Since we have f ''' () c > 0 , we know that f ' () x at x = 1 and not a local max or min.

48. Since we have f ''' () c > 0 , we know that f ' () x at x = 1 and not a local max or min.

is concave up in a neighborhood around x =. c

() means that F(x) is c = 0 , we then know that the graph of 51. a. Not possible; Fx () > 0

Since f '

increasing. Fx ′′ () > 0 means that the rate at

f ' () x must be positive in that neighborhood.

which F(x) is increasing never slows down. This means that the graph of f must be increasing

Thus the values of F must eventually to the left of c and increasing to the right of c.

become positive.

Therefore, there is a point of inflection at c.

b. Not possible; If F(x) is concave down for all

x , then F(x) cannot always be positive.

c.

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52. a.

53. a.

No global extrema; inflection point at (0, 0)

ππ fx ′ () = 0 when x = – , 22

f 2 ′′ () x = –2 sin – 2 cos x x + 2 sin 2 x

4 sin = 2 x – 2 sin – 2; ( ) x f ′′ x = 0 when

sin x = – or sin x = 1 which occur when

No global maximum; global minimum at

(0, 0); no inflection points

⎛ Global minimum π

f ⎜ – ⎞= ⎟ –2; global ⎝ 2 ⎠

maximum f ⎛⎞= ⎜⎟ 2; inflection points ⎝⎠ 2

6 ⎟ = –, – f = – Global minimum ⎝ ⎠ 4 ⎜ 6 ⎝ ⎟ ⎠ 4

b.

f(– π ) = –2π + sin (–π ) = –2π ≈ –6.3;

global maximum

f () π = 2 π + sin π = 2 π ≈ 6.3 ; inflection point at (0, 0)

d.

fx ′ () = 2 cos x + 2 sin cos x x = 2 cos (1 sin ); x + x fx ′ () = 0 when

Global minimum

= –4 sin 2 x – 2 sin x + 2; ( ) f ′′ x = 0 when

which occur when

sin π

maximum f () π=π+

=π≈ 3.1; πππ x 5 = – ,,

inflection point at (0, 0).

⎛ π Global minimum f ⎜ – ⎞= –1; 2 ⎟

global

maximum f ⎛⎞= ⎜⎟ 3; inflection points ⎝⎠ 2

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fx ′ () = 2 cos 2 x + 3sin 3 x Using the graphs, f(x) has a global minimum

at f(2.17) ≈ –1.9 and a global maximum at

f (0.97) ≈ 1.9

f ′′ () x = –4 sin 2 x + 9 cos 3 ; ( ) xf ′′ x = 0 when

ππ x = – ,

fx ′ () = –2 sin 2 x + 2 sin x 22

and when

= –4 sin cos x x + 2 sin x = 2sin (1 – 2 cos ); x x x ≈ –2.469, –0.673, 0.413, 2.729.

Inflection points: ⎛ π ⎞ ⎛ π ⎞

fx ′ () = 0 when x =π –,– , 0, , π

f ′′ () x = –4 cos 2 x + 2 cos ; x f ′′ () x = 0 when

x ≈ –2.206, –0.568, 0.568, 2.206

Global minimum ⎛ π – ⎞

⎟ = ⎛⎞ f ⎜⎟ = –1.5;

Global maximum f(– π ) = f(π ) = 3;

Inflection points: ≈ ( − 2.206, 0.890 ) , ( − 0.568, 1.265 − ) , ( 0.568, 1.265 − ) ,

d.

55. y

fx ′ () = 3cos 3 – cos ; ( ) x xfx ′ = 0 when

3cos 3x = cos x which occurs when ππ

and when

22 x ≈ –2.7, –0.4, 0.4, 2.7

f ′′ () x = –9 sin 3 x + sin x which occurs when

x =– π , 0, π and when x ≈ –2.126, –1.016, 1.016, 2.126

a. f is increasing on the intervals ( −∞ − ,3 ]

π Global minimum f ⎛⎞= –2; and [ − ] 1, 0 ⎜⎟ .

⎝⎠ 2 f is decreasing on the intervals [ −− 3, 1 ]

global maximum f ⎜ – ⎟ ⎞= 2; and [ 0, ∞ ) .

Inflection points: ≈ ( − 2.126, 0.755 ) ,

b. f is concave down on the intervals

( −∞ − ,2 ) and ( 2, ∞ ) .

