§ 4.1 Maximum and Minimum Values (極大與極小值）

§ 4.1 Maximum and Minimum Values (極大與極小值）

Def

(1) (a) A function f has an absolute maximum ^{絕對極大值}

(or global maximum ^{全區極大值}) at x = c if for all x in The number is called the of f on

(b) f has an absolute minimum (or global minimum) at x = c if for all x in

The number is called the of f on

(2) (a) A function f has a local maximum ^{局部極大值}

(or relative maximum ^{相對極大值}) at x = c if when x is near c (for all x in some open interval containing c)

The number is called the of f (b) f has a local minimum (or relative minimum) at x = c

if when x is near c

The number is called the of f maximum value

minimum value

local maximum value

local minimum value

Df

( ) ( )

f c ≥ f x D_f

( ) ( ) f c ≥ f x ( ) ( )

f c ≤ f x D_f

( ) ( ) f c ≤ f x

Df

( ) f c

( ) f c

( ) f c

( ) f c

The maximum and minimum values of af re called the extreme values ^極值of f

The absolute minimum is not a local minimum because it occurs at an endpoint absolute maximum absolute maximum value ( ) absolute minimum absolute minimum

( ) has an at , the

( ) has an at , the

( ) has a local m

val aximum at , the local maximu

(

ue (

m v

) alue ( )

x e f e

x f x

f x

f x x c

a

f x

c

f a

= = f

= =

= =

) has a at , the

( ) has a at ,

local maximum local maximum value ( ) local minimum local minimum value ( ) local m

the

( ) has a inimum at , the local minimum value ( )

x e f e

x d f d

x f

f l l

f x x

= =

= =

= =

[Ex1] The function takes on its (local and absolute) maximum value

of 1 infinitely ^無限地 many times. It also takes on its (local and absolute) minimum value of -1 infinitely many times.

[Ex2] has an absolute (and local) minimum value , and it has no maximum value.

( ) 2

f x = x = f (0) = 0

( ) cos

f x = x

If is continuous on a closed interval [ , ], then attains an absolute maximum value ( ) and an absolute minimum value ( ) at some numbers and in [ , ]

f a b f

f c f d c d a b

The Extreme Value Theorem （極值定理）

達到

such that either f c′( ) = 0 or f c′( ) does not exist ( ) ,

= f b

[ ]

If is not continuous on the closed interval , , it may not have extreme values. (as show below)

f a b

A of critical number ^臨界數 a function is a numbf er c ∈ D_f Def

f has an absolute minimum value but no maximum value.

This continuous function g has no extreme values.

[Ex]

[Ex]

0 is the critical number of ( ) since (0) does not exist.

x = f x = x f ′

3 2 2

If ( ) then ( ) 3 . ( ) 0 3 0 0 0 is a critical number of

f x x f x x f x x x

x f

′ ′

= = = ⇔ = ⇒ =

∴ =

[Ex7] Find the critical numbers of [Sol]:

3

( ) 5(4 ) f x = x − x

2 3

5 5

2 2

5 5

3 3(4 ) 5 12 8

( ) (4 ) ( 1)

5 5 5

( ) 0 12 8 0 3

2

( ) D.N.E 0

Thus, the critical numbers are = 3 and =0 2

x x x

f x x x x

x x

f x x x

f x x

x x

− − − −

′ = − + − = =

′ = ⇔ − = ⇔ =

′ ⇔ =

( ) ( )

The critical numbers of are , , '( ) 0 and ( ) D.N.E Note that the local maximum and minimum occur at these points.

f x = c x = e x = l f x = x = d f x′ [Ex]

The theorem asserts that every local maximum or minimum occur at a critical number.

But, be careful! The converse ^反敘述 is false!! That means it may happen that c is a critical number of f , but f has no local maximum or minimum at x = c.

Thm

If f has a local maximum or minimum at x = c, then c is a critical number of f .

( ) 3

f x = x

3

0 is a critical number of since (0) 0,

but there's no local maximum or minimum at 0 (0) 0 simply means that the curve

has a horizontal tangent at 0.

x f f

x

f y x

x

= ′ =

=

′ = =

= [Ex]

[Ex]

1

( ) 3

f x = x

1 3

0 is a critical number of since (0) D.N.E, but has no local maximum or minimum at 0.

Here, " (0) D.N.E" simply means that the curve has a vertical tangent at 0.

x f f

f x

f

y x x

= ′

=

′

= =

The Closed Interval Method （閉區間法）

To find the absolute maximum and minimum values of a continuous function f 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 ^端點of the interval.

3. The largest of the values from step 1 and 2 is the absolute maximum value;

the smallest of these values is the absolute minimum value.

