Download L. 2.4 Solutions.tst

Calculus
Lesson 2.4 Solutions
2-4, 8-12, 15-16
In Exercises 2-4, find the average rate of change of the function over each interval.
2) f(x) = 4x + 1
(a) [0, 2] (b) [10, 12]
3) f(x) = ex
(a) [-2, 0] (b) [1, 3]
4) f(x) = ln x
(a) [1, 4] (b) [100, 103]
In Exercise 8, a distance-time graph is shown.
(a) Estimate the slopes of the secants PQ1, PQ2, PQ3, and PQ4, arranging them in order in a
table. What is the appropriate unit for these slopes?
(b) Estimate the speed at point P.
8)
The accompanying figure shows a distance-time graph for a wrench that fell from
the top platform of a communication mast on the moon to the station roof 80
meters below.
P
80
Q4
Q3
60
Q2
40
20
Q1
5
10
1
In Exercises 9-12, at the indicated point, find
(a) the slope of the curve
(b) an equation of the tangent, and
(c) an equation of the normal line
(d) Then draw a graph of the curve, tangent line, and normal line in the same square
viewing window.
9) y = x2 at x = -2
10)
y = x2 - 4x, at x = 1
11)
y=
12)
y = x2 - 3x - 1, at x = 0
1
, at x = 2
x-1
In Exercises 15-16, determine whether the curve has a tangent at the indicated point. If it
does, give its slope. If not, explain why not.
2
15) f(x) = 2 - 2x - x ,
2x + 2,
16) f(x) =
-x,
x2 - x,
x<0
x≥0
x<0
x≥0
at x = 0
at x = 0
2
Answer Key
Testname: L. 2.4 SOLUTIONS
2)
9- 1
=1
2
(a)
f(2) - f(0)
=
2-0
(b)
f(12) - f(10)
=
12 - 10
49 - 41 7 - 41
‘ 0.298
=
2
2
3)
(a)
f(0) - f(-2) e0 - e(-2) 1 - e(-2)
‘ 0.432
=
=
2
2
0 - (-2)
(b)
f(3) - f(1) e3 - e(1)
‘ 8.684
=
3-1
2
4)
(a)
f(4) - f(1) ln 4 - ln 1 ln 4
‘ 0.462
=
=
3
3
4-1
(b)
f(103) - f(100) ln 103 - ln 100
‘ 0.010
=
103 - 100
3
8)
80 - 20
‘ 12.000
(a) PQ1 =
10 - 5
80 - 39
‘ 13.667
PQ2 =
10 - 7
80 - 57
‘ 15.333
PQ3 =
10 - 8.5
80 - 71
‘ 18.000
PQ4 =
10 - 9.5
What is the appropriate unit for these slopes? meters per second
(b) Estimate the speed at point P. Approximately 18 meters per second
3
Answer Key
Testname: L. 2.4 SOLUTIONS
9)
a) m = lim
h¬0
f(-2 + h) - f(-2)
h
m = lim
h¬0
(-2 + h)2 - (-2)2
h
m = lim
h¬0
4 + -4h + h 2 - 4
h
m = lim
h¬0
-4h + h 2
= lim -4 + h = -4
h
h¬0
b) when x = -2, y = 4
y = mx + b
4 = (-4)(-2) + b
4=8+b
b = -4
y = -4x - 4
c) m =
1
4
y-4=
1
(x + 2)
4
y-4=
1
1
x+
4
2
y=
1
1
x+4
4
2
d)
7
y
6
5
4
3
2
1
-7 -6 -5 -4 -3 -2 -1
-1
1
2
3 x
-2
-3
4
Answer Key
Testname: L. 2.4 SOLUTIONS
10)
a) m = lim
h¬0
f(1 + h) - f(1)
h
m = lim
h¬0
(1 + h)2 - 4(1 + h) - ((1)2 - 4(1))
h
m = lim
h¬0
1 + 2h + h2 - 4 - 4h - 1 + 4
h
m = lim
h¬0
2h + h2 - 4h
= lim 2 + h - 4 = -2
h
h¬0
b) when x = 1, y = -3
y = mx + b
-3 = (-2)(1) + b
b = -1
y = -2x - 1
c) m =
1
2
y+3=
1
(x - 1)
2
y+3=
1
1
x2
2
y=
1
1
x-3
2
2
d)
6
y
5
4
3
2
1
-5 -4 -3 -2 -1
-1
1
2
3
4
5
x
-2
-3
-4
-5
-6
5
Answer Key
Testname: L. 2.4 SOLUTIONS
6
y
6
Answer Key
Testname: L. 2.4 SOLUTIONS
11)
a) m = lim
h¬0
m = lim
h¬ 0
f(2 + h) - f(2)
h
1
1
2+h-1 2-1
h
1
-1
1+h
m = lim
h¬0
h
1
1+h
1
1
m = lim
= lim
h¬0 (1 + h)h h
h¬0 (1 + h)h (1 + h)h
-h
-1
= lim
= lim
= -1
h¬0 (1 + h)h
h¬0 (1 + h)
b) when x = 2, y = 1
y = mx + b
1 = (-1)(2) + b
b=3
y = -1x + 3
c) m = 1
y - 1 = 1(x - 2)
y - 1 = 1x - 2
y=x-1
d)
6
y
5
4
3
2
1
-5 -4 -3 -2 -1
-1
1
2
3
4
5
x
-2
-3
-4
-5
-6
7
Answer Key
Testname: L. 2.4 SOLUTIONS
6
y
8
Answer Key
Testname: L. 2.4 SOLUTIONS
12)
a) m = lim
h¬0
f(0 + h) - f(0)
h
m = lim
h¬0
(h2 - 3(h) - 1) - (02 - 3(0) - 1)
h
m = lim
h¬0
h2 - 3h - 1 + 1
h
m = lim
h¬0
h2 - 3h
= lim h - 3 = -3
h
h¬0
b) when x = 0, y = -1
y = mx + b
-1 = (-3)(0) + b
b = -1
y = -3x - 1
c) m =
1
3
y+1=
1
(x - 0)
3
y+1=
1
x
3
y=
1
x-1
3
d)
6
y
5
4
3
2
1
-5 -4 -3 -2 -1
-1
1
2
3
4
5
x
-2
-3
-4
-5
-6
9
Answer Key
Testname: L. 2.4 SOLUTIONS
6
y
5
(2 - 2(h + 0) - (h + 0)2) - (2 - 2(0) - (0)2)
= lim
h
h¬0=
lim
h ¬ 0-
=
lim
h¬0+
=
lim
h ¬ 0+
(2 - 2h - h2) - 2
= -2 - h = -2
h
(2(h + 0) + 2) - (2(0) + 2)
h
(2h + 2) - 2)
=2
h
The tangent line does not exist. The slope from the left is different than the slope from the
right.
6
y
5
4
3
2
1
-5 -4 -3 -2 -1
-1
1
2
3
4
5
x
-2
-3
-4
-5
-6
10
Answer Key
Testname: L. 2.4 SOLUTIONS
16)
=
lim
h ¬ 0-
=
lim
h ¬ 0+
=
lim
h¬0+
-(h + 0) - (-0)
= -1
h
((h + 0)2 - (h + 0)) - ((0)2 - 0)
h
h2 - h
=
h
6
lim h - 1 = -1
h¬0+
y
5
4
3
2
1
-5 -4 -3 -2 -1
-1
1
2
3
4
5
x
-2
-3
-4
-5
-6
11