( 2.126, 0.755 − ) f is concave up on the intervals ( − 2, 0 ) and () 0, 2 .

e.

c. f attains a local maximum at x =− 3 and

f attains a local minimum at x =− 1 .

d. f has a point of inflection at x =− 2 and

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; () fx ′ is never 0,

2 x –6 x + 40 and always positive, so f(x) is increasing for

a. f is increasing on the interval [ −∞ 1, ) .

all x. Thus, on [–1, 7], the global minimum is

f is decreasing on the interval ( −∞ − ,1 ] f (–1) ≈ –6.9 and the global maximum if

f (7) ≈ 48.0.

b. is concave up on the intervals − 2, 0 2 x − 18 x + 147 – 240 f x ( ) f ′′ () x = ; () f ′′ x = 0

and ( 2, ∞ ) .

( x 2 –6 x + 40) 3/2

when x ≈ 2.02; inflection point

f is concave down on the interval () 0, 2 .

f (2.02) ≈ 11.4

b.

c. f does not have any local maxima.

f attains a local minimum at x =− 1 .

d. f has inflection points at x = 0 and x = 2 .

Global minimum f (0) = 0; global maximum

f (7) ≈ 124.4; inflection point at x ≈ 2.34,

No global minimum or maximum; no inflection points

d.

Global minimum f (3) ≈ –0.9; global maximum f(–1) ≈ 1.0 or f(7) ≈ 1.0; Inflection points at x ≈ 0.05 and x ≈ 5.9,

f (0.05) ≈ 0.3, f(5.9) ≈ 0.3.

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60. a.

3.6 Concepts Review

1. continuous; (a, b); ( ) – ( ) fb fa = fcb ′ ( )( – ) a

2. f′ (0) does not exist.

3. F(x) = G(x) + C fx ′ () = 3 x 2 – 16 x + 5; ( ) fx ′ = when 0

1 4 x = , 5. 4. x + C

3 Global minimum f(5) = –46;

global maximum f ⎜⎟ ⎛⎞≈ 4.8

1 Problem Set 3.6

⎝⎠ 3 x

8 1. fx ′ () =

f ′′ () x = x 6 – 16; ( ) f ′′ x = when 0 x = ; x

f (2) – (1) f 2–1

inflection point: 8

= for all c > 0, hence for all c in (1, 2) 1

b.

Global minimum when x ≈ –0.5 and

x ≈ 1.2, f(–0.5) ≈ 0, f(1.2) ≈ 0; global maximum f(5) = 46

Inflection point: ( − ) 0.5, 0 , ( ) 1.2, 0 ,

2. The Mean Value Theorem does not apply because g′ (0) does not exist.

c.

No global minimum or maximum; inflection point at

f (2) – (–2) f = 6–2 =

2c + 1 = 1 when c = 0

No global minimum, global maximum when x ≈ 0.26, f(0.26) ≈ 4.7 Inflection points when x ≈ 0.75 and x ≈ 3.15, f(0.75) ≈ 2.6, f(3.15) ≈ –0.88

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4. gx ′ () = 3( x + 1) 2 1 2 2 7. 1 fz ′ () = (3 z += 1) z +

g (1) – (–1) g 8–0

2 f (2) – (–1) f 2 – (–2) = 4 =

c + 1) = 4 when c = –1 +

3 2 1 c 4 + = when c = –1, 1, but 1 − is not in

( 1, 2) − so c = is the only solution. 1

5. Hs ′ () = 2 s + 3

H (1) – H (–3)

1 8. The Mean Value Theorem does not apply

because F(t) is not continuous at t = 1. 2c + 3 = 1 when c = –1

F (2) – F (–2) 3 –– () 3 4 =

c 2 = when c =±

≈± 1.15 ( – 3) c

c = 3–3 ≈ 1.27 (3 +

3 is not in (0, 2).)

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14. ′ () = because f(x) is not continuous at x = 3. 2/3 gx x

10. The Mean Value Theorem does not apply

3 g (1) – (–1) g = 1 – (–1) =

c = when 1 c ⎛⎞ =± ⎜⎟

1/ 3 = 2 when c =

π cos c = 0 when c =± .

12. The Mean Value Theorem does not apply because (0) h′ does not exist.

16. The Mean Value Theorem does not apply

because C () θ is not continuous at θ =− ππ , 0, .