[Ex8] Find the absolute maximum and minimum values of the function

[Sol]: Since is continuous on the closed interval , we can use the Closed Interval Method

( )

f x 1

2 , 4

⎡− ⎤

⎢ ⎥

⎣ ⎦

3 2 1

( ) 3 1 , 4

f x = x − x + − ≤ ≤2 x

1. ( ) 3 2 6 3 ( 2), ( ) 0 0 or 2 critical numbers (0) 1 , (2) 3

1 1

2. ( ) , (4) 17

2 8

3. the absolute maximum value (4) 17 the absolute minimum value (2) 3

f x x x x x f x x x

f f

f f

f f

′ = − = − ′ = ⇔ = = ←

= = −

− = =

= =

= = −

[Sol]:

[Ex9] Find the absolute maximum and minimum values of the function

1. ( ) 1 2 cos ,

1 5

( ) 0 cos or

2 3 3

( ) 2 sin 3 0

3 3 3 3

5 5 5 5

( ) 2 sin 3 6.96

3 3 3 3

2. (0) 0 ; (2 ) 2 6.28

5 5

3. the absolute maximum value ( ) 3

3 3

the absolute mi

f x x

f x x x x

f f

f f

f

π π

π π π π

π π π π

π π

π π

′ = −

′ = ⇔ = ⇒ = =

= − = − <

= − = + ≈

= = ≈

= = +

nimum value ( ) 3

3 3

f π π

= = −

( ) 2 sin , 0 2 f x = −x x ≤ ≤x π

[Ex] Find the absolute maximum and minimum of

[Sol]:

4 , when 3

2 3

1. ( ) and ( ) D.N.E.

3 2 4 , when

2

3 3

is a critical number. ( ) 0

2 2

2. ( 3) 18 , (3) 6

3. the absolute maximum ( 3) 18 the absolute minimum ( )3 0

2 x

f x f

x

f

f f

f f

⎧− <

′ = ⎨⎪⎪ ′

⎪ >

⎪⎩

∴ =

− = =

= − =

= =

[ ]

( ) 6 4 on 3, 3

f x = − x −

§ 4.2 The Mean Value Theorem (平均值定理）

Rolle’s Theorem （洛爾定理）

Let f be a function that satisfies the following three hypothesis:

1. f is continuous on the closed interval [a, b].

2. f is differentiable on the open interval (a, b).

3.

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

( ) f c ( ) ( ) f a = f b

( ) 0 f c′ = ( ) ( )

f a = f b

Since f is continuous on a closed interval, by the Extreme Value Theorem, f has a maximum value Somewhere in [a, b]. Because , the maximum value must occurs at a number c in (a, b).

That is, is a local maximum value.

Since f is differentiable at c by hypothesis 2,we have

[Fig.

Case1: If ( ) (a constant), then ( ) 0 ( , ) Case2: If ( ) ( ) for some

1]

[Fig.2 and 3]

( , ).

f x k f x x a b

f x f a x a b

= ′ = ∀ ∈

> ∈

[Proof]:

( ) s = f t

( ( )v c = 0) ( ) 0

f c′ =

Similarly, has a minimum value in [ , ].

Since ( ) ( ) , the minimum value must occurs at a number in ( , ). And therefore ( ) is a local minimum value.

Again, ( ) 0 since is differentiable

f a b

f a f b

c a b

f c

f c f

=

′ = at .c

Let stand for the position function of a moving object. If the object is in the same place at two different instants

Rolle’s Theorem says there is some instant of time t = c between a and b such that , that is, the velocity is 0.

[Ex1]

and , then ( ) ( ).

t a t b f a f b

Case3: If ( )f x < f a( ) for some x∈( , ).a b [Fig.4]

= = =

[Ex2] Prove that the has exactly one real root.

[Sol]:

(1)

Since f is a polynomial, f is continuous on [0, 1]

By the Intermediate Value Theorem, there is a number Thus, the equation has a root.

(2) To show that the equation has exactly one root, we use Rolle’s Theorem and argue by contradiction. Suppose that the equation had two roots

a and b, Besides, since f is a polynomial, it is

differentiable on (a, b) and continuous on [a, b]. Thus, by Rolle’s Theorem, there exists a

can never be 0.

This gives a contradiction. Therefore, the equation can’t have two real roots.

That is, it has exactly one root.

equation x3 + − =x 1 0

so, ( )f x′

such that ( )f c = 0.

( , ) c ∈ a b

then ( )f b = f a( ) = 0.

Let ( )f x = x3 + −x 1. Then (0)f = − <1 0 and (1)f = >1 0

number c ∈( , ) s.t. ( )a b f c′ = 0. But ( )f x′ = 3x2 + ≥1 1 for all ,x

The Mean Value Theorem

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 b f a or ( ) ( ) ( )( )

f c f b f a f c b a

b a

′ = − − = ′ −

−

( ) ( )

AB

m f b f a

b a

−

= −

secant line AB ( ) ( )

( ) ( )

( ) ( )

The equation of the is

or Let

( ) ( )

( ) ( ) (

( ) (

) )

( )

h f

f b f a

y f a x a

b a f b f a y f

x

a x a

b a

f b f a

f a x a

x b

a

− = − −

−

= + − −

−

⎡ − ⎤

= − ⎢⎣ + − − ⎥⎦

[pf]:

Since is the sum of and ,

both of which are continuous on [ , ] and differentiable on ( , ), we

( )

( )

know that is also conti

a first-degree polynom

nuous on [ , ] and differentiable on

ial

(

f

a b a b

a x

h a

h

x b

一次多項式

( )

( ) 0

( ) 0 (

, ) and ( )

Besides, ( ) and

( ) i.e.

