HOMEWORK 9
SHUANGLIN SHAO
1. P 175, Ex. 1
Proof. For any x and h,
(x + h)n = xn + nxn−1 h +
n n−2 2
n
n n
x
h + ··· +
xhn−1 +
h .
2
n−1
n
Then
n n−2
n
n n−1
(x + h)n − xn
= nxn−1 +
x
h + ··· +
xhn−2 +
h
.
h
2
n−1
n
Hence
(x + h)n − xn
= nxn−1 .
h→0
h
This shows that f 0 (x) = nxn−1 .
lim
2. P 175, Ex. 4
Proof.
cos(x + h) − cos x
cos x(cos h − 1) − sin x sin h
=
.
h
h
Recall the formula that
h
cos h − 1 = −2 sin2 ,
2
and
sin h
lim
= 1.
h→0 h
Then
sin2 h2
cos(x + h) − cos x
sin h
= −2 lim cos x(h/2)2
−sin x lim
= − sin x.
2
h→0
h→0
h→0 h
h
(h/2)
lim
Hence
(cos x)0 = − sin x.
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3. P175, Ex. 7
Proof. We know
f (0) = 0.
For h 6= 0, consider
f (0 + h) − f (0)
f (h)
|h|3
=
=
.
h
h
h
3
2
Since limh→0 |h|
|h| = limh→0 |h| = 0,
|h|3
f (0 + h) − f (0)
= 0, ⇒ lim
= 0.
h→0 h
h→0
h
lim
This proves that f 0 (0) = 0.
4. P 176 Ex. 15
Proof. f is differentiable at 0.
h2 sin 1/h − 0
f (0 + h) − f (0)
= lim
= lim h sin 1/h = 0.
h→0
h→0
h→0
h
h
The last equation follows that
lim
|h sin 1/h| ≤ |h|
and the squeezing theorem.
5. P 176. Ex. 17
Proof. Given x. f is differentiable at x. Then the limit exists:
f 0 (x) = lim
h→0
f (x + h) − f (x)
.
h
We take h > 0; since f is increasing, we have x + h > x implies that
f (x + h) ≥ f (x), and so
f (x + h) − f (x)
≥ 0, for all h > 0,
h
which yields that
f 0 (x) ≥ 0.
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6. P 176 Ex. 22
Proof. By using the chain rule,
f (x2 ) − f (0)
= f 0 (0)(x2 )0 |x=0 = f 0 (0)2x|x=0 = 0.
x→0
x
lim
7. P 181. Ex. 2
Proof. For any x1 < x2 in (a, b), by Rolle’s theorem, there exists c in (x1 , x2 )
such that
f (x2 ) − f (x1 ) = f 0 (c)(x2 − x1 ).
Since f 0 (c) ≥ 0 and x2 − x1 ≥ 0,
f (x1 ) ≤ f (x2 ).
This shows that f is increasing.
8. P 181. Ex. 3
Proof. For any x1 < x2 in (a, b), by Rolle’s theorem, there exists c in (x1 , x2 )
such that
f (x2 ) − f (x1 ) = f 0 (c)(x2 − x1 ).
Since f 0 (c) ≤ 0 and x2 − x1 ≥ 0,
f (x1 ) ≥ f (x2 ).
This shows that f is decreasing.
9. P 181. Ex. 4
Proof. By Rolle’s theorem, for any x and y in (a, b), there exists c ∈ (x, y),
f (x) − f (y) = f 0 (c)(x − y).
Then since |f 0 (c)| ≤ M ,
|f (x) − f (y)| ≤ M |x − y|.
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10. P 181. Ex. 12
Proof. Given x ∈ [0, π/6]. There exists c ∈ (0, x) such that
x
cos x − cos 0 = − sin c × (x − 0) = −x sin c ≥ − .
2
The last inequality follows from that, x ∈ [0, π/6] implies that 0 ≤ sin c ≤
1/2. Hence
x
cos x − 1 ≥ − .
2
11. P 181. Ex. 13
Proof. We just need to verify that, for x > 0,
xf 0 (x) − f (x)
f (x) 0 f 0 (x) f (x)
=
− 2 =
≥0
x
x
x
x2
which is equivalent to showing that
f 0 (x) ≥
f (x)
.
x
We write
f (x) − f (0)
f (x)
=
x
x−0
By the mean value theorem, there exists c ∈ (0, x) such that
f (x)
= f 0 (c).
x
Since f 0 is increasing for x > 0,
f 0 (c) ≤ f 0 (x).
We conclude that
f (x)
x
is monotone increasing for x > 0.
12. P181. Ex. 14
Proof. We use the Exercise 17 on Page 182, the generalized mean value
theorem. For h 6= 0 there exists 0 < θ < h or h < θ < 0 such that
f (t + h) − 2f (t) + f (t − h)
f 0 (t + θ) − f 0 (t − θ)
=
.
2
h
2θ
Write
f 0 (t + x) − f 0 (t − x)
1 f 0 (t + x) − f 0 (x) f 0 (x) − f 0 (t − x)
=
+
.
2x
2
x
x
4
Since f 00 exists at t, by the definition of derivatives,
f 0 (t + x) − f 0 (t − x)
00
lim
= f (t).
x→0
2x
If h → 0 and θ is between 0 and h, θ → 0 as well. So
f (t + h) − 2f (t) + f (t − h)
f 0 (t + θ) − f 0 (t − θ)
lim
=
lim
= f 00 (t).
h→0
θ→0
h2
2θ
13. P 181. Ex. 15
Proof. Fix x ∈ (a, b). Then
|f (x + h) − f (x)| ≤ ch2 .
Hence
|
f (x + h) − f (x)
| ≤ ch.
h
Hence
f 0 (x) = lim
h→0
f (x + h) − f (x)
= 0.
h
So f is constant on (a, b).
14. P 182. Ex. 17
Proof. Define
h(t) = (f (b) − f (a))g(t) − (g(b) − g(a))f (t).
Then h is continuous on [a, b] and is differentiable on (a, b). Also h(a) = h(b).
So by the mean value theorem, there exists c ∈ (a, b) such that
h0 (c) = 0.
Hence
(f (b) − f (a))g 0 (c) − (g(b) − g(a))f 0 (c) = 0.
This equation implies the conclusion if g(a) 6= g(b) and g 0 (c) 6= 0.
Department of Mathematics, KU, Lawrence, KS 66045
E-mail address: [email protected]
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