MATH 110, Assignment 2
Make sure you justify all your work and include complete arguments and explanations. Please include your name and student number on the top of the first page of your assignment.
Problem 1.
Let f ( x ) = x
1+ x
.
(i) Find the rates of change of f ( x ) in the following intervals:
[0 .
5 , 1] , [0 .
9 , 1] , [0 .
99 , 1] , [1 , 1 .
01] [1 , 1 .
1] , [1 , 1 .
5] .
(You can use calculator for this equation.)
Solution.
Note that the rate of change of f ( x ) in some interval [ a, b ] is given by f ( b ) f ( a )
.
b a
By this formula, the answers are followings: f (1) f (0 .
5)
0 .
5
= 0 .
333 · · · .
f (1) f (0 .
9)
0 .
1 f (1) f (0 .
99)
= 0 .
263 · · · .
0 .
01
= 0 .
251 · · · .
f (1 .
01) f (1)
0 .
01
= 0 .
248 · · · .
f (1 .
1) f (1)
0 .
1
= 0 f (1 .
5) f (1)
.
238 · · · .
= 0 .
2 .
0 .
5
(ii) Using the result of (i), guess the rate of change of f ( x ) at x = 1.
Solution.
Just guess any number between 0 .
248 and 0 .
251.
(iii) Sketch the graph of f ( x ) and interpret the geometric meanings of the result of (i) and
(ii) by describing on the graph.
Solution.
First, let’s rewrite f : f ( x ) = x
1 + x
=
1 + x 1
= 1
1 + x
1
1 + x
.
By the rewritten form, we can notice that the graph of f is given by shifting the graph of 1 x left by 1 and up by 1. Thus, the graph is the following:
1

The geometric meaning of the rate of change in some interval [ a, b ] is the slope of the secant line passing through ( a, f ( a )) and ( b, f ( b )). Also, the rate of change at some point c is interpreted as the slope of the tangent line at c . By drawing the secant lines and the tangent line, we can see the slope of the secant line through ( b, f ( b )) and
(1 , f (1)) approaches the slope of the tangent line at x = 1 as b approaches 1.
(iv) Express the rate of change of f ( x ) at 1 by using limit and find the limit.
Solution.
the rate of change of f ( x ) at 1 = lim h !
0 f (1 + h ) f (1) h
2(1+ h ) (2+ h )
2(2+ h )
= lim h !
0 h
= lim
= lim h !
0 h !
0 h
2(2+ h ) h
1+ h
2+ h h
1
2
= lim h !
0
=
1
1
2(2 + h )
4
= 0 .
25
Problem 2.
Find each limit, or explain why it does not exist. (If it is either infinity or negative infinity,
(a) lim x !
3
(2 x + | x 3 | ) .
Solution.
The right-hand limit: lim x !
3+
(2 x + | x 3 | ) = lim x !
3+
(2 x + x 3) = lim x !
3+
(3 x 3) = 3 ⇥ 3 3 = 6 .
The left-hand limit: lim x !
3
(2 x + | x 3 | ) = lim x !
3
(2 x x + 3) = lim x !
3
( x + 3) = 3 + 3 = 6 .
Thus, the limit is given by lim x !
3
(2 x + | x 3 | ) = 6.
2

(b) lim x !
2 x 2 x 2 x 2 4 x +4
.
Solution.
By direct substitution ( x = 2): x 2 x 2 x 2 4 x + 4
=
4 2 2
4 8 + 4
=
0
0
.
Thus, factor the numerator and denominator: x
2 x 2 = ( x 2)( x + 1) , x
2
4 x + 4 = ( x 2)
2
.
and cancel the common factors: lim x !
2 x x 2
2
4 x x
2
+ 4
= lim x !
2
( x 2)( x + 1)
( x 2) 2
= lim x !
2 x + 1 x 2
.
Now the problem is reduced to find lim x !
2 x +1 x 2
. Again, by direct substitution ( x = 2): x + 1 x 2
=
3
0
.
Thus, the left-hand and right-hand limits are either 1 or 1 . Since x !
2 means that x is very close to 2 and x < 2, the sign of the left-hand limit is sign
✓ x + 1
◆ x 2
= positive negative
= negative .
As the result, lim x !
2 x + 1 x 2
= 1 .
Similarly, since x !
2+ means that x is very close to 2 and x > 2, the sign of the right-hand limit is
✓ x + 1 sign
◆
= positive
= positive .
x 2 positive
As the result, lim x !
2+ x + 1 x 2
= + 1 .
Problem 3.
Is there a number a such that lim x !
2
3 x 2 x
+ ax + a + 3
2 + 3 x + 2 exists? If so, find the value of a and the value of the limit.
3

Solution.
By direct substitution ( x = 2), we get
3 x 2 + ax + a + 3 x 2 + 3 x + 2
=
12 2 a + a + 3
4 6 + 2
=
15 a
.
0
To make the limit exist, 15 a = 0 and hence a = 15.
When a = 15, we get lim x !
2
3 x 2 x
+ ax + a + 3
2 + 3 x + 2
=
=
= lim x !
2 lim x !
2
3 ⇥ 1
1
3
( x x
2 + 15 x + 15 + 3
3( x 2 x 2 + 3
+ 5 x x + 2
+ 6)
+ 1)( x + 2)
= lim
= lim x !
2 x !
2
3(
( x x
3 x x
2
2
+ 2)(
+ 1)(
+ 15
+ 3 x x x x
+ 3)
+ 2)
+ 18
+ 2
= lim x !
2
= 3 .
3( x + 3) x + 1
Problem 4.
Sketch the graph of a function f ( x ) defined on [ 5 , 5] which satisfies the following conditions:
• f ( 5) = 1
5 and f (5) = 15.
• lim x !
0 f ( x ) = 1 .
• The x -intercepts of f ( x ) are 0 and 2.
• The rate of change of f ( x ) 1 and 0 at -1 and 1, respectively.
• f is continuous on the domain except at x = 0.
Solution.
4

The LHS picture is for the corrected problem. The RHS picture is for the original problem.
Problem 5.
For what value of the constant c is the function f continuous on ( 1 , 1 )?
f ( x ) =
( cx 2 x 3
+ 2 x x < 2 c | x | x 2 .
Solution.
• The function value at 2: f (2) = 2 3
• The left-hand limit as x !
2 ( x < 2): c | 2 | = 8 2 c.
lim x !
2 f ( x ) = lim x !
2 cx 2 + 2 x = 4 c + 4 .
• The right-hand limit as x !
2 ( x > 2): lim x !
2+ f ( x ) = lim x !
2+ x
3 c | x | = 8 2 c.
To make f continuous, we demand
8 2 c = 4 c + 4 .
Thus, 6 c = 4 and hence c = 2
3
.
Problem 6.
Solution.
Since f
Prove that there is a root of
(1) = 1
Let
2 f ( x ) = x 2
1 + 1 = 1 p
+ 1. Then,
2 < intermediate value theorem, there exists x 2 =
0 and x 2 (1 , f p f ( x x
+ 1 in (1
(2) = 2 2
2) such that
, p f
2).
(
2 + 1 = 4 x ) = 0. i,e,.
p x 2
3 >
=
, 1
3, by the x + 1.
).
5