Therefore, by Rolle's Theorem,

( ) ( ) ( ) ( )

( ) ( )

( )

there exist

( ) ( ) ( )

s a numbe

) 0 r

) ( f b f a

b a

f b f a

f a a a

b a f

h x h a

h b b f a

f a b a

b

b f x

f a

h a h b

c

f a b

= − −

−

⎡ − ⎤

= − ⎢⎣ + − − ⎥⎦

⎡ − ⎤

= − ⎢⎣ + − − ⎥

′

′

= = =

⎦

=

in ( , ) such that

that is, ( ) ( ) ( ) i.e. ( ) ( ) ( )

( ) 0 ( ) 0

a b f

h c h c c f b f a c f b f a

f a

b a b

′ ′ =

′ = ′ = − − = −

− −

Consider ( )f x = x3 − x , a = 0 , b = 2

Since is a polynomial, is continuous on [0, 2] and

differentiable on (0, 2). Therefore, by Mean Value Theorem, there is a number (0, 2) s.t. (2) (0) ( )(2 0).

Substitute (2) 6, (0) 0 and (

f f

c f f f c

f f f x

∈ − = ′ −

= = ′ ²

2

2 2

) 3 1 into the equation, we get

6 0 (3 1)(2 0)

4 4 2

6 8

3 3 3

But must lie in (0, 2), so 2

3

x

c

c c c

c c

= −

− = − −

⇒ = ⇒ = ⇒ = ± = ±

=

The main significance ^重要性 of the Mean Value Thm is that it enables us to obtain information about a function from information about its derivative.

[Ex3]

[Sol]:

Since exists for all x, that is, f is differentiate and therefore continuous everywhere. In particular, we can apply the Mean Value Theorem on the interval [0, 2].

( ) f x′

Suppose that (0) 3 and ( ) 5 for all values of . How large can (2) possibly be?

f f x x

f

= − ′ ≤

There exists a number (0, 2) s.t. (2) (0) ( )(2 0).

(2) (0) 2 ( ) 3 2 ( ) 3 2 5 7

The largest possible value for (2) is 7.

c f f f c

f f f c f c

f

∈ − = ′ −

′ ′

⇒ = + = − + ≤ − + ⋅ =

[Ex5]

[Sol]:

Theorem 5

If for all x in an interval (a, b), then f is constant on (a, b) [Proof]:

Let be any two numbers in (a, b) with

Since f is differentiable and therefore continuous on (a, b), it must be differentiable on and continuous on . By applying the Mean Value Theorem to f on the interval , we know that there is a number c

Since , we have

Therefore, f has the same value ay any two numbers in (a, b).

This means f is constant on (a, b).

[

] [

]

2 1 2 1

such that (f x ) − f x( ) = f c x′( )( − x ).

2 1 2 1

( ) ( ) 0 i.e. ( ) ( ) f x − f x = f x = f x ( ) 0

f c′ =

1 and 2

x x x1 < x2.

(

)

( ) 0 f x′ =

Corollary 7

If for all x in an interval (a, b), then is constant on (a, b);

that is, where C is a constant.

( ) ( )

f x^′ = g x^′ f − g

( ) ( ) f x = g x +C

Let ( ) ( ) ( ), then ( ) ( ) ( ) 0 for all in ( , ) Thus, by theorem5, we conclude that is constant. i.e. is constant.

F x f x g x F x f x g x x a b

F f g

′ ′ ′

= − = − =

−

1 1

tan cot

x x π2

− + − =

1 1

2 2

1 1

Let ( ) tan cot

1 1

then ( ) 0 for all

1 1

Therefore, ( ) where is a constant.

To determine the value of , we substitute 1 for into the equation (1) tan (1) cot (1)

4 4

f x x x

f x x

x x

f x C C

C x

C f π π π

− −

− −

= +

′ = − =

+ +

=

= = + = + =

1 1

2 Thus, tan cot

x x π2

− + − =

[Sol]:

[Ex6] Prove the identity ^等式 [Proof]:

§ 4.3 How Derivatives Affect the Shape of a Graph

Increasing / Decreasing Test ( I / D Test) 遞增.遞減檢驗法

(a) If on an interval, then f is increasing on that interval.

(b) If on an interval, then f is decreasing on that interval.

x1 x2

2 1 2 1 2 1

( ) ( ) ( )( ) 0 ( ( ) 0 and 0)

f x − f x = f c x′ − x > ∵ f c′ > x − >x

2 1

i.e. (f x ) > f x( ) ( ) 0

f x′ <

( ) 0 f x′ >

[

] (

)

1 2. x < x [Proof]:

(1) Let and be any two numbers in the interval with

Since f is differentiable (and therefore continuous) on that interval, we know f is differentiable on and continuous on . So by the Mean Value Theorem, there is a number c∈

(

)

(2) Part (b) is proved similarly.

導數如何影響圖形的形狀

[Ex1] Find where the function is increasing and where it is decreasing?

[Sol]:

3 2 2

( ) 12 12 24 12 ( 2)

( ) 0 0, 2, 1

f x x x x x x x

f x x

′ = − − = − −

′ = ⇔ = −

4 3 2

( ) 3 4 12 +5

f x = x − x − x

So is increasing ( ) on ( 1, 0) and (2, ) and it is decreasing ( ) on ( , 1) and (0, 2)

f − ∞

−∞ −

Interval x<-1 -1<x<0 0<x<2 x>2

f ’(x) – + – +

f (x)

The First Derivative Test

Suppose that x= c is a of a continuous function f.

