MA119-A Applied Calculus for Business
2006 Fall
Homework 4 Solutions
Due 9/29/2006 10:30AM
2.4 #30 Find the limit
lim 4x5
20x2 + 2x + 1 .
x!0
[Solution]
20x2 + 2x + 1 = 4 (0)5
lim 4x5
x!0
2.4 #38 Find the limit
lim
x!2
[Solution]
lim
x!2
r
2x3 + 4
=
x2 + 1
2.4 #46 Find the limit
lim
x!a
r
s
20 (0)2 + 2 (0) + 1 = 1.
2x3 + 4
.
x2 + 1
2 (2)3 + 4
=
(2)2 + 1
r
20
= 2.
5
p
3
5f (x) + 3g (x)
by given lim f (x) = 3 and lim g (x) = 4.
x!a
x!a
[Solution]
lim
x!a
p
3
5f (x) + 3g (x) =
q
3
5 lim f (x) + 3 lim g (x) =
x!a
x!a
p
3
5
3+3
2.4 #52 Find the limit
lim
2x2
3x
x
x!0
if it exists.
[Solution]
lim
x!0
2x2
3x
x
= lim (2x
x!0
1
3) = 2 (0)
3=
3.
4 = 3.
2
2.4 #56 Find the limit
x+2
x!2 x
2
lim
if it exists.
[Solution]
According to the graph,
y
10
8
6
4
2
-5
-4
-3
-2
-1
1
2
3
-2
-4
-6
-8
-10
the limit does not exist (DNE).
2.4 #74 Find the limit
4x2 1
x!1 x + 2
lim
if it exists.
[Solution]
Since
lim 4 lim x12
4 x12
4x2 1
4
x!1
x!1
lim
= lim 1
=
= ,
2
1
2
x!1 x + 2
x!1
0
+ x2
lim x + lim x2
x
x!1
the limit does not exist (DNE).
2.4 #76 Find the limit
2x2 + 3x + 1
x!1
x4 x2
lim
if it exists.
[Solution]
x!1
4
5
x
3
The limit is
2x2 + 3x + 1
lim
=
x!1
x4 x2
lim
2x2
x4
+
x4
x4
x!1
1
2
x!1 x
2 lim
=
3x
x4
+
1
x4
x2
x4
=
1
3
x!1 x
+ 3 lim
lim 1
x!1
lim 22
x!1 x
+
3
3
x!1 x
+ lim
lim 1
x!1
lim 14
x!1 x
=
lim 12
x!1 x
1
4
x!1 x
lim 12
x!1 x
+ lim
2 (0) + 3 (0) + 0
= 0.
1 0
2.4 #78 Find the limit
4x4 3x2 + 1
x!1 2x4 + x3 + x2 + x + 1
lim
if it exists.
[Solution]
The limit is
4
4x4 3x2 + 1
= lim
1
4
3
2
x!1 2x + x + x + x + 1
x!1 2 +
+
x
lim
3
+ x14
x2
1
+ x13 + x14
x2
=
4
= 2.
2
2.4 #84 Concentration of a Drug in the Bloodstream The concentration of a certain drug
in a patient’s bloodstream t hr after injection is given by
0:2t
C (t) = 2
t +1
3
mg/cm . Evaluate lim C (t) and interpret your result.
x!1
[Solution]
We have
0:2
0:2t
0
t
=
lim
=
= 0.
1
x!1 t2 + 1
x!1 1 + 2
1
+
0
t
lim C (t) = lim
x!1
So, the concentration of a certain drug in a patient’s bloodstream goes to zero when
time goes in…nity. This means that the drug will eventually disappear from the patient’s
blood.
2.5 #24 Find the one-side limit
lim+
x!1
if it exists.
[Solution]
The limit is
lim+
x!1
x+2
x+1
x+2
1+2
3
=
= .
x+1
1+1
2
2.5 #32 Find the one-side limit
lim + x 1 +
x! 5
if it exists.