⎛⎞ 3 17. The Mean Value Theorem does not apply

c = when 1 c =±⎜⎟ ,

because T () θ is not continuous at θ = .

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22. By the Mean Value Theorem because f(x) is not continuous at x = 0.

18. The Mean Value Theorem does not apply

fb () − fa ()

= fc ′ () for some c in (a, b).

Since f(b) = f(a),

There are three values for c such that

fc 1 ′ () =− . 4

19. fx ′ () = 1–

They are approximately 1.5, 3.75, and 7.

5 24. fx ′ () = 2 α x + β

f (2) – (1) f 2 –2 1

fb ()–() fa 1

when c =± 2, c = 2 ≈ 1.41 = α ( a ++ b ) β

a + b ( c = – 2 is not in (1, 2).) 2 α c += βα ( a ++ b ) β when c = which is

the midpoint of [a, b].

25. By the Monotonicity Theorem, f is increasing on the intervals ( , ) and ( ax 0 xb 0 ,) . To show that fx ( 0 ) > fx () for x in (, ) ax 0 ,

consider f on the interval (, ] ax 0 .

f satisfies the conditions of the Mean Value Theorem on the interval [, ] xx 0 for x in (, ) ax 0 .

So for some c in ( , ), xx

fx ( 0 ) − fx () = fcx ′ ( )( 0 − x ) . because f(x) is not continuous at x = 2.

20. The Mean Value Theorem does not apply

so ( fx 0 ) > fx () . Similar reasoning shows that fx () > fx ( 0 ) for in ( x xb 0 , ) . Therefore, f is increasing on (a, b).

26. a.

fx ′ () = 3 x 2 > 0 except at x = 0 in (– ∞ , ∞ ). fx () = x 3 is increasing on (– ∞ , ∞ ) by

21. The Mean Value Theorem does not apply

Problem 25.

because f is not differentiable at x = 0 .

b. fx ′ () = 5 x 4 > 0 except at x = 0 in (– ∞ , ∞ ). fx () = 5 x is increasing on (– ∞ , ∞ ) by

Problem 25.

c. fx () =⎨

> 0 except at x = 0 in

is increasing on

⎪⎩ x

(– ∞ , ∞ ) by Problem 25.

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27. s(t) is defined in any interval not containing t = 0.

34. ′ () 6 fx 2 = x – 18 x = xx 6 ( – 3); ( ) fx ′ = 0 when ′

sc () = – 2 < 0 for all c ≠ 0. For any a, b with

x = 0 or x = 3.

f (–1) = –10, f(0) = 1 so, by Problem 33, f(x) = 0

a < b and both either positive or negative, the

has exactly one solution on (–1, 0).

Mean Value Theorem says

f (0) = 1, f(1) = –6 so, by Problem 33, f(x) = 0 has

sb ()–() sa = scb ′ ( )( – ) a for some c in (a, b).

exactly one solution on (0, 1).

Since a < b, b – a > 0 while sc ′ () < 0, hence

f (4) = –15, f(5) = 26 so, by Problem 33, f(x) = 0 has exactly one solution on (4, 5).

s (b) – s(a) < 0, or s(b) < s(a). Thus, s(t) is decreasing on any interval not

35. Suppose there is more than one zero between containing t = 0.

successive distinct zeros of f′ . That is, there are 2

a and b such that f(a) = f(b) = 0 with a and b

28. sc ′ () = – 3 < 0 for all c > 0. If 0 < a < b, the

between successive distinct zeros of f ′ . Then by c

Mean Value Theorem says Rolle’s Theorem, there is a c between a and b

sb ()–() sa = scb ′ ( )( – ) a for some c in (a, b).

fc supposition that a and b lie between successive

such that ′ () = 0 . This contradicts the

Since a < b, b – a > 0 while sc ′ () < 0, hence

distinct zeros.

s (b) – s(a) < 0, or s(b) < s(a). Thus, s(t) is decreasing on any interval to the right of the

36. Let x 1 , x 2 , and x 3 be the three values such that origin.

gx () 1 = gx ( 2 ) = gx ( 3 ) = 0 and

a ≤ x 1 < x 2 < x 3 ≤ b . By applying Rolle’s By Theorem B,

29. Fx ′ () =

0 and Gx () = 0; Gx ′ () = 0 .