(a) If changes from positive to negative at x= c, then f has a local maximum at x= c.

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

(c) If does not change sign at x= c , then f has no local maximum or minimum at x= c.

f ′ f ′

f ′

critical number

[Ex2] Find the local maximum and minimum values of the function f in Ex1.

[Sol]:

(1) Since 0 when 1 and 0 when 1 0

( 1) 0 is a local minimum value (2) Since 0 when 1 0 and 0 when 0 2

(0) 5 is a local maximum value (3) Since 0 when 0 2

f x

f x

f

f x

f x

f

f x

′ < < −

′ > − < <

− =

′ > − < <

′ < < <

=

′ < < < and 0 when 2

(2) 27 is a local minimum value

f x

f

′ > >

= −

[Ex3] Find the local maximum and minimum values of the function

[Sol]:

( ) 2 sin , 0 2

g x = +x x ≤ ≤x π

( ) 1 2 cos

1 2 4

( ) 0 cos ,

2 3 3

g x x

g x x x π π

′ = +

′ = ⇔ = − ⇒ =

the local maximum value the local minimum value

2 2 2 2 3 2

( ) 2 sin 2( ) 3

3 3 3 3 2 3

g π π π π π

= = + = + = +

4 4 4 4 3 4

( ) 2 sin 2( ) 3

3 3 3 3 2 3

g π π π π π

= = + = + − = −

← critical numbers

By the First Derivative Test Interval

g ’(x) + – +

g (x)

0 2

x 3π

< < 2 4

3π _< x _< 3π 4

3π < x < 2π

What Does Say about ?f ′′ f Def

(1) If the graph of f lies above all of its tangent lines on an interval I, then it is called ^上凹 (CU) on I .

(2) If the graph of f lies below all of its tangent lines on an interval I, then it is called ^下凹 (CD) on I .

concave upward concave downward

Concavity Test ^{凹性檢驗法}

(a) If for all x in I, then the graph of f is concave upward (CU) on I.

(b) If for all x in I, then the graph of f is concave downward (CD) on I.

( ) 0 f ′′ x >

( ) 0 f ′′ x <

Def

A point P on the curve is called an ^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.

( )

y = f x inflection point

[Ex5] Sketch a possible graph of a function f that satisfies the following conditions

By (i), we know that

By (ii), we know that f is CU on and and f is CD on

By (iii), we know that are

horizontal asymptotes of

(−∞ −, 2) (2,∞) on ( ,1) and on (1, )

f −∞ ∞

(i) ( ) 0 on ( ,1), ( ) 0 on (1, )

(ii) ( ) 0 on ( , 2) and (2, ), ( ) 0 on ( 2, 2) (iii) lim ( ) 2 , lim ( ) 0

x x

f x f x

f x f x

f x f x

→−∞ →∞

′ > − ∞ ′ < ∞

′′ > −∞ − ∞ ′′ < −

= − =

( 2, 2)− ( ) y = f x [Sol]:

2 and 0 y = − y =

The Second Derivative Test Suppose is continuous near

(a) If ( ) 0 and ( ) 0, then has a local minimum at . (b) If ( ) 0 and ( ) 0, then has a local maximum at .

f c

f c f c f x c

f c f c f x c

′′

′ = ′′ > =

′ = ′′ < =

[Ex6] Discuss the curve w.r.t. concavity, points of inflection,

and local maximum or minimum Use this information to sketch the curve.

4 3

4 y = x − x

[Sol]:

Therefore, (critical numbers) Since is a local minimum

Since , the Second Derivative Test gives no information about the critical number 0.

But, by the First Derivative Test, since

f has no local maximum or minimum at 0.

Set

( ) 0 0, 3

f x′ = ⇒ =x x = (3) 36 0, (3) 27 f ′′ = > f = −

4 3 3 2 2

2

If ( ) 4 , then ( ) 4 12 4 ( 3) and ( ) 12 24 12 ( 2).

f x x x f x x x x x

f x x x x x

= − ′ = − = −

′′ = − = −

(0) 0 f ′′ =

( ) 0 for 0 and 0 3, f x′ < x < < <x

( ) 0 0, 2

f ′′ x = ⇔ =x x =

Therefore, the inflection points

Note: (1) The Second Derivative Test is inconclusive ^{沒有結論的} when It gives no information about the critical number c if

So when

( ) 0.

f ′′ c = ( ) 0.

f ′′ c = ( ) 0 Use the First

( ) 0 and Derivative Test.

f ′ c = f ′′ c = →

(2) The Second Derivative Test fails when f c′( ) D.N.E Use the First Derivative Test.