[Solution]
p
5+x
4
p Since we approach 5 from the right side (all more then 5), we are able to calculate
5 + x. Thus, the limit is
p
p
lim+ x 1 + 5 + x = ( 5) 1 + 5 + ( 5) = 5.
x!5
2.5 #38 Find the one-side limits lim+ f (x) and lim f (x), where
x!0
x!0
f (x) =
x + 1 if x 0
2x + 3 if x > 0
if it exists.
[Solution]
We have
lim f (x) = lim+ (2x + 3) = 2 (0) + 3 = 3
x!0+
x!0
and
lim f (x) = lim+ ( x + 1) =
x!0
x!0
(0) + 1 = 1.
x 1
2.5 #52 Find the values of x for which the function f (x) = x2 +2x
is continuous.
3
[Solution]
x 1
If x2 + 2x 3 = 0, then f is not de…ned. Otherwise, x2 +2x
is a real number. So, f is
3
continuous except x = 3 or 1. (x2 + 2x 3 = (x + 3) (x 1).)
2.5 #56 Find the values of x for which the function f (x) = jxx 11j is continuous.
[Solution]
Note that jx 1j is continuous everythere (even x = 1) and x 1 is continuous everywhere. So, f (x) = jxx 1j
is continuous everywhere except x 1 = 0, or, x = 1.
1
1
2.5 #58 Determine all values of x at which the function f (x) = (x 1)(x
is discontinuous.
2)
[Solution]
Since f is not de…ned when x = 1 or 2, f is not continuous there.
1
For x 6= 1 and 2, (x 1) (x 2) is continuous. Thus, f (x) = (x 1)(x
is continuous
2)
except x = 1 or 2.
So, f is discontinuous only at x = 1 or 2.
2.5 #74 Let
x2 4
x+2
if x 6= 2
.
k
if x = 2
For what value of k will f be continuous on ( 1; 1)?
[Solution]
If a 6= 2, f (a) is de…ned and lim f (x) = f (a). Thus, f is continuous at x = a.
x!a
Since
x2 4
(x + 2) (x 2)
lim f (x) = lim
= lim
= lim x 2 = 4,
x! 2
x! 2 x + 2
x! 2
x! 2
x+2
we know that the limit exists at x = 2.
f (x) =
5
If f is continuous at x =
Therefore, if k =
2, lim f (x) must be equal to f (2) = k. Thus, k =
x! 2
4.
4, then f is continuous on ( 1; 1).
2.5 #78 Let f (x) = 2x3 3x2 36x + 14. Also, let a = 0 and b = 1.
(a) Show that the function f is continuous for all values of x in the interval [a; b].
(b) Prove that f must have at least one zero in the interval (a; b) by showing that f (a)
and f (b) have opposite signs.
[Solution]
(a) Since f is a polynomial, f is continuous in [0; 1].
(b) Note that f (0) = 14 and f (1) = 23. Since f (1) = 23 < 0 < 14 = f (0), by the
Intermediate Value Theorem, there is a real number c 2 (0; 1) such that f (c) = 0.
So, f must have at least one zero in the interval (a; b).
2.5 #82 Use the Intermediate Value Theorem to …nd the value of c such that f (c) = M , where
f (x) = x2 x + 1 on [ 1; 4] and M = 7.
[Solution]
We have that f ( 1) = 3 and f (4) = 13. Since f ( 1) = 3 < M < 3 = f (4), by the
Intermediate Value Theorem, there is a real number c 2 ( 1; 4) such that f (c) = M = 7.
Thus, we have 7 = f (c) = c2 c + 1, that is, c2 c 6 = 0. This tells us that c = 3 or
2 (since 0 = c2 c 6 = (c 3) (c + 2).)
2.6 #12 Use the four-step process to …nd the slope of the tangent line of the graph of the function
1 2
f (x) =
x
2
at any point.
[Solution]
By the four-step process,
(1) f (x + h) = 12 (x + h)2 = 12 (x2 + 2xh + h2 ) = 21 x2 xh 12 h2 .
1 2
1 2
(2) f (x + h) f (x) =
x
xh 21 h2
x = xh 21 h2 .
