SOLUTION OF HW2
MINGFENG ZHAO
September 25, 2011
1. [Page 56, Problem 4] Let f (x) = x2 for all real x. Verify each of the following formulas. In each
case describe the set of real x, y, t, etc., for which the given formula is valid.
a. f (−x) = f (x).
b. f (y) − f (x) = (y − x)(y + x).
c. f (x + h) − f (x) = 2xh + h2 .
e. f (t2 ) = f (t)2 .
Proof. a. Since f (x) = x2 for all x ∈ R. Then for all x ∈ R, we have
f (−x) = (−x)2 = x2 = f (x).
So f (−x) = f (x) for all x ∈ R.
b. For any x, y ∈ R, we know that
f (x) = x2 ,
and f (y) = y 2 .
Then we get
f (y) − f (x) = y 2 − x2 = (y − x)(y + x).
Therefore, we conclude that f (y) − f (x) = (y − x)(y + x) for all x, y ∈ R.
c. For any x, h ∈ R, we know that
f (x) = x2 ,
and f (x + h) = (x + h)2 = x2 + 2xh + h2 .
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2
MINGFENG ZHAO
Then we get
f (x + h) − f (x) = x2 + 2xh + h2 − x2 = 2xh + h2 .
Therefore, we have that f (x + h) − f (x) = 2xh + h2 for all x ∈ R.
e. For any t ∈ R, we know that
f (t2 ) = (t2 )2 = t4 ,
and f (t)2 = [f (t)]2 = [t2 ]2 = t4 .
Therefore, we get that f (t2 ) = f (t)2 for all x ∈ R.
2. [Page 56, Problem 5] Let g(x) =
p
4 − x2 for |x| ≤ 2. Verify each of the following formulas and
tell for which values of x, y, s, and t the given formula is valid.
a. g(−x) = g(x).
p
b. g(2y) = 2 1 − y 2 .
e. g
s
2
=
1p
16 − s2 .
2
Proof. a. Since the domain of g is [−2, 2]. Then for all x ∈ [−2, 2], w know that −2 ≤ −x ≤ 2, so we
have
g(x) =
p
4 − x2 ,
and g(−x) =
p
p
4 − (−x)2 = 4 − x2 = g(x).
Therefore, we get that g(−x) = g(x) for all x ∈ [−2, 2].
b. Since the domain of g is [−2, 2], so if we want to consider g(2y), then −2 ≤ 2y ≤ 2, that is,
−1 ≤ y ≤ 1. So for any y ∈ [−1, 1], we know that −2 ≤ 2y ≤ 2, and we have
g(2y) =
p
4 − (2y)2 =
p
p
p
4 − 4y 2 = 4(1 − y 2 ) = 2 1 − y 2 .
p
Therefore, we get that g(2y) = 2 1 − y 2 for all y ∈ [−1, 1].
e. Since the domain of g is [−2, 2], so if we want to consider g
−4 ≤ s ≤ 4. So for any s ∈ [−4, 4], we have −2 ≤
g
s
2
r
=
4−
s 2
2
r
=
s
2
s
2
, then −2 ≤
≤ 2, and
s2
4−
=
4
r
1
1p
· [16 − s2 ] =
16 − s2 .
4
2
s
2
≤ 2, that is,
SOLUTION OF HW2
Therefore, g
s
2
=
1
2
√
3
16 − s2 for all s ∈ [−4, 4].
3. [Page 56, Problem 6] Let f be defined as follows: f (x) = 1 for 0 ≤ x ≤ 1; f (x) = 2 for 1 < x ≤ 2.
The function is not defined if x < 0 or if x > 2.
a. Draw the graph off.
b. Let g(x) = f (2x). Describe the domain of g and draw its graph.
Proof. a. By the definition of f , we know that the graph of f should be:
b. Since the domain of f is [0, 2], now we want to consider g(x) = f (2x), then 0 ≤ 2x ≤ 2, that is,
0 ≤ x ≤ 1. So the domain of g is [0, 1]. And also for all x ∈ [0, 1], we know that 0 ≤ 2x ≤ 2, and
g(x)
=
=
f (2x)




 1, if 0 ≤ 2x ≤ 1



 2, if 1 < 2x ≤ 2.
=




 1, if 0 ≤ x ≤



 2, if
1
2
1
2
< x ≤ 1.
Hence the graph of g should be:
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MINGFENG ZHAO
4. [Page 57, Problem 9] This exercise develops some fundamental properties of polynomial of degree
n. Let f (x) =
n
X
ck xk be a polynomial of degree n. Prove each of the following:
k=0
a. If n ≥ 1 and f (0) = 0, then f (x) = xg(x), where g is a polynomial of degree n − 1.
b. For each real a, the function p given by p(x) = f (x + a) is a polynomial of degree n.
Proof. a. Since f (x) =
n
X
ck xk = c0 +
n
X
ck xk , and f (0) = 0, that is,
k=1
k=0
f (0) = c0 +
n
X
ck 0k = c0 = 0.
k=1
Then we have c0 = 0, so
f (x) =
n
X
ck xk = x ·
Pn−1
i=0
ck xk−1 = x ·
k=1
k=1
If we let g(x) =
n
X
n−1
X
ci+1 xi .
i=0
ci+1 xi , then g is a polynomial of degree n − 1, and f (x) = xg(x).
b. Recall the binomial formula, the Exercise 4, in Page 44, we know that for any k ≥ 1, we have
k X
k k−i i
(x + a) =
a x.
i
i=0
k
So we have
p(x)
= f (x + a)
= c0 +
n
X
k=1
ck (x + a)k
SOLUTION OF HW2
=
c0 +
n
X
5
ck (a + x)k
k=1
=
c0 +
n
X
k=1
= c0 +
"
k X
k k−i i
ck ·
a x
i
i=0
n X
k
X
k−i
ck · a
k=1 i=0
k
·
· xi
i
k
· xi
i
i=0 k=i
!
n
n
X
X
k
c0 +
· xi
ck · ak−i ·
i
i=0
= c0 +
=
n X
n
X
#
ck · ak−i ·
k=i
Therefore, p is a polynomial of degree n.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu