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13.2.12: Chapter 12

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    117287
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    Try It

    12.1 Finding Limits: Numerical and Graphical Approaches

    1.

    a=5, a=5, f( x )=2 x 2 4, f( x )=2 x 2 4, and L=46. L=46.

    2.

    a. 0; b. 2; c. does not exist; d. 2; 2; e. 0; f. does not exist; g. 4; h. 4; i. 4

    3.

    lim x0 ( 20sin(x) 4x )=5 lim x0 ( 20sin(x) 4x )=5

    Table showing that f(x) approaches 5 from either side as x approaches 0 from either side.
    4.

    does not exist

    12.2 Finding Limits: Properties of Limits

    1.

    26

    2.

    59

    3.

    10

    4.

    64 64

    5.

    3 3

    6.

    1 50 1 50

    7.

    1 8 1 8

    8.

    2 3 2 3

    9.

    −1 −1

    12.3 Continuity

    1.
    1. removable discontinuity at x=6; x=6;
    2. jump discontinuity at x=4 x=4
    2.

    No. The function is not continuous at x=2x=2 because the left hand limit is 1212 and the right hand limit is 6.56.5.

    3.

    No, the function is not continuous at x=3. x=3. There exists a removable discontinuity at x=3. x=3.

    4.

    x=6 x=6

    12.4 Derivatives

    1.

    3

    2.

    f (a)=6a+7 f (a)=6a+7

    3.

    f (a)= 15 ( 5a+4 ) 2 f (a)= 15 ( 5a+4 ) 2

    4.

    3 2 3 2

    5.

    0

    6.

    2 , 2 , 0, 0, 3 3

    7.
    1. After zero seconds, she has traveled 0 feet.
    2. After 10 seconds, she has traveled 150 feet east.
    3. After 10 seconds, she is moving eastward at a rate of 15 ft/sec.
    4. After 20 seconds, she is moving westward at a rate of 10 ft/sec.
    5. After 40 seconds, she is 100 feet westward of her starting point.
    8.

    The graph of f f is continuous on ( ,1 )( 1,3 )( 3, ). ( ,1 )( 1,3 )( 3, ). The graph of f f is discontinuous at x=1 x=1 and x=3. x=3. The graph of f f is differentiable on ( ,1 )( 1,3 )( 3, ). ( ,1 )( 1,3 )( 3, ). The graph of f f is not differentiable at x=1 x=1 and x=3. x=3.

    9.

    y=19x16 y=19x16

    10.

    –68 ft/sec, it is dropping back to Earth at a rate of 68 ft/s.

    12.1 Section Exercises

    1.

    The value of the function, the output, at x=a x=a is f( a ). f( a ). When the lim xa f( x ) lim xa f( x ) is taken, the values of x x get infinitely close to a a but never equal a. a. As the values of x x approach a a from the left and right, the limit is the value that the function is approaching.

    3.

    –4

    5.

    –4

    7.

    2

    9.

    does not exist

    11.

    4

    13.

    does not exist

    15.

    Answers will vary.

    16.

    Answers will vary.

    17.

    Answers will vary.

    18.

    Answers will vary.

    19.

    Answers will vary.

    20.

    Answers will vary.

    21.

    Answers will vary.

    23.

    7.38906

    25.

    54.59815

    27.

    e 6 403.428794, e 6 403.428794, e 7 1096.633158, e 7 1096.633158, e n e n

    29.

    lim x2 f(x)=1 lim x2 f(x)=1

    31.

    lim x3 ( x 2 x6 x 2 9 )= 5 6 0.83 lim x3 ( x 2 x6 x 2 9 )= 5 6 0.83

    33.

    lim x1 ( x 2 1 x 2 3x+2 )=2.00 lim x1 ( x 2 1 x 2 3x+2 )=2.00

    35.

    lim x1 ( 1010 x 2 x 2 3x+2 )=20.00 lim x1 ( 1010 x 2 x 2 3x+2 )=20.00

    37.

    lim x 1 2 ( x 4 x 2 +4x+1 ) lim x 1 2 ( x 4 x 2 +4x+1 ) does not exist. Function values decrease without bound as x x approaches –0.5 from either left or right.

    39.

    lim x0 7tanx 3x = 7 3 lim x0 7tanx 3x = 7 3

    Table shows as the function approaches 0, the value is 7 over 3 but the function is undefined at 0.
    40.
    Table shows as the function approaches 4, the value does not exist since approaching the limit value from the left is negative infinity and approaching the limit value from the right is positive infinity.
    43.

    lim x0 e e 1 x 2 =1.0 lim x0 e e 1 x 2 =1.0

    45.

    lim x 1 | x+1 | x+1 = (x+1) (x+1) =1 lim x 1 | x+1 | x+1 = (x+1) (x+1) =1 and lim x 1 + | x+1 | x+1 = (x+1) (x+1) =1; lim x 1 + | x+1 | x+1 = (x+1) (x+1) =1; since the right-hand limit does not equal the left-hand limit, lim x1 | x+1 | x+1 lim x1 | x+1 | x+1 does not exist.

