Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Solved Assignment 5 - Introduction to Real Analysis 2 | MATH 5152G, Assignments of Mathematical Methods for Numerical Analysis and Optimization

Material Type: Assignment; Class: Intro to Real Analysis 2; Subject: Mathematics; University: Columbus State University; Term: Unknown 1989;

Typology: Assignments

Pre 2010

Uploaded on 08/04/2009

koofers-user-y4a
koofers-user-y4a 🇺🇸

10 documents

1 / 3

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
1
Section 5.1
1. Use the definition to find the derivative of each of the following functions.
f. g(x) = x
x2+1 , x R
g0(x) = lim
h0
g(x+h)g(x)
h
= lim
h0
x+h
(x+h)2+1 x
x2+1
h
= lim
h0
h(x21+xh)
((x+h)2+1)(x2+1)
h
= lim
h0x21 + xh
((x+h)2+ 1) (x2+ 1) =1x2
(x2+ 1)2
5. For each of the following, determine whether the given function is differentiable at the indicated
point x0. Justify your answer!
a. f(x) = x|x|, at x0= 0.
fis differentiable at x= 0. Since for x0, f(x) = x2and for x < 0, f(x) = x2,
f0(0) = lim
h0
f(0 + h)f(0)
h
= lim
h0
h2
h= lim
h0h= 0
and
f0
+(0) = lim
h0+
f(0 + h)f(0)
h
= lim
h0+
h2
h= lim
h0+h= 0.
Therefore f0(0) = f0(0) = f0
+(0) = 0. 2
b. f(x) = x2xQ
0, x /Qat x0= 0
fis differentiable at x= 0. For h6= 0, define function
g(h) = f(0 + h)f(0)
h=f(h)
h
We show that limh0g(h) = 0. For any ε > 0, let δ=ε. When 0 <|h|< δ, if hQ, then
|g(h)0|=
f(h)
h
=
h2
h
=|h|< δ =ε,
if h /Q,
|g(h)0|=
f(h)
h
=
0
h
= 0 < ε.
Therefore f0(0) = limh0g(h) = 0. 2
c. g(x) = (x2)[x], at x0= 2,
pf3

Partial preview of the text

Download Solved Assignment 5 - Introduction to Real Analysis 2 | MATH 5152G and more Assignments Mathematical Methods for Numerical Analysis and Optimization in PDF only on Docsity!

Section 5.

  1. Use the definition to find the derivative of each of the following functions.

f. g(x) =

x x^2 +

, x ∈ R

g

′ (x) = lim h→ 0

g(x + h) − g(x)

h

= lim h→ 0

x+h (x+h)^2 +

x x^2 +

h

= lim h→ 0

−h(x^2 −1+xh) ((x+h)^2 +1)(x^2 +1)

h

= lim h→ 0

x

2 − 1 + xh

((x + h)^2 + 1) (x^2 + 1)

1 − x

2

(x

2

2

  1. For each of the following, determine whether the given function is differentiable at the indicated

point x 0. Justify your answer!

a. f (x) = x|x|, at x 0 = 0.

f is differentiable at x = 0. Since for x ≥ 0, f (x) = x

2 and for x < 0, f (x) = −x

2 ,

f

′ (0) = lim h→ 0 −

f (0 + h) − f (0)

h

= lim h→ 0 −

−h

2

h

= lim h→ 0 −^

−h = 0

and

f

′ +(0)^ =^ lim h→ 0 +

f (0 + h) − f (0)

h

= lim h→ 0 +

h

2

h

= lim h→ 0 +^

h = 0.

Therefore f

′ (0) = f

′ (0) = f

′ +(0) = 0.^2

b. f (x) =

x

2 x ∈ Q

0 , x /∈ Q

at x 0 = 0

f is differentiable at x = 0. For h 6 = 0, define function

g(h) =

f (0 + h) − f (0)

h

f (h)

h

We show that limh→ 0 g(h) = 0. For any ε > 0, let δ = ε. When 0 < |h| < δ, if h ∈ Q, then

|g(h) − 0 | =

f (h)

h

h

2

h

= |h| < δ = ε,

if h /∈ Q,

|g(h) − 0 | =

f (h)

h

h

= 0 < ε.

Therefore f

′ (0) = limh→ 0 g(h) = 0. 2

c. g(x) = (x − 2)[x], at x 0 = 2,

For 1 < x < 2, g(x) = (x − 2). For 2 < x < 3, g(x) = 2(x − 2). So g

′ −(2) = 1 and^ g

′ +(2) = 2. Therefore

g is not differentiable at x = 2. 2

d. h(x) =

x + 2.

h is not differentiable at x = −2. Note that Dom h = [− 2 , ∞). So there is no left derivative h

′ −(−2).

In fact there is not right derivative h

′ +(−2) either.

h

′ +(−2)^ =^ lim x→− 2 +

h(x) − h(−2)

x − (−2)

= lim x→− 2 +

x + 2

x + 2

= lim x→− 2 +

x + 2

e. f (x) =

sin x x ∈ Q

x, x /∈ Q

at x 0 = 0

f is differentiable at x = 0. For h 6 = 0, define function

g(h) =

f (0 + h) − f (0)

h

f (h)

h

We show that limh→ 0 g(h) = 1. For any ε > 0, since

lim h→ 0

sin h

h

There exists a δ 1 > 0 such that when 0 < |h| < δ 1 ,

sin h

h

− 1 | < ε.

Let δ = min{ε, δ 1 }. When 0 < |h| < δ, if h ∈ Q, then 0 < |h| < δ 1 ,

|g(h) − 1 | =

f (h)

h

sin h

h

< ε,

if h /∈ Q,

|g(h) − 1 | =

f (h)

h

h

h

∣ = 0^ < ε.

Therefore f

′ (0) = limh→ 0 g(h) = 1. 2

  1. Let g be defined by g(x) =

x

2 sin

1 x x^6 = 0 0 , x = 0

a. Prove that g is differentiable at 0 and that g

′ (0) = 0.

Proof. Sine

lim x→ 0

g(x) − g(0)

x − 0

= lim x→ 0

x

2 sin

1 x

x

= lim x→ 0

x sin

x

g is differentiable at 0 and that g

′ (0) = 0. 2

b. Show that g

′ (x) is not continuous at 0.

Proof. Using the Product Rule and Chain Rule, we have

g

′ (x) = 2 x sin

x

  • x

2

cos

x

x

2

= 2 x sin

x

− cos

x