Theorem
a
ant
is
ca b
on
roots
f is
cont
diff
apply the MVT oof
Now since f has 2 roots
b Assure acb then
ft
1
1 root
If flexi gtx for all
constant
2
f
If fEx7
on
to
denoted these
MVT
ftp.fcbaI
contradict to
and apply MVT
on
i.e
by
Sit
C E Ca b
function
IR adf has
has atleast 1 root
and
fkx
Show
appropriate
photofit bipepcotty
EXI
then there
Such that
prove by Contradiction
so this
at least
ca b
on
fcay
ftp.fcb
b a
Set up an
Suppose
f is diff
c E Cab
port
keytechinguesi
EX
of
Nov 16th
MVT
If f
exists
men valuetheorem week
of
Some extra execrises
to
z
Axel R
as
there exists
Ea
o
the IR
So
nd
a
a
thx has
D
C Ca b
ca b
C for all XEIR show
then f g is
fax
a
CXtd for
Some
RFhF
DEIR using port 1
from lecture
set h f g and check the theorem had
0 for
then
h
1 Se
ft
constant
f
function
tht about Zero deviatre and
all XE ca b
hex C where C is a Consent
2 sfefaglfxkE.ca DaPbPgHthEorttheovim about the constant
function ad zero denature
fix gtx C on ca b
1
2
Set g
f
x
CX
So
C
C Now since
g
f 4 7 54 7 for all XGIR
so
d on
fix g
by port 1
some DEIR so
fast CX d for xHR
then
EX
I
pHoohf
Is.hn
Show
ShylElX bl for
all
x
y
IR
for
D
c
IR
ExShi8ant.daPPldtiff.thenMRVTthfus Tag MVT
BSyetfaif.es
Consider
hayeelekists
S't
the ybyowdecdtteaesufoG.ca
fk
six
Cos c
71
Knx
xay
Song
x
Cost 1st for all telR
1
By
15M
e
y
thus
I
single IX yl
D
EXIL Let f
g be cut
for fcb
suppose
exists
and
f
diff
Ca b
on
Show there
gas gcb
Sit
C e Ca b
a
and
on Cgb
Cc
a
h f g and apply MVT
set
Hinti
his ant on cats and
proofi Set hast fix gox
diff on cab by Sanneh rule of ant ad diff
Now use
MVT there exists
Set
htt ftc gta
a
constant
Ex
tptnhf
Cab
hch
hay
b a
fcb gCb
IT
a C Cab
c c
gars
0
fKc Skc
So
there exists
show
all the
cubic polynomials have
byfcxgetrado.ie
otnbxktowcxmfyd roots
sit
fca
y
at most
3 roots in IR
PEE
polynomials
Suppose
as
Now
f an
know
could have at most
of
quadratic
has 4 roots in IR we dented these
Xi Xz Xz and X4
by f is ant and diff on IR and he have
fkxt 3 at 4zbXtC
fix
is
quadratic polynomial
quadratic polynmel
and from precalculus
has utmost
z roots
we
NOW
byMIT
S't
three exists
GECK XD
C 713,74
ftp.fkqj
with a Cz f Cz
fkcs
0
So this contradict to
the fact tht
at most 2roots So
at most 3 roots
qaadnhcpdynm.nl
has
a C CX XD
has
Ex5extrai show nthdegree polