3.2 Rolle’s Theorem and the Mean Value Theorem 3.3 Increasing and Decreasing Functions and the First Derivative Test

3.4 Concavity and the Second Derivative Test 3.5 Limits at Infinity

3.7 Differentials

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Section 3.1

FExtrema on an Interval

(區間上的極值)

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Def. of Relative Extrema (相對極值)

Let f be a real-valued function defined on D⊆ R with c ∈ D.

(1) f has a relative maximum (相對極大值; rel. max.) at the point (c, f(c)) if ∃ open interval I s.t. f(x) ≤ f(c) ∀ x ∈ I.

(2) f has a relative minimum (相對極小值; rel. min.) at the point (c, f(c)) if∃ open interval I s.t. f(x) ≥ f(c) ∀ x ∈ I.

(3) Rel. max. and rel. min. are called the relative extrema off.

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Some Questions

Let f be a real-valued function defined on D⊆ R.

Does f always have a relative extremum on D?

How to find the relative extrema of f?

What is f(c) if f(c) is a relative extremum?

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Example 1 (極值發生處的導數)

(a) The rational function

f(x) = 9(x2− 3)

x3 with f(x) = 9(9− x2) x4

has a rel. max. at the point (3, 2), and f(3) = 0 in this case.

(b) The function f(x) =|x| has a rel. min. value f(0) = 0 at the origin (0, 0), but f (0)@. (Why?)

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Example 1 的示意圖 (承上頁)

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Def. of Critical Numbers (臨界數)

Iff(c) = 0 orf (c)@ for some c∈ dom(f), then the value c is called a critical number of f.

Thm 3.2 (發生相對極值的必要條件)

If f has a relative extremum at the point (c, f(c)) with c∈ D = dom(f), then

f(c) = 0 or f(c) @, i.e. x = c must be a critical number of f.

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Example (Thm 3.2 的反例)

For f(x) = x3, x = 0 is the only critical number of f, since f(x) = 3x2 = 0⇐⇒ x = 0. But, f(0) = 0 is NOT a relative extremum of f.

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Proof of Thm 3.2

Suppose that f(c) is a relative extremum with f (c)̸= 0 ∃.

Without loss of generality, we may assume that f(c) > 0.

For ε = f (c)

2 > 0, ∃ δ > 0 s.t. if 0 < |x − c| < δ, then f(x)− f(c)

x− c − f (c) < f(c)

2 or f(x)− f(c)

x− c > f (c) 2 > 0.

Thus, we know that

f(x) > f(c) ∀ x ∈ (c, c+δ) and f(x) < f(c) ∀ x ∈ (c−δ, c).

This contradicts to the assumption and hence completes the proof.

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Def. of Absolute Extrema (絕對極值)

Let f be a real-valued function defined on D⊆ R with c ∈ D.

(1) f(c) is the absolute maximum (絕對極大值; abs. max.) of f on D if f(x)≤ f(c) ∀ x ∈ D.

(2) f(c) is the absolute minimum (絕對極小值; abs. min.) of f on D if f(x)≥ f(c) ∀ x ∈ D.

(3) Abs. max. and abs. min. are called the absolute extrema off on D.

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Thm 3.1 (Extreme Value Theorem; E.V.T. 極值定理)

If f isconti. on I = [a, b], then ∃ c1, c2∈ I s.t.

f(c1)≤ f(x) ≤ f(c2) ∀ x ∈ I,

i.e., f(c1) is the abs. min. value of f on I and f(c2) is the abs. max.

value of f on I, respectively.

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How to find the points c

1

and c

2

in Thm 3.1?

Step 1 find all critical numbers c1, c2, . . . , ck of f in the open interval (a, b), where k∈ N.

Step 2 evaluate f(a), f(b) and f(ci) for i = 1, 2, . . . , k.

Step 3 compare the function values obtained in Step 2.

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Section 3.2

Rolle’s Theorem and the Mean Value Theorem

(洛爾定理與均值定理)

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Thm 3.3 (Rolle’s Theorem)

Suppose that f isconti. on [a, b]anddiff. on (a, b). If f(a) = f(b), then∃ c ∈ (a, b) s.t. f(c) = 0.

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Proof of Thm 3.3

Since f is conti. on [a, b] and f(a) = f(b), it follows from E.V.T. that ∃ c ∈ (a, b) s.t.f(c) is a relative extremum.