    47.

    lim x1 1 ( x+1 ) 2 lim x1 1 ( x+1 ) 2 does not exist. The function increases without bound as x x approaches 1 1 from either side.

    49.

    lim x0 5 1 e 2 x lim x0 5 1 e 2 x does not exist. Function values approach 5 from the left and approach 0 from the right.

    51.

    Through examination of the postulates and an understanding of relativistic physics, as vc, vc, m. m. Take this one step further to the solution,

    lim v c m= lim v c m o 1( v 2 / c 2 ) = lim v c m= lim v c m o 1( v 2 / c 2 ) =

    12.2 Section Exercises

    1.

    If f f is a polynomial function, the limit of a polynomial function as x x approaches a a will always be f( a ). f( a ).

    3.

    It could mean either (1) the values of the function increase or decrease without bound as x x approaches c, c, or (2) the left and right-hand limits are not equal.

    5.

    10 3 10 3

    7.

    6

    9.

    1 2 1 2

    11.

    6

    13.

    does not exist

    15.

    12 12

    17.

    5 10 5 10

    19.

    108 108

    21.

    1

    23.

    6

    25.

    1

    27.

    1

    29.

    does not exist

    31.

    6+ 5 6+ 5

    33.

    3 5 3 5

    35.

    0

    37.

    3 3

    39.

    does not exist; right-hand limit is not the same as the left-hand limit.

    41.

    2

    43.

    Limit does not exist; limit approaches infinity.

    45.

    4x+2h 4x+2h

    47.

    2x+h+4 2x+h+4

    49.

    cos(x+h)cos(x) h cos(x+h)cos(x) h

    51.

    1 x(x+h) 1 x(x+h)

    53.

    1 x+h + x 1 x+h + x

    55.

    f( x )= x 2 +5x+6 x+3 f( x )= x 2 +5x+6 x+3

    57.

    does not exist

    59.

    52

    12.3 Section Exercises

    1.

    Informally, if a function is continuous at x=c , x=c , then there is no break in the graph of the function at f( c ), f( c ), and f( c ) f( c ) is defined.

    3.

    discontinuous at a=3 a=3 ; f(3) f(3) does not exist

    5.

    removable discontinuity at a=4 a=4 ; f(4) f(4) is not defined

    7.

    Discontinuous at a=3 a=3 ; lim x3 f(x)=3 , lim x3 f(x)=3 , but f(3)=6 , f(3)=6 , which is not equal to the limit.

    9.

    lim x2 f(x) lim x2 f(x) does not exist.

    11.

    lim x 1 f(x)=4; lim x 1 + f(x)=1 lim x 1 f(x)=4; lim x 1 + f(x)=1 . Therefore, lim x1 f(x) lim x1 f(x) does not exist.

    13.

    lim x 1 f(x)=5 lim x 1 + f(x)=1 lim x 1 f(x)=5 lim x 1 + f(x)=1 . Thus lim x1 f(x) lim x1 f(x) does not exist.

    15.

    lim x 3 f(x)=6 lim x 3 f(x)=6 , lim x 3 + f(x)= 1 3 lim x 3 + f(x)= 1 3

    Therefore, lim x3 f(x) lim x3 f(x) does not exist.

    17.

    f( 2 ) f( 2 ) is not defined.

    19.

    f( 3 ) f( 3 ) is not defined.

    21.

    f( 0 ) f( 0 ) is not defined.

    23.

    Continuous on (,) (,)

    25.

    Continuous on (,) (,)

    27.

    Discontinuous at x=0 x=0 and x=2 x=2

    29.

    Discontinuous at x=0 x=0

    31.

    Continuous on (0,) (0,)

    33.

    Continuous on [4,) [4,)

    35.

    Continuous on (,) (,) .

    37.

    1, but not 2 or 3

    39.

    1 and 2, but not 3

    41.

    f( 0 ) f( 0 ) is undefined.

    43.

    (,0)(0,) (,0)(0,)

    45.

    At x=1, x=1, the limit does not exist. At x=1, x=1, f( 1 ) f( 1 ) does not exist.

    At x=2, x=2, there appears to be a vertical asymptote, and the limit does not exist.

    47.

    x 3 +6 x 2 7x ( x+7 )( x1 ) x 3 +6 x 2 7x ( x+7 )( x1 )

    49.

    The function is discontinuous at x=1 x=1 because the limit as x x approaches 1 is 5 and f( 1 )=2. f( 1 )=2.

    12.4 Section Exercises

    1.

    The slope of a linear function stays the same. The derivative of a general function varies according to x. x. Both the slope of a line and the derivative at a point measure the rate of change of the function.

    3.

    Average velocity is 55 miles per hour. The instantaneous velocity at 2:30 p.m. is 62 miles per hour. The instantaneous velocity measures the velocity of the car at an instant of time whereas the average velocity gives the velocity of the car over an interval.

    5.

