Like prof said, it will be enough if we cover the quiz 1, quiz 2 and quiz 3 and the additional material ,So here we will solve some problems straightly for each.

Quiz 3

**1.6.**

Compactness: If a set S is closed and bounded, it’s compact.

Characterization of compactness (BW): S is compact if and only if every sub-sequence has a convergent sub-sequence whose limit is in S.

Extreme Value theorem: if f is continuous ans S is compact then f has an absolute minimum value and an absolute maximum value on S.

**1.7.**

Disconnection: if S = S1 U S2 and S1 ∩ S2 closure = empty and S1 closure ∩ S2 = empty

connected set: if the set is not disconnected, then the set is connected.

Intermediate value theorem: If f: S -> R and V is a subset of S and if a, b are elements in V, and f(a)< t < f(b) then, there is c included in V such that f(c) is = t.

**2.1.**

Differentiability in R: there should me a number m, such that f(a+h) = f(a) + mh + E(h), where E(h) /h->0

Role’s theorem: if f is continuous on the [a,b] and f(a) = f(b) and f is differentiable on (a,b) then there should be c, such that f ‘(c) = 0.

Mean Value theorem: if f is continuous [a,b] and differentiable on (a,b) then there is c such that

f ‘(c) = f(a) – f(b)/ a – b

1.6. compactness

a. S = [1,3] then f(x) = 1/x-2 (x != 2) and 1 when x = 2.

b. f(x) = c, c for some constant

The answer of the text book:

Because, R is both open and closed. ( I forgot about it!)

a. 1/x

b.

1.7.

a. x >= 0 and x < 0 because the graph is like below.

b. a included in R, R/a

c. x + , x- case

For my mind, there is a vague answer

which is just getting two points and make a plane and cut the sphere with the plane and make the function that goes around the arc.

There is an answer for this problem! I got almost right except for thing that I forgot to add origin to make a plane(How come I came up with the idea that I want to make a plane with just two points!)

I understand this proof finally! Yeah! This is glorious for me!

But not by the proof that is given from the text I think there’s something wrong..

Because, f(0) = (2,0) I think it should be f(0) = (2,1) because there’s no (2,0) in S.

However, I found more conceivable prf

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