→

are (0, 0) and (2, 16)−

x<0 0<x<2 x>2

+ – +

f (x) CU CD CU

( ) f ′′ x

[Ex7] Sketch the graph of the function

2 1

3 3

( ) (6 )

f x = x − x [Sol]:

1 2

3 3

( ) 4

(6 )

( ) 0 4

( ) D.N.E 0, 6

0, 4, 6 are critical numbers f x x

x x

f x x

f x x x

x

′ = −

−

′ = ⇔ =

′ ⇔ = =

∴ =

4 5

3 3

( ) 8

(6 )

( ) D.N.E 0, 6

f x

x x

f x x x

′′ = −

−

′′ ⇔ = =

5 3

(0) 0 is a local minimum (4) 2 is a local maximum ( ) has no local maximum or minimum at 6.

f f f x

x

∴ =

=

=

∴ the point of inflection is (6,0) x<0 0<x<4 4<x<6 x>6

f '(x) – + – –

f (x)

x<0 0<x<6 x>6 f "(x) – – +

f (x) CD CD CU

[Ex8] Use the first and second derivative of , together with asymptotes, to sketch its graph.

1

( ) ^x f x = e

[Sol]:

1

( ) 2

( ) D.N.E 0 ex

f x x

f x x

′ = −

′ ⇔ =

1

4

(2 1) ( )

( ) 0 1

2

( ) D.N.E 0

ex x f x

x

f x x

f x x

′′ = +

′′ = ⇔ = −

′′ ⇔ =

has no local maximum or minimum

∴ f

(1) (2)

x<0 x>0

f '(x) – –

f (x)

x>0

f "(x) – CD

+ +

f (x) CU CU

1 2

the inflection point is , 2 e⁻

⎛ ⎞

∴ ⎜⎝ − ⎟⎠

1 0

2 x

− < <

1 x < − 2

(3)

1

0 1

0 1

0

lim 0 is a vertical asymptote

lim 0

lim 1 1 is a horizontal asymptote

x x

x x

x x

e x

e

e e y

+

−

→

→

→±∞

= ∞ ∴ =

⎛ ⎞

⎜ = ⎟

⎝ ⎠

= = ∴ =

∵

§ 4.4 Indeterminate Forms and L’Hospital’s Rule

indeterminate form of typ 0

If lim ( ) 0 and lim ( ) 0, then the limit lim ( ) is called an ( )

If lim ( ) (or ) and lim ( ) (or ), then the limit lim ( )

( ) is called an

e 0

inde

x a x a x a

x a x a x a

f x g x f x

g x

f x g x f x

g x

→ → →

→ → →

= =

= ∞ − ∞ = ∞ − ∞

terminate form of type ∞

∞

L’Hospital’s Rule

Suppose f and g are differentiable and near a (except possibly at a).g x′( ) ≠ 0

( ) 0

Suppose that lim is an indeterminate form of type or

( ) 0

( ) ( ) ( )

Then lim = lim if the limit lim exist (or is or )

( )

( ) ( )

x a

x a x a x a

f x g x

f x f x f x

g x g x g x

→

→ → →

∞

∞

′ ′

∞ − ∞

′ ′

不定型與洛必達法則

Vote 1: It is especially^特別 important to verify ^檢驗the conditions regarding the limits of f and g before using L’Hospital’s Rule.

Vote 2: L’Hospital’s Rule also valid for one-sided limit and for limits at infinity or negative infinity; that is, can be replace by"x → a" x → a⁺,x → a⁻,

[Ex1] Find

1

lim ln

1

x

x

→ x − [Sol]:

The limit is an indeterminate form of type , we can apply L’Hospital’s Rule:0 0

( )

( )

1 1 1

ln 1

lim ln lim lim 1

1 1 1

x x x

d x

x dx x

x d x

dx

→ = → = → =

− −

or .

x → ∞ x → −∞

[Ex2] Calculate lim ₂

x

x

e

→∞ x [Sol]:

The limit is an indeterminate form of type , we can apply L’Hospital’s Rule:∞

( )

( )

( )

2

( )

2

lim lim lim lim lim

2 2 2

x x

x x x

x x x x x

d d

e e

e dx e dx e

d d

x x x x

dx dx

→∞ = →∞ = →∞ = →∞ = →∞ = ∞

lim 2

x

x

e

→∞ x is still an indeterminate form of type ∞

∞ so use the L’Hospital’s Rule again

[Ex3] Calculate

3

lim ln

x

x

→∞ x [Sol]:

Apply L’Hospital’s Rule to it because it’s an indeterminate form of type ∞

( )

1 2

3 3

3 3

ln 1

lim ln 3lim lim lim 0

1 3

x x x x

d x

x dx x

x d x

x x dx

→∞ = →∞ = →∞ − = →∞ =

⎛ ⎞

⎜ ⎟

⎝ ⎠

[Ex4] Find

0 3

lim tan

x

x x

→ x

−

[Sol]:

It’s of the type , so we can apply L’Hospital’s Rule:0 0

2

2

3 2

0 0 0 0

tan sec 1 2 sec sec tan 1 sin 1

lim lim lim lim sec

3 6 3 cos

x x x x

x x x x x x x

x x x x x x

→ → → →

− = − = ⋅ = ⋅ ⋅

[Ex5] Find lim

sin

x

x

x x

→∞ + [Sol]:

type , use L’Hospital’s Rule again0 0

2 0

1 sin 1 1 1

lim sec 1 1 1

3 ^x cos 3 3

x x

x x

= → ⋅ ⋅ = ⋅ ⋅ ⋅ =

lim lim 1 the limit does not exist.