2
2
xh
1
h2
2
(3) f (x+h)h f (x) =
= x
h
f (x+h) f (x)
0
(4) f (x) = lim
= lim
h
h!0
h!0
1
h.
2
x
Thus, the slope of the tangent line is
1
h
2
=
x.
x.
2.6 #16 Use the four-step process to …nd the slope of the tangent line of the graph of the function
f (x) = 2x2 + 5x
at any point.
[Solution]
By the four-step process,
(1) f (x + h) = 2 (x + h)2 + 5 (x + h) = 2 (x2 + 2xh + h2 ) + 5 (x + h) = 2x2 + 5x + 4xh +
5h + h2 .
(2) f (x + h) f (x) = (2x2 + 5x + 4xh + 5h + h2 ) (2x2 + 5x) = 4xh + 5h + h2 .
2
= 4x + 5 + h.
(3) f (x+h)h f (x) = 4xh+5h+h
h
f (x+h) f (x)
0
(4) f (x) = lim
= lim (4x + 5 + h) = 4x + 5.
h
h!0
h!0
6
Thus, the slope of the tangent line is 4x + 5.
2.6 #22 Find the slope of the tangent line of the graph of the function
f (x) =
3
2x
at the point 1; 32 and determine an equation of the tangent line.
[Solution]
By the four-step process at x = 1,
3
(1) f (1 + h) = 2(1+h)
.
(2) f (1 + h)
(3)
f (1+h) f (1)
h
(4) f 0 (1) =
3
2(1+h)
f (1) =
3
2(1)
3
2(1+h)
3
2
=
3 3(1+h)
2(1+h)
=
3
2(1+h)
=
3
.
2
Thus, the slope of the tangent line at the point 1; 23 is
of the tangent line is
3
(x
2
3
=
2
y
3
.
2
Therefore, the equation
1) .
2.6 #26 Let f (x) = x 1 1 .
(a) Find the derivative f 0 of f .
(b) Find an equation of the tangent line to the curve at the point
(c) Sketch the graph of f .
[Solution]
(a) By the four-step process,
1
(1) f (x + h) = (x+h)
= x+h1 1 .
1
(2) f (x + h)
(3)
3h
.
2(1+h)
3h
2(1+h)
3
= 2(1+h)
.
h
lim f (1+h)h f (1) = lim
h!0
h!0
=
=
f (x+h) f (x)
h
f (x) =
=
(4) f 0 (x) = lim
h!0
(b) The slope at x =
1
x+h 1
h
(x+h 1)(x 1)
h
f (x+h) f (x)
h
0
=
=
(x 1) (x+h 1)
(x+h 1)(x 1)
=
1
2
.
h
.
(x+h 1)(x 1)
1
.
(x+h 1)(x 1)
1
1
= (x 1)(x
=
(x+h 1)(x 1)
1)
1
1
= 4 . Thus, the equation
(( 1) 1)2
= lim
1 is f (1) =
1
is y
= 14 (x
2
(c) The graph of f is
1
x 1
1;
h!0
( 1)), or, x + 4y + 3 = 0.
1
.
(x 1)2
of the tangent
7
y
2
1
-5
-4
-3
-2
-1
1
2
3
4
5
x
-1
-2
2.6 #30 Velocity of a Ball Thrown into the Air A ball is thrown straight up with an initial
velocity of 128 ft/sec, so that its height (in feet) after t sec is given by s (t) = 128t 16t2 .
(a) What is the average velocity of the ball over the time intervals [2; 3], [2; 2:5], and
[2; 2:1].
(b) What is the instantaneous velocity at time t = 2?
(c) What is the instantaneous velocity at time t = 5? Is the ball rising or falling at this
time?
(d) When will the ball hit the ground?
[Solution]
(a) First, we have s (2) = 128 (2) 16 (2)2 = 192, s (3) = 240, s (2:5) = 220, and
s (2:1) = 198:24.
