Exercise 2.5 P.128 a Sketch the graph of the function.

Exercise 2.5 P.128 a Sketch the graph of the function.

Exercise 2.5
P.128
20. Explain why the function is discontinuous at the given number a .
Sketch the graph of the function.
{ x2 −x
, if x ̸= 1
x2 −1
f (x) =
a=1
1
, if x = 1
< pf >
∵ limx→1+ f (x) = limx→1− f (x) = limx→1
∴ f (x) is discontinuous at x = 1
x2 −x
x2 −1
= limx→1
x
x+1
=
1
2
̸= f (1) = 1
30. Explain, using Theorems 4, 5, 7, and 9, why the function is continuous at every number in its
domain. State the domain.
tan x
B(x) = √
4 − x2
< pf >
(2k−1)π (2k+1)π
),∀k∈Z
(1) tan
2
√ x is continuous in any interval ( 2 ,
(2) x is continuous at every x ≥ 0
(3) polynomial 4 − x2 is continuous at every x ∈ R
By (1),(2),(3) B(x) is continuous at every x ∈ (0, π2 )
38. Use continuity to evaluate the limit
lim arctan(
x→2
x2 − 4
)
3x2 − 6x
< pf >
4
2
−4
lim arctan( 3xx2 −6x
) = arctan( lim (x−2)(x+2)
3x(x−2) ) = arctan( 6 ) = arctan( 3 )
2
x→2
x→2
P.129
42. Find the numbers at which f is discontinuous. At which of these numbers is f continuous from
the right, from the left, or neither? Sketch the graph of f .

, if x ≤ 1
 x+1
1
,
if 1 < x < 3
f (x) =
x
 √
x − 3 , if x ≥ 3
< pf >
∵(1) x + 1 is continuous everywhere
(2) x1 is continuous at every x ̸= 0
√
(3) x − 3 is continuous at every x ≥ 3
∴we only need to check x = 1 ∧ 3
lim f (x) = lim+ x1 = 1 ̸= lim− f (x) = lim− (x + 1) = 2
x→1+
x→1
x→1
x→1 √
lim f (x) = lim x1 = 13 ̸= lim f (x) = lim x − 3 = 0
x→3−
x→3−
x→3+
⇒ f (x) is discontinuous at x = 1 ∧ 3
x→3+
1
46. Find the values of a and b that make f continuous everywhere.
 2
−4
, if x < 2
 xx−2
f (x) =
ax2 − bx + 3 , if 2 ≤ x < 3

2x − a + b
, if x ≥ 3
< pf >
It’s easy to check f (x) continuous at every x ̸= 2 ∧ 3
So we only need to check x = 2 ∧ 3
if f (x) is continuous at everywhere
then lim f (x) = f (2)∧ lim f (x) = f (3)
x→2 
x→3
 lim f (x) = lim ax2 − bx + 3 = 4a − 2b + 3
x→2+
x→2+
2
lim f (x) =
−4
x→2
 lim f (x) = lim xx−2
= lim− (x−2)(x+2)
=4
x−2
x→2−
x→2−
x→2
{
lim f (x) = lim+ 2x − a + b = 6 − a + b
x→3+
x→3
lim f (x) =
lim− f (x) = lim− ax2 − bx + 3 = 9a − 3b + 3
x→3
x→3
x→3
{
{
4a − 2b + 3 = 4
a = 12
⇒
⇒
9a − 3b + 3 = 6 − a + b
b = 12
54. Use the Intermediate Value Theorem to show that there is a root of the given equation in the
specified interval.
sin x = x2 − x, (1, 2)
< pf >
sin x and x2 − x are both continuous at (1, 2)
Define f (x) = sin x − (x2 − x)
then f (1) = sin 1 − (1 − 1) = sin 1 > 0
f (2) = sin 2 − (4 − 2) = sin 2 − 2 < 0
∵ f (1)f (2) < 0
∴By I.V.T there exist a c ∈ (1, 2) s.t f (c) = 0
64. For what values of x is g continuous
{
g(x) =
0 , if x is rational
x , if x is irrational
< pf >
Claim : g(x) is continuous only at x = 0
Given ε > 0 take δ = ε > 0
if 0 <{|x − 0| < δ = ε
|g(x) − 0| = 0 < δ = ε
, if x is rational
then
|g(x) − 0| = |x| < δ = ε , if x is irrational
⇒ lim g(x) = 0 = g(0)
x→0
For any c ̸= 0
(1) if c is irrational
Take ε = 2c then∀ δ > 0
if 0 <
{ |x − c| < δ
|g(x) − g(c)| = c > 2c = ε
then
|g(x) − g(c)| = |x − c| < δ
(2) if c is rational
, if x is rational
, if x is irrational
2
Given ε > 0 take δ = ε
if 0 <
{ |x − c| < δ
|g(x) − g(c)| = 0 < ε
, if x is rational
then
|g(x) − c| = |x − c| < δ = ε , if x is irrational
⇒ lim g(x) = 0 ̸= lim g(x) = c
x→c, x∈Q
/
x→c, x∈Q
By (1) , (2) lim g(x) does not exist at every c ̸= 0
x→c
⇒ g(x) is discontinuous at every x ̸= 0
65. Is there a number that is exactly 1 more than its cube?
< pf >
Yes,
Define f (x) = x − (1 + x3 )
f (−1) = −1 ∧ f (−2) = 5
∵ f (x) is continuous at everywhere
∴By I.V.T ∀ t ∈ (−1, 5) ∃c ∈ (−2, −1) s.t f (c) = t
i.e ∃ c ∈ (−2, −1) s.t f (c) = 0
⇒ c − (1 + c3 ) = 0 ⇒ c = 1 + c3
66. If a and b are positive numbers, prove that the equation
x3
a
b
+ 3
=0
2
+ 2x − 1 x + x − 2
has at least one solution in the interval (−1, 1).
< pf >
x3 + 2x2 − 1 = (x + 1)(x2 + x − 1) ∧ x3 + x − 2 = (x − 1)(x2 + x + 2)
a
b
Define f (x) = x3 +2x
2 −1 + x3 +x−2
b
b
lim f (x) = a2 + lim x3 +x−2
⇒ lim x3 +x−2
= −∞
x→1−
x→1−
lim
√
x→(
5−1 +
)
2
f (x) =
lim
√
x→(
5−1 +
)
2
x→1−
a
x3 +2x2 −1 +
b
√
3 5−9
2
⇒
lim
√
x→(
5−1 +
)
2
Given ε > 0 ∃ x1 , x2 √
∈ (−1, 1)
s.t 1 − x1 < ε ∧ x2 − 5−1
<ε
2
satisfy f (x1 ) < 0 ∧ f (x2 ) > 0
⇒ f (x) has at least one solution in the interval (−1, 1).
3
a
x3 +2x2 −1
=∞