sin 1 cos

lim lim lim 1 1

sin sin

sin 1

x x

x x x

x

x x x

x

x x

x x x

x x

x x

→∞ →∞

→∞ →∞ →∞

+ = +

= = =

+ + +

If lim ( ) 0 and lim ( ) (or ), then the limit lim ( ) ( ) is call indeterminate form of type 0

ed an

x a f x x a g x x a f x g x

→ = → = ∞ − ∞ →

⋅∞

Indeterminate Products 不定乘積

lim ( ) 1 ( ) lim ( )

1 ( ) lim ( ) ( )

x

x a x

a

f x g x g x

f x f x g x

→

→

∞

→

⎧⎪

⎪⎪

= ⎨⎪

⎪⎪⎩

← indeterminate form of type

← indeterminate form of type 0 0

∞

∞ [Ex6] Evaluate

0

lim ln

x

x x

→ + ⋅ [Sol]:

0 0

Since lim 0 and lim ln ,

x x

x x

+ +

→ = → = −∞ the limit is an indeterminate form of type 0⋅∞

Using L’Hospital’s Rule, (but converting the limit into the form of type or

0 0 0 0

2

ln 1

lim ln lim ln lim lim lim 0

1 1 1 ^x

x x x x

d x

x dx x

x x x

d

x dx x x

+ + + + →∞

→ ⋅ = → = → = → = − =

⎛ ⎞ −

⎜ ⎟⎝ ⎠

0 0 first), we have

∞

∞

[ ]

If lim ( ) and lim ( ) , then the limit lim ( ) ( ) is called an indeterminate form of type

x a f x x a g x x a f x g x

→ = ∞ → = ∞ → −

∞ − ∞ Indeterminate Differences 不定差

0

0 ∞.

∞

[Ex7] Compute

( )

2

lim sec tan

x

x x

π ⁻

→⎜ ⎟⎛ ⎞⎝ ⎠

− [Sol]:

This is an indeterminate form of . We’ll try to convert ^轉換 the difference into a quotient ^商 .

In this case, we try to convert the difference into a quotient so that we have an indeterminate form of type or

∞ − ∞

( )

2 2 2 2

1 sin 1 sin cos

lim sec tan lim lim lim 0

cos cos cos sin

x x x x

x x x

x x

x x x x

π ⁻ π ⁻ π ⁻ π ⁻

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞

→⎜ ⎟ →⎜ ⎟ →⎜ ⎟ →⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

− −

⎛ ⎞

− = ⎜⎝ − ⎟⎠= = − =

(indeterminate form of type )0 0

[ ]

0

If lim ( ) 0 and lim ( ) 0 , then the limit lim ( ) ( ) is called an indeterminate form of typ 0e

g x

x a f x x a g x x a f x

→ = → = →

Indeterminate Powers 不定次方

[ ]

0

If lim ( ) and lim ( ) 0 , then the li indeterminat

mit lim ( ) is called an e form of type

g x

x a f x x a g x x a f x

→ = ∞ → = →

∞

[ ]

If lim ( ) 1 and lim ( ) , then the limit lim ( ) is called indeterminate form of typ

an e 1

g x

x a f x x a g x x a f x

→

∞

→ = → = ±∞

In these cases, we’ll write the function as an exponential

[

]

[

]

0

[Ex9] Find lim ^x

x

+ x

→

It’s an indeterminate form of type 0⁰

0

lim ln

ln ln 0

0 0 0

lim ^x lim ^xx lim ^{x x x x x} 1

x x x

x e e e ^→ e

+ + →

→ = → = = = =

( is continuous and lim0 ln 0 exists b [Sol]:

y ex6)

x

e x x x

→ ⋅ =

∵

[Ex8] Calculate

( )

0

lim 1 sin 4 ^x

x

+ x

→ +

[Sol]:

This is an indeterminate form of type1^∞

( ) ( )

0 0 0 2

cos 4 4 ln 1 sin 4 1 sin 4

Since lim cot ln 1 sin 4 lim lim 4,

tan sec

x x x

x

x x

x x

x x

+ + +

→ → →

+ + ⋅

⋅ + = = =

( )

0 0

we have that lim 1 sin 4 ^x lim ^{x x}

x x

x e

+ +

⋅ +

→ + = →

( )

0

lim cot ln 1 sin 4

4

x

x x

e e

→ + ⋅ +

=

=

(∵e^x is a continuous function)

( )

0 0 0

lim 1 sin 4 lim lim

x x x x x

x x x

x e e

+ + +

+ ⋅ +

→ + = → = →

§ 4.5 Summary of Curve Sketching

Domain Intercept

Guidelines for Sketching a Curve : ( A.

B.

C.

s

Symmetry : (i) (f − =x) f x( ) ⇔ f is an even function 描繪曲線的方法）

⇔ the graph of f is symmetric about the y-axis.

(ii) (f − = −x) f x( ) ⇔ f is an odd function

⇔ the graph of f is symmetric about the origin.