Otherwise, f must be a constant function on [a, b] and hence f (x) = 0 ∀ x ∈ (a, b).

Next, we shall claim that f(c) = 0. If not, say f(c) > 0, then it fowws from the ε-δ Def. of a limit that∃ δ > 0 s.t.

f(x) < f(c) for x∈ (c − δ, c) and f(c) < f(x) for x ∈ (c, c + δ). Thus, f(c) is NOT a relative extremum and this gives a contradiction!

Similarly, we can deduce that f(c) < 0 =⇒ f(c) is NOT a relative extremum. Consequently, we must have f (c) = 0 for some c∈ (a, b).

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Thm 3.4 (Mean Value Theorem; M.V.T. 均值定理)

If f is conti. on [a, b] and diff. on (a, b), then∃ c ∈ (a, b) s.t.

f (c) = f(b)− f(a)

b− a or f(b)− f(a) = f(c)(b− a).

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Proof of Thm 3.4

Let g : [a, b]→ R be a function defined by g(x) = f(x)−f(b)− f(a)

b− a (x− a) − f(a) ∀ x ∈ [a, b].

Since f is conti. on [a, b] and diff. on (a, b), we know that g is conti. on [a, b], diff. on (a, b) and g(a) = 0 = g(b).

Since g(x) = f(x)−f(b)− f(a)

b− a ∀ x ∈ (a, b), it follows from Thm 3.3 (Rolle’s Thm) that∃ c ∈ (a, b) s.t.

0 = g(c) = f (c)−f(b)− f(a) b− a or, equivalently, we prove that ∃ c ∈ (a, b) s.t.

f (c) = f(b)− f(a) b− a .

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Example (M.V.T. 的補充題)

For any a, b∈ R, prove the following inequality

| sin a − sin b| ≤ |a − b|.

Proof: Let a, b

∈ R. Without loss of generality, we may assume that a < b. Since f(x) = sin x is conti. on [a, b] and diff. on (a, b), it follows from M.V.T. that∃ c ∈ (a, b) s.t.

sin b− sin a = f (c)· (b − a) = (cos c) · (b − a).

So, we immediately see that

| sin a − sin b| = | cos c| · |a − b| ≤ |a − b|

because| cos c| ≤ 1, and hence this completes the proof.

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Section 3.3

Increasing and Decreasing Functions

and the First Derivative Test

(遞增、遞減函數與一階導數測試)

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Def (單調函數的定義)

Let f be a real-valued function defined on an interval I.

(1) f is increasing (遞增; ↗) on I if f(x1) < f(x2) whenever x1, x2∈ I with x1 < x2.

(2) f is decreasing (遞減;↘) on I if f(x1) > f(x2) whenever x1, x2∈ I with x1 < x2.

(3) The increasing or decreasing functions are called monotonic functions (單調函數).

Note: Monotonic functions are one-to-one, but one-to-one

functions are NOT necessarily monotonic!

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單調函數的示意圖

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Thm 3.5 (單調函數的充分條件)

(1) f (x) > 0 ∀ x ∈ (a, b) =⇒ fis increasing (↗)on [a, b].

(2) f (x) < 0 ∀ x ∈ (a, b) =⇒ fis decreasing (↘)on [a, b].

(3) f (x) = 0 ∀ x ∈ (a, b) =⇒ fis constant on [a, b].

Example (Thm 3.5 的反例)

The function f(x) = x1/3 isincreasing on R, but its first derivative satisfiesf (x) = 3x12/3 > 0 ∀ x ∈ R\{0}.

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Proof of Thm 3.5

For any x1, x2 ∈ (a, b) withx1 < x2, since f is conti. on [x1, x2] and diff. on (x1, x2), it follows from M.V.T. that ∃ c ∈ (x1, x2) s.t.

f(x2)− f(x1) = f(c)(x2− x1).

(1) If f (x) > 0 ∀ x ∈ (a, b), thenf(c) > 0 and hence f(x2)− f(x1) > 0 orf(x2) > f(x1). This implies that f is increasing (↗) on (a, b).

(2) If f (x) < 0 ∀ x ∈ (a, b), thenf(c) < 0 and hence f(x2)− f(x1) < 0 orf(x2) < f(x1). This implies that f is decreasing (↘) on (a, b).