So, the average velocity over [2; 3] is f (3)3 2f (2) = 2403 2192 = 48, the average velocity
f (2)
192
over [2; 2:5] is f (2:5)
= 220
= 56, and the average velocity over [2; 2:1] is
2:5 2
2:5 2
f (2:1) f (2)
198:24 192
= 2:1 2 = 62:4.
2:1 2
(b) s0 (2) = lim s(2+h)h s(2) = 128 32 (2) = 64.
h!0
(c) s0 (5) = lim
h!0
s(5+h) s(5)
h
= 128
32 (5) =
32. So, the ball is falling.
(d) The ground is height 0. So, let s (t) = 0. We can solve t from 128t 16t2 = 0. This
tells us that t = 0 or 8. Therefore, we know that after 8 seconds, the ball hits the
ground.
2.6 #34 Cost of Producing Surfboards The total cost C (x) (in dollars) incurred by Aloha
Company in manufacturing x surfboards a day is given by
C (x) =
10x2 + 300x + 130
(0
x
15)
(a) Find C 0 (x).
(b) What is the rate of change of the total cost when the level of production is ten
turfboards a day?
[Solution]
8
(a) By the four-step process,
(1) C (x + h) = 10 (x + h)2 + 300 (x + h) + 130 = 10x2 20xh 10h2 + 300x +
300h + 130.
(2) C (x + h) C (x) = ( 10x2 20xh 10h2 + 300x + 300h + 130) ( 10x2 + 300x + 130)
= 20xh + 300h 10h2 .
10h2
= 20x + 300 10h.
(3) C(x+h)h C(x) = 20xh+300h
h
C(x+h) C(x)
0
(4) C (x) = lim
= lim ( 20x + 300 10h) = 20x + 300.
h
(b) C 0 (10) =
h!0
h!0
20 (10) + 300 = 100.
2.6 #48 The graph of a function is shown at page 151. Whether or not
(a) f (x) has a limit at x = a,
(b) f (x) is continuous at x = a,
(c) f (x) is di¤erentiable at x = a.
Justify your answer.
[Solution]
(a) By the graph, the right-side limit at x = a is positive and the left-side limit at x = a
is negative. They cannot be equal. Thus, f does not have a limit at x = a.
(b) Since f does not have a limit at x = a, f is not continuous at x = a.
(c) Since f is not continuous at x = a, f is not di¤erentiable at x = a.
2.6 #50 The graph of a function is shown at page 151. Whether or not
(a) f (x) has a limit at x = a,
(b) f (x) is continuous at x = a,
(c) f (x) is di¤erentiable at x = a.
Justify your answer.
[Solution]
(a) By the graph, the right-side limit at x = a is in…nity and the left-side limit at x = a
is negative in…nity. They cannot be equal. Thus, f does not have a limit at x = a.
(b) Since f does not have a limit at x = a, f is not continuous at x = a.
(c) Since f is not continuous at x = a, f is not di¤erentiable at x = a.
2
2.6 #58 Sketch the graph of the function f (x) = x 3 . Is the function continuous at x = 0? Does
f 0 (0) exist? Why or why not?
[Solution]
The graph of f is
9
y
3
2
1
-5
-4
-3
-2
-1
1
2
3
4
5
x
-1
-2
-3
From the graph, we can see that lim f (x) = 0 and lim+ f (x) = 0. So, we have
x!0
x!0
lim f (x) = 0 = f (0). Thus, f is continuous at x = 0.
x!0
f (x) f (0)
.
x 0
x!0
2
f (x) f (0)
x3
=
lim
x
x
x!0
For the derivative, we need to calculate lim
First, let us see the left-side limit lim
p
3
p
3
x!0
x < 0. So the limit is
1.
Similarly, the right-side limit lim+
x!0
f (x) f (0)
x
= lim
x!0
2
= lim+
x!0
x3
x
= lim+
x!0
1
p
3 x.
1
p
3 x.
When x < 0,
When x > 0,
x > 0. So the limit is 1.
Therefore, lim f (x)x 0f (0) does not exists. This tells us that f 0 (0) does not exist.
x!0