(iii) ( ) ( ) for all , is a periodic function

f x p f x x Df

f

+ = ∈

⇔

where p is a positive integer

Asymptotes : find vertical asymptotes or horizontal asymptotes or slant asymp

D. totes

( )

If lim ( ) 0 , then the line is call slant asymptot

ed e.

a

x f x ax b y ax b

→∞⎡⎣ − + ⎦⎤ = = +

[ ]

1 1

[Ex] If ( ) , then lim ( ) lim 0.

Therefore is a slant asymptote of ( )

x x

f x x f x x

x x

y x y f x

→∞ →∞

= + − = =

= =

(對稱性)

(定義域) (截距)

(漸近線)

(週期函數)

(斜漸近線)

Intervals of Increase or Decrease

Local Maximum and Minimum Values Concavity and Points of Inflection

Sketch E.

th F

e C .

G.

H. urve

[Ex1] Sketch the curve

2 2

2 1 y x

= x

−

[Sol]: ²

2

Let ( ) 2

1 f x x

= x

−

{

} ⁽

^{) (}

^{) ( )}

. 1

A D_f = x x ≠ ± = −∞ − ∪ − ∪ ∞, B. The x-intercept = 0 and the y-intercept = 0 C. Symmetry

Since , f is even . The curve is symmetric about the y-axis.

D. Asymptotes

( ) ( )

f − =x f x

1 1 1 1

2 2

lim ( ) , lim ( ) , lim ( ) , lim ( ) 1 and 1 are vertical asymptotes of the curve 2

1 lim ( ) 2 2 is the horizontal asymptotes of the curve.

x x x x

x

f x f x f x f x

x x y x

x

f x y

+ − + −

→ → →− →−

→±∞

= ∞ = −∞ = −∞ = ∞

∴ = = − =

−

= ∴ =

∵

∵

(遞增或遞減區間)

(局部極大或極小值) (凹性與反曲點)

E. Intervals of Increase or Decrease

( )

( ) ( )

2 2

2 2

2 2

4 1 2 2 4

( ) , ( ) 0 0 (critical number)

1 1

x x x x x

f x f x x

x x

− − ⋅ −

′ = = ′ = ⇔ =

− −

F. The local maximum value

G. Concavity and Points of Inflection (0) 0

= f =

( ) ^{( )} ( )

( ) ( )

2 2 2 2

4 3

2 2

2

4 1 4 2 1 2 12 4

( )

1 1

( ) D.N.E. 1 0 1

x x x x x

f x

x x

f x x x

− − − − ⋅ − ⋅ +

′′ = =

− −

′′ ⇔ − = ⇔ = ±

Interval

f ’(x) + –

f (x)

0, 1

x < x ≠ − x > 0,x ≠1

There is no inflection point because = 1 are not in the domain of .x f

∴ ±

H.

x<-1 -1<x<1 x>1 f "(x) +

CU

– +

f (x) CD CU

[Ex2] Sketch the graph of ^{f x( ) = 5}

(

)

(

)

[Sol]:A. D_f = ℜ

C. Symmetry: None D. Asymptote: None

E. Intervals of Increase or Decrease B. Intercepts:

( ) ( )

( ) ( )

2 5

3 3

2 3

5 1 2 1 when 0 , 7

when 0, 1 5 2 1 0 1 or 7 2

y x x x y

y x x x

= − − − = =

⎡ ⎤

= − ⎣ − − ⎦ = ⇒ =

( )

( )

( ) 10 2 , ( ) 0 2

3 1

( ) D.N.E 1.

f x x f x x

x

f x x

′ = − ′ = ⇔ =

−

′ ⇔ =

F. The local maximum value The local minimum value

(2) 3

= f =

(1) 0

= f =

x<1 1<x<2 x>2 f '(x) –

CU

+ –

f (x) CD CU

( ) ( )

10 1 2 1

( ) , ( ) 0 9 1 2

( ) D.N.E. 1

f x x f x x

x

f x x

′′ = − ′′ = ⇔ =

−

′′ ⇔ =

G. Concavity and Points of Inflection

1 3

the inflection point is , 3 2 2

⎛ ⎞

∴ ⎜⎝ ⎟⎠

H.

x>1 f "(x) +

CU

– –

f (x) CD CD

1 1

2 < <x 1

x 2

−∞ < <

[Ex3] Sketch the graph of ( )f x = xe^x [Sol]:

A. D_f = ℜ

B. The x-intercept and y-intercept are both 0.

C. Symmetry: None D. Asymptotes:

F. The local min. value 1

( 1)

f e

= − = −

E. Intervals of Increase or Decrease lim ^x ; lim ^x 0

x xe x xe

→∞ = ∞ →−∞ =

0

∴ =y is the horizontal asymptote

( )

( ) 1

( ) 0 1

x x x

f x xe e x e

f x x

′ = + = +

′ = ⇔ = −

There’s no local maximum

( ) ( )

( ) 1 2

( ) 0 2

x x x

f x x e e x e

f x x

′′ = + + = +

′′ = ⇔ = − G.