(3) If f (x) = 0 ∀ x ∈ (a, b), then f(x2)− f(x1) = 0 or f(x1) = f(x2) ∀ x1, x2 ∈ (a, b), i.e., f is constant on (a, b).

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Example 1 的示意圖

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Thm 3.6 (First Derivative Test; 一階導數測試)

Let f be diff. on an open interval except possibly at c. If x = c is a critical number of f, then

(1) sign of f changes from (+) to (−) at c =⇒ f(c) is a rel. max.

value of f.

(2) sign of f changes from (−) to (+) at c =⇒ f(c) is a rel. min.

value of f.

(3) sign of f does notchange on both sides of c =⇒ f(c) isnota relative extremum.

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示意圖 (承上頁)

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Example 3 的示意圖

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So, x =−1 and x = 1 are critical numbers of f.

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Example 4 的示意圖

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Section 3.4

Concavity and the Second Derivative Test

(凹性與二階導數測試)

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Def. of Concavity (凹性)

Let f be diff. on an open interval I = (a, b).

(1) The graph of f is concave upward (凹向上; C.U.) on I if its first derivative f is ↗ on I.

(2) The graph of f is concave downward (凹向下; C.D.) on I if its first derivative f is ↘ on I.

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凹性的示意圖 (承上頁)

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Thm 3.7 (Test for Concavity; 凹性測試法)

Suppose that f′′(x) exists on an open interval I.

(1) f ′′(x) > 0 ∀ x ∈ I =⇒ the graph of f is C.U. on I.

(2) f ′′(x) < 0 ∀ x ∈ I =⇒ the graph of f is C.D. on I.

pf: It follows immediately from the Def. of Concavity and

f′′(x) =dxd[f(x)] that

(1) f ′′(x) > 0 ∀ x ∈ I =⇒ f is increasing on I =⇒ the graph of f is C.U. on I.

(2) f ′′(x) < 0 ∀ x ∈ I =⇒ f is decreasing on I =⇒ the graph of f is C.D. on I.

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Example 1 的示意圖

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Def. of Points of Inflection (反曲點; P.I.)

Let f be conti. on an open interval containing c. If the graph of f

1 has a (vertical) tangent line at (c, f(c)), and

2 its concavity changes on both sides of c,

then (c, f(c)) is called a point of inflection of the graph of f.

(函數圖形凹性改變的轉折點即為反曲點!)

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反曲點的示意圖 (1/2)

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反曲點的示意圖 (2/2)

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Thm 3.8 (反曲點的必要條件)

Suppose thatf ′′(x) exists on an open interval containing c. If (c, f(c)) is a point of inflection of the graph of f, then

f ′′(c) = 0 or f′′(c)@.

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Proof of Thm 3.8

Without loss of generality, we assume that f ′′(c) > 0 ∃. Since f ′′ exists at c, we know that,

xlim→c

f(x)− f(c)

x− c = f ′′(c).

Thus, for ε = f′′2(c) > 0, ∃ δ > 0 s.t. if 0 < |x − c| < δ, then

f(x)− f(c)

x− c − f′′(c) <f′′(c)

2 or f(x)− f(c)

x− c >f′′(c) 2 > 0.

Thenf (x) < f (c) for x∈ (c − δ, c) and f(c) < f (x) for x∈ (c, c + δ). Thus, it follows from the Def. of concavity that the graph of f is C.U. on I = (c− δ, c + δ). This contradicts to our assumption that (c, f(c)) is a point of inflection of f, and hence we complete the proof.

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Thm 3.8 的反例

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Example (Thm 3.8 的另一個反例)

Consider f(x) =√3

x2= x2/3 ∀ x ∈ R. Then its first and second derivatives are given by

f(x) = 2

3x−1/3 and f′′(x) = −2

9 x−4/3< 0

for all x̸= 0. In this case, we see that f ′′(0) @ and the origin (0, 0) is NOT a point of inflection!

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示意圖 (承上例)

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Example 3 的示意圖

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Thm 3.9 (Second Derivative Test; 二階導數測試)

Suppose thatf (c) = 0and f ′′ ∃ on an open interval containing c.

(1) f ′′(c) > 0=⇒ f(c) is a rel. min. value.

(2) f ′′(c) < 0=⇒ f(c) is a rel. max. value.

(3) f ′′(c) = 0=⇒ the test is inconclusive.

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示意圖 (承上頁)

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Example 4 (Thm 3.9 的例子)

Find all relative extrema of the (polynomial) function f(x) =−3x5+ 5x3.