Theorem (see Problem 22) there is at least one

F (x) = G(x) + C, so F(x) = 0 + C = C. number x 4 in (, ) xx 1 2 and one number x 5 in ( x 2 , x 3 ) such that gx ′ ( 4 ) = gx ′ ( 5 ) = 0 . Then by

30. Fx 2 () = cos + sin 2 x xF ; (0) = 1 2 + 0 2 = 1 applying Rolle’s Theorem to gx ′ () , there is at Fx ′ () = 2 cos ( sin ) 2 sin (cos ) x − x + x x = 0 least one number x 6 in ( x 4 , x 5 ) such that By Problem 29, F(x) = C for all x.

gx ′′ ( 6 ) = 0 .

Since F(0) = 1, C = 1, so sin 2 x + cos 2 x = 1 for

all x.

37. f(x) is a polynomial function so it is continuous on [0, 4] and f ′′ () x exists for all x on (0, 4).

31. Let Gx () = Dx F x ; () ′ = D and ( ) Gx ′ = D .

f (1) = f(2) = f(3) = 0, so by Problem 36, there are By Theorem B, F(x) = G(x) + C; F(x) = Dx + C.

at least two values of x in [0, 4] where fx ′ () = 0

32. Fx ′ () = 5; (0) F = 4 and at least one value of x in [0, 4] where

f ′′ () x = 0.

F (x) = 5x + C by Problem 31.

F (0) = 4 so C = 4.

38. By applying the Mean Value Theorem and taking

F (x) = 5x + 4

the absolute value of both sides,

33. Since f(a) and f(b) have opposite signs, 0 is

fx ( 2 ) − fx () 1

= fc ′ () , for some c in (, ) xx . between f(a) and f(b). f(x) is continuous on [a, b],

since it has a derivative. Thus, by the Intermediate Value Theorem, there is at least one

Since fx ′ () ≤ M for all x in (a, b),

point c,

fx ( 2 ) − fx () 1

a < c < b with f(c) = 0. ≤ M ;( fx 2 ) − fx () 1 ≤ Mx 2 − x 1 .

Suppose there are two points, c and cc ′ , < c ′ in

(a, b) with fc () = () f c′ = 0. Then by Rolle’s

39. fx ′ () = 2 cos 2 ; ( ) xfx ′ ≤ 2

Theorem, there is at least one number d in (, ) c c′

fx ( 2 ) − fx () 1 fx ( 2 ) − fx () 1

with fd ′ () = 0. This contradicts the given

= fx ′ ();

x 2 − x 1 x 2 − x 1 information that fx ′ () ≠ 0 for all x in [a, b], thus

fx ( 2 ) − fx () 1 ≤ 2 x 2 − x 1 ;

there cannot be more than one x in [a, b] where

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40. a.

44. Let fx () = x so fx ′ () =

. Apply the Mean 2 x

Value Theorem to f on the interval [x, x + 2] for x > 0.

Thus x +− 2 x =

(2) = for some c in 2 c c

(x, x + 2). Observe

Thus as x →∞ ,

b.

x →∞ (

x +− 2 x ) = lim = 0 .

Therefore lim

x →∞ c

45. Let f(x) = sin x. fx ′ () = cos x, so

fx ′ () = cos x ≤ 1 for all x.

By the Mean Value Theorem, fx () − fy () =′ fc () for some c in (x, y).

41. Suppose fx ′ () ≥ 0 . Let a and b lie in the interior fx () − fy ()

of I such that b > a. By the Mean Value Theorem,

=′ fc () ≤ 1; there is a point c between a and b such that

Since a < b, f(b) ≥ f(a), so f is nondecreasing.

46. Let d be the difference in distance between horse

A and horse B as a function of time t. of I such that b > a. By the Mean Value Theorem,

Suppose fx ′ () ≤ 0. Let a and b lie in the interior

Then d′ is the difference in speeds. there is a point c between a and b such that

Let t 0 and t 1 and be the start and finish times of

the race.

b − a b − a dt () 0 = dt () 1 = 0

a < b, f(a) ≥ f(b), so f is nonincreasing.

By the Mean Value Theorem,

dt () 1 − dt ()

42. [ f ( )] x ′

= dc () for some c in (, ) t t . = 2() fxfx () t 01 1 − t 0

01 t . on I.

