The Real And Complex Number Systems

Proof: By Generalized Mean Value Theorem, we complete it.
(b) There are many proofs about that e x 1 x for all x R. We list them as a
reference.
(b-1) Let fx e x 1 x, and thus consider the extremum.
(b-2) Use Mean Value Theorem.
(b-3) Since e x 1 0forx 0ande x 1 0forx 0, we then have
0
x
et 1dt 0and
x
0
et 1dt 0.
So, e x 1 x for all x R.
(iii) Let f be continuous function on a, b, and differentiabel on a, b. Prove that there
exists a c a, b such that
f fc fa
c .
b c
Proof: (STUDY) Since f c fcfa , we consider f
bc
cb c fc fa.
Hence, we choose gx fx fab x, then by Rolle’s Theorem,
ga gb g ca b where c a, b
which implies thatf c fcfa .
bc
(iv) Let f be a polynomial of degree n, iff 0onR, then we have
f f ..f n 0onR.
Proof: Let gx f f ..f n , then we have
g g f 0onR since f is a polynomial of degree n. *
Consider hx gxe x , then h x g x gxe x 0onR by (*). It means that h
is a decreasing function on R. Since lim x hx 0bythefactg is still a polynomial,
then hx 0onR. Thatis,gx 0onR.
(v) Suppose that f is continuous on a, b, fa 0 fb, and
x 2 f x 4xf x 2fx 0 for all x a, b. Prove that fx 0ona, b.
Proof: (STUDY) Since x 2 f x 4xf x 2fx x 2 fx by Leibnitz Rule, let
gx x 2 fx, then claim that gx 0ona, b.
Suppose NOT, there is a point p a, b such that gp 0. Note that since fa 0,
and fb 0, So, gx has an absolute maximum at c a, b. Hence, we have g c 0.
By Taylor Theorem with Remainder term, wehave
gx gc g cx c g
x c 2 ,where x, c or c, x
2!
gc since g c 0, and g x 0 for all x a, b
0sincegc is absolute maximum.
So,
x 2 fx c 2 fc 0 **
which is absurb since let x a in (**).
(vi) Suppose that f is continuous and differentiable on 0, , and
lim x f x fx 0, show that lim x fx 0.
Proof: Since lim x f x fx 0, then given 0, there is M 0 such that as
x M, wehave