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Example 4 的示意圖

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Section 3.5

Limits at Infinity

(無窮遠處的極限)

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Def (Limits at Infinity)

(1) lim

x→∞f(x) = L⇐⇒ ∀ ε > 0,∃ M > 0s.t. ifx > M, then

|f(x) − L| < ε.

(2) lim

x→−∞f(x) = L⇐⇒ ∀ ε > 0,∃ N < 0s.t. if x < N, then

|f(x) − L| < ε.

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示意圖 (承上頁)

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Thm 3.10 (重要的極限法則)

(1) If r > 0 is arational number andc∈ R, then

xlim→∞

c

xr = 0 = lim

x→−∞

c xr. (2) lim

x→∞e−x= 0 = lim

x→−∞ex.

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Proof of (1) in Thm 3.10

Let ε > 0 be given arbitrarily. Choose M, N∈ R with M >|c|

ε

1/r

> 0 and N <−|c|

ε

1/r

< 0.

Thus we have the following inequalities

|c|

Mr < ε and |c|

(−N)r < ε. (Check!) If x > M(> 0) or x < N(< 0), then

c

xr−0 = |c|

xr < |c|

Mr < ε or c

xr−0 = |c|

|x|r = |c|

(−x)r < |c|

(−N)r < ε.

So, it follows from the Def. that

xlim→∞

c

xr = 0 = lim

x→−∞

c xr.

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Example 5 的示意圖

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Def (水平漸近線的定義)

A line y = L is callled a horizontal asymptote (水平漸近線) of the graph of f if

xlim→∞f(x) = L or lim

x→−∞f(x) = L.

 

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Infinite Limits at Infinity

Def (無窮遠處的無窮極限值)

(1) lim

x→∞f(x) =∞ (−∞)⇐⇒ ∀ M > 0 (M < 0), ∃ N > 0 s.t. if x > N, then f(x) > M (f(x) < M).

(2) lim

x→−∞f(x) =∞ (−∞)⇐⇒ ∀ M > 0 (M < 0), ∃ N < 0s.t. if x < N, then f(x) > M (f(x) < M).

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Slant Asymptotes

Def (斜漸近線的定義)

A line y = mx + b with slopem̸= 0 is called a slant (or oblique) asymptote (斜漸近線) of the graph of f if

xlim→∞[f(x)− (mx + b)] = 0 or lim

x→−∞[f(x)− (mx + b)] = 0.

Note: If p(x)/q(x) is a rational function with

deg(p) = deg(q) + 1, applying the method of long division (長除法), we obtain

p(x)

q(x) =(mx + b)+ r(x) q(x), where r(x) is a polynomial withdeg(r) < deg(q).

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Example 8 的示意圖

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Section 3.7

Differentials

(全微分或是微分)

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Def. of Differentials

Let f be diff. on an open interval I with x∈ I.

(1) The differential of x, denoted by dx, is any nonzeroreal number.

(2) The differential of y = f(x) is defined by dy = f(x)dx.

Main Question

If dx = ∆x≈ 0 is sufficiently small, how to estimate

∆y = f(x + ∆x)− f(x) = f(x + dx) − f(x) using the differential dy directly?

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Thm (利用 dy 估計 ∆y)

If dx = ∆x≈ 0 is sufficiently small, then (1) f(x + ∆x)− f(x) =∆y≈ dy= f (x)dx.

(2) f(x + dx) = f(x + ∆x)≈ f(x) + dy = f(x) + f(x)dx.

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∆y 與 dy 的示意圖 (承上頁)

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Equivalent Def. of Differentiability (可微分性的等價定義)

Let f be a real-valued function defined on D = dom(f). Then

f is diff. at x∈ D.

=⇒ lim

∆x→0

f(x + ∆x)− f(x)

∆x = f(x) ∃.

=⇒∃ a function ε1 = ε1(∆x) with lim

∆x→0ε1(∆x) = 0 s.t.

∆y = f(x + ∆x)− f(x) = f (x)· ∆x + ε1(∆x)· ∆x.

Note: In Example 2, f(x) = x

2 is diff. at any x∈ R, since we have

∆y = f(x + ∆x)− f(x) =(2x)· ∆x +ε1(∆x)· ∆x, whereε1(∆x)≡ ∆x → 0 as ∆x → 0.

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Thank you for your attention!

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