Because f(x) 2 ≥ 0 and fx ′ () ≥ 0 on , [ I f ( )] x ′ ≥ 0 Therefore dc ′ () = 0 for some c in (, ) t

As a consequence of the Mean Value Theorem,

47. Let s be the difference in speeds between horse A

f ( x 2 ) − f () x 1 ≥ 0 for all x 2 > x 1 on I.

and horse B as function of time t.

Then s′ is the difference in accelerations.

Therefore 2 f is nondecreasing.

Let t 2 be the time in Problem 46 at which the horses had the same speeds and let t

be the

43. Let f(x) = h(x) – g(x).

fx ′ () =′ hx () −′ gx ( ); ( ) fx ′ ≥ 0 for all x in

finish time of the race.

st () 2 = st () 1 = 0

(a, b) since gx ′ () ≤ hx ′ () for all x in (a, b), so f is By the Mean Value Theorem,

nondecreasing on (a, b) by Problem 41. Thus

st () 1 − st () 2

x 1 < x 2 ⇒ fx () 1 ≤ fx ( 2 ); = sc ′ () for some c in (, ) t 21 t .

t 1 − t 2 hx () 1 − gx () 1 ≤ hx ( 2 ) − gx ( 2 );

Therefore sc ′ () = 0 for some c in (, ) t 21 t .

gx ( 2 ) − gx () 1 ≤ hx ( 2 ) − hx () 1 for all x 1 and x 2

in (a, b).

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48. Suppose x > c. Then by the Mean Value

52. s is differentiable with s (0) = and 0 s (18) = 20 so Theorem,

fx () − fc () = fax ′ ( )( − c ) for some a in (, ) cx .

we can apply the Mean Value Theorem. There

exists a c in the interval ( 0,18 ) such that

Since f is concave up,

≈ ( 1.11 miles per minute −

Monotonicity Theorem f′ is increasing.

Therefore fa ′ () > fc ′ () and

66.67 miles per hour

fx () − fc () = fax ′ ( )( −> c ) fcx ′ ( )( − c ) fx () > fc () + fcx ′ ( )( − cx ), > c

Suppose x < c. Then by the Mean Value Theorem,

fc () − fx () = fac ′ ( )( − x ) for some in ( , ) a xc .

Since f is concave up,

f ′′ > 0 , and by the

Monotonicity Theorem f′ is increasing.

Therefore, fc ′ () > fa ′ () and

fc () − fx () = fac ′ ( )( − x ) < fcc ′ ( )( − x ) . − fx () <− fc () + fcc ′ ( )( − x ) fx () > fc () − fcc ′ ( )( − x )

fx () > fc () + fcx ′ ( )( − cx ), < c Therefore fx () > fc () + fcx ′ ( )( − cx ), ≠ c .

49. Fix an arbitrary x.

fy ()() − fx

() x = lim = 0 , since

fy ()() − fx

53. Since the car is stationary at t = , and since v is 0

≤ My − x .

continuous, there exists a δ such that vt () < 2

So, f ' ≡→= 0 f constant .

for all t in the interval [0, ] δ. vt () is therefore

and δ < ⋅ = . By the Mean δ s () δ

less than

50. fx () = x 1/ 3 on [0, a] or [–a, 0] where a is any

Value Theorem, there exists a c in the interval positive number. f′ (0) does not exist, but f(x)

( , 20) δ such that

has a vertical tangent line at x = 0.

vc () = sc '( ) = ⎝

51. Let f(t) be the distance traveled at time t.

By the Mean Value Theorem, there is a time c

20 − δ such that fc ′ () = 56. = 1 mile per minute

At some time during the trip, Johnny must have

60 miles per hour

gone 56 miles per hour.

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54. Given the position function

st 2 () =

at ++, bt c

Problem Set 3.7

the car’s instantaneous velocity is given by the

function st ' () = 2 at + b .

1. Let fx () = x 3 + x 2 – 6.

The midpoint of the interval [ AB , ] is

f (1) = –3, f(2) = 6

Thus, the car’s instantaneous velocity at the

midpoint of the interval is given by n fm ( n )

2 0.25 1.25 –1.546875 = aA ( + B ) + b 3 0.125 1.375 –0.650391

The car’s average velocity will be its change in

4 0.0625 1.4375 –0.154541 position divided by the length of the interval.