2

−∞ < < −x

(

)

the inflection point is 2, 2e⁻

∴ − −

H.

f "(x) – CD

+

f (x) CU

2 x

− < < ∞

f '(x) – +

f (x)

1

−∞ < < −x − < < ∞1 x

[Ex4] Sketch the graph of ( )f x = 2 cos x +sin 2x [Sol]: A. D_f = ℜ

C. f is neither odd nor even, but for all x. Therefore f is a periodic function with period 2π. We may consider only

B. the y-intercepts is the x-intercepts:

( )

( ) ^{( )( )}

2

2

( ) 2sin 2 cos 2 2sin 2 1 2 sin

2 2 sin sin 1 2 sin 1 2 sin 1

f x x x x x

x x x x

′ = − + = − + −

= − + − = − + −

(0) 2

f =

( )

[ ]

( )

2 cos sin 2 0 2 cos 1 sin 0 cos 0 or sin 1 or 3 in 0, 2

2 2

x x x x

x x

x π x π π

+ = ⇒ + =

⇒ = = −

⇒ = =

(

) ( )

f x+ π = f x

0 ≤ ≤x 2 .π

D. Asymptote: None E.

[ ]

( ) 0 sin 1 or sin 1

2

5 3

in 0, 2 , , ,

6 6 2

f x x x

x π π π π

′ = ⇔ = − =

⇒ =

F.

3 3 5 3 3

the local maximum value the local minimum value

6 2 6 2

f _{⎛ ⎞}π f _⎛ π _⎞

∴ = ⎜ ⎟ = = ⎜ ⎟ = −

⎝ ⎠ ⎝ ⎠

( )

1 2

1 1

1 2

( ) 2 cos 1 4 sin

1 3

( ) 0 cos 0 or sin , , ,

4 2 2

where sin 1

4 2 sin 1

4

f x x x

f x x x x π π α α

α π α π

−

−

′′ = − +

′′ = ⇔ = = − ⇒ =

= + ⎛ ⎞⎜ ⎟⎝ ⎠

= − ⎛ ⎞⎜ ⎟⎝ ⎠

+ – +

f '(x) +

f (x)

x π6

−∞ < < 5

6 ^x 6

π π

< < 5 3

6 ^x 2

π π

< < 3

2 ^x

π < < ∞

H. We draw the curve on first, then extend ^延展 the curve by translation ^轉換.

(

( )

)

(

^{( )}

)

The inflection points are , 0 , , , , 0 , ,

2 f 2 f

π α α π α α

⎛ ⎞ ⎛ ⎞

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

[

]

+ +

CU –

CU

– CD CD

f "(x) –

f (x) CD

0 _{< <}x π2

2 x 1

π < <α 3 ₂

2π < <x α

1

3 x 2π

α < < α₂ < <x 2π

[Ex5] Sketch the graph of [Sol]:

{

} ^{

^{} ( )}

A. D_f = x − x > = x − < <x = −2, 2 B. the y-intercepts is

the x-intercepts:

(0) ln 4

f =

D. Asymptote:

(

)

ln 4 y = −x

(

)

Let ( )f x =ln 4−x

(

)

ln 4−x = ⇒ −0 4 x = ⇒ = ±1 x 3 C. Symmetry:

( ) ( ) is an even function f − =x f x ∴ f

∵

The curve is symmetric about y-axis.

(

) (

)

2 2

lim ln 4 ; lim ln 4

2 and 2 are vertical asymptotes.

x x x x

x x

− +

→ − = −∞ →− − = −∞

∴ = = −

E. Intervals of Increase or Decrease

2

( ) 2 4

( ) 0 0

f x x

x

f x x

′ = −

−

′ = ⇔ =

-2<x<0 0<x<2

f '(x) + –

f (x)

F. The local maximum value = f (0) = ln 4 G. Concavity and points of inflection

( ) ^{( )( )}

( ) ( )

( )

2 2

2 2

2 2

2 4 2 2 8 2

( )

4 4

x x x x

f x

x x

− − − − − − +

′′ = =

− −

( )

Since ( )f ′′ x < 0 for all in x −2, 2 .

H.

( )

The curve is CD on −2, 2 and there is no point of inflection.

[Ex6] Sketch the graph of

3

( ) 2

1 f x x

= x [Sol]: +

A. D_f = ℜ

B. The x-intercept and y-intercept are both 0.

C. Symmetry:

F. There’s no local maximum or minimum E. Intervals of Increase or Decrease

( )

( )

2 2

2 2

( ) 3

1 x x f x

x

′ = +

+ D. Asymptotes:

Since (f − = −x) f x( ), f is odd and its graph is symmetric about the origin.

( )

2 2

( ) lim ( ) lim 0

1 1

is a slant asymptote.

x x

x x

f x x f x x

x x

y x

→±∞ →±∞

⎛ ⎞

= − + ∴ − = ⎜⎝− + ⎟⎠ =

∴ =

∵

Since ( )f x′ > 0 for all x∈ℜ ≠,x 0. is increasing on .f ℜ

G. Concavity and points of inflection

( )

( )

2 2 3

2 3

( ) , ( ) 0 0 , 3

1

x x

f x f x x x

x

′′ = − ′′ = ⇔ = = ±

+

H.

the inflection points are

∴ ^{3, 3 3} , 0, 0 ,

( )

4 4

⎛ ⎞ ⎛ ⎞

− −

⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

+ –

CD CU

+ CU

f "(x) –

f (x) CD

3

−∞ < < −x − 3 < <x 0 0 < <x 3 3 < < ∞x