    The average rate of change of the amount of water in the tank is 45 gallons per minute. If f( x ) f( x ) is the function giving the amount of water in the tank at any time t ,t, then the average rate of change of f( x ) f( x ) between t=a t=a and t=b t=b is f(a)+45(ba). f(a)+45(ba).

    7.

    f (x)=2 f (x)=2

    9.

    f (x)=4x+1 f (x)=4x+1

    11.

    f (x)= 1 (x2) 2 f (x)= 1 (x2) 2

    13.

    16 ( 3+2x ) 2 16 ( 3+2x ) 2

    15.

    f (x)=9 x 2 2x+2 f (x)=9 x 2 2x+2

    17.

    f (x)=0 f (x)=0

    19.

    1 3 1 3

    21.

    undefined

    23.

    f (x)=6x7 f (x)=6x7

    25.

    f (x)=9 x 2 +4x+1 f (x)=9 x 2 +4x+1

    27.

    y=12x15 y=12x15

    29.

    k=10 k=10 or k=2 k=2

    31.

    Discontinuous at x=2 x=2 and x=0. x=0. Not differentiable at –2, 0, 2.

    33.

    Discontinuous at x=5. x=5. Not differentiable at -4, –2, 0, 1, 3, 4, 5.

    35.

    f( 0 )=2 f( 0 )=2

    37.

    f( 2 )=6 f( 2 )=6

    39.

    f ( 1 )=9 f ( 1 )=9

    41.

    f ( 1 )=3 f ( 1 )=3

    43.

    f ( 3 )=9 f ( 3 )=9

    45.

    Answers vary. The slope of the tangent line near x=1 x=1 is 2.

    47.

    At 12:30 p.m., the rate of change of the number of gallons in the tank is –20 gallons per minute. That is, the tank is losing 20 gallons per minute.

    49.

    At 200 minutes after noon, the volume of gallons in the tank is changing at the rate of 30 gallons per minute.

    51.

    The height of the projectile after 2 seconds is 96 feet.

    53.

    The height of the projectile at t=3 t=3 seconds is 96 feet.

    55.

    The height of the projectile is zero at t=0 t=0 and again at t=5. t=5. In other words, the projectile starts on the ground and falls to earth again after 5 seconds.

    57.

    36π 36π

    59.

    $50.00 per unit, which is the instantaneous rate of change of revenue when exactly 10 units are sold.

    61.

    $21 per unit

    63.

    $36

    65.

    f'(x)=10a1 f'(x)=10a1

    67.

    4 ( 3x ) 2 4 ( 3x ) 2

    Review Exercises

    1.

    2

    3.

    does not exist

    5.

    Discontinuous at x=1( lim xa f(x)x=1( lim xa f(x) does not exist ),x=3(),x=3(jump discontinuity)), andx=7( lim xa f(x)x=7( lim xa f(x) does not exist). ).

    7.

    lim x2 f(x)=0 lim x2 f(x)=0

    9.

    Does not exist

    11.

    -35-35

    13.

    11

    15.

    66

    17.

    500500

    19.

    -67-67

    21.

    At x=4, x=4, the function has a vertical asymptote.

    23.

    At x=3, x=3, the function has a vertical asymptote.

    25.

    Removable discontinuity at a=9a=9

    27.

    Removable discontinuity at x=5x=5

    29.

    Removable discontinuity at x=5x=5, discontinuity at x=1x=1

    31.

    Removable discontinuity at x=-2x=-2, discontinuity at x=5x=5

    33.

    33

    35.

    1 (x+ 1)(x+h+1) 1 (x+ 1)(x+h+1)

    37.

    e2x+2h-e2x h e2x+2h-e2x h

    39.

    10x-3 10x-3

    41.

    The function would not be differentiable at however, 0 is not in its domain. So it is differentiable everywhere in its domain.

    Practice Test

    1.

    3

    3.

    0

    5.

    −1 −1

    7.

    lim x 2 f(x)= 5 2 a lim x 2 f(x)= 5 2 a and lim x 2 + f(x)=9 lim x 2 + f(x)=9 Thus, the limit of the function as x x approaches 2 does not exist.

    10.

    -150 -150

    12.

    1 1

    14.

    Removable discontinuity at x=3 x=3

    16.

    f'(x)=- 3 2a32 f'(x)=- 3 2a32

    18.

    Discontinuous at −2, 0, not differentiable at −2, 0, 2

    20.

    Not differentiable at x=0 x=0 (no limit)

    22.

    The height of the projectile at t=2 t=2 Seconds

    24.

    The average velocity from t=1 t=1 t=2 t=2

    26.

    13 13

    28.

    0 0

    29.

    2

    30.

    x=1 x=1

    32.

    y=14x18 y=14x18

    34.

    The graph is not differentiable at x=1 x=1 (cusp).

    36.

    f ' (x)=8x f ' (x)=8x

    38.

    f ' (x)= 1 ( 2+x ) 2 f ' (x)= 1 ( 2+x ) 2

    40.

    f ' (x)=3 x 2 f ' (x)=3 x 2

    42.

    f'(x)= 1 2 x1 f'(x)= 1 2 x1


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