5 0.03125 1.46875 0.105927 That is,

sB ()() − sA ( aB

2 + bB +− c )( aA 2 + bA c +

B − A B − A 8 0.00390625 1.45703 0.00725670

aB 2 − aA 2 + bB bA −

B − A r ≈ 1.46

( − A ) + bB ( − A )

aB ( − A )( B + A )( + bB − A )

f (–1) = –3, f(0) = 1

aB + A + b h n m n fm ( ( ) n ) = aA ( + B ) + b 1 0.5 –0.5 0.4375

This is the same result as the instantaneous

2 0.25 –0.75 –0.792969 velocity at the midpoint.

3.7 Concepts Review

1. slowness of convergence

2. root; Intermediate Value

r ≈ –0.61

3. algorithms

3. Let fx () = 2 cos x − sin x .

4. fixed point

f () 1 ≈ 0.23913 ; f () 2 ≈− 1.74159

fm () n

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4. Let fx () =−+ x 2 2 cos x n x n

f () 1 =−+ 1 2 2 cos 1 () ≈ 0.080605 11

f () 2 =−+ 2 2 2 cos 2 () ≈− 0.832294 2 0.8636364

fm () n

7. Let fx () =−+ x 2 2 cos x .

f ' () x =− 1 2 sin x n

fx ′ () = 3 x 2 + 12 x + 9 1 4 2 3.724415

n x n 3 3.698429

8. Let fx () = 2 cos x − sin x .

f ' () x =− 2 sin x − cos x n

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9. Let f(x) = cos x – 2x. fx ′ () = 4 3 x 2 – 24 x + x 44 – 24 Note that f(2) = 0.

r = 2, r ≈ 0.58579, r ≈ 3.41421 r ≈ 0.45018

12. Let fx () = 4 x + 6 3 x 2 + 2 x + x 24 – 8.

10. Let fx () = 2 x − sin x − 1 .

f ' () x =− 2 cos x n x

r ≈ 0.88786 n x n

5 0.3166248 r ≈ –6.31662, r ≈ 0.31662

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13. Let

fx 2 () = 2

f ' () x = 4 x − cos x 47 ≈ 2.61833

17. fx = x 4 + () 3 x + x 2 + x is continuous on the

given interval.

() From the graph of f, we see that the maximum

value of the function on the interval occurs at the right endpoint. The minimum occurs at a stationary point within the interval. To find

where the minimum occurs, we solve f ' () x = 0

on the interval [ − 1,1 ] .

= 4 3 + 3 () 2 x x x + 2 x += 1 gx ()

Using Newton’s Method to solve gx () = 0 , we

get:

f ' x =− 2 csc 2 () x − 1 1 0 2 − 0.5

r ≈ 1.07687 Minimum: f ( − 0.60583 ) ≈− 0.32645

Maximum: f () 1 = 4

Section 3.7 Instructor’s Resource Manual

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18. fx =

() x

is continuous on the given

Minimum: f ( 4.493409 ) ≈− 0.21723

From the graph of f, we see that the maximum

Maximum: f ( 7.725252 ) ≈ 0.128375

and minimum will both occur at stationary points within the interval. The minimum appears to

20. fx () = x sin is continuous on the given

occur at about x =− 1.5 while the maximum

appears to occur at about x = 0.8 . To find the

interval.

stationary points, we solve f ' () x = 0 .

f ' () x =

= gx

Using Newton’s method to solve gx () = 0 on

From the graph of f, we see that the minimum value and maximum value on the interval will

the interval, we use the starting values of − 1.5 occur at stationary points within the interval. To

and 0.8 .

find these points, we need to solve f ' () x = 0 on

the interval.

() 2 x =

2 = gx ()

Using Newton’s method to solve gx () = 0 on

3 the interval, we use the starting values of π and

Maximum: f ( 0.692505 ) ≈ 1.08302

Minimum: f ( − 1.78436 ) ≈− 0.42032 n

19. fx () =

is continuous on the given interval.

Minimum: f ( 10.173970 ) ≈− 96.331841 Maximum: f ( 4.577859 ) ≈ 15.78121

From the graph of f, we see that the minimum value and maximum value on the interval will occur at stationary points within the interval. To

find these points, we need to solve f ' () x = 0 on

the interval.

f ' () x =

x cos x − sin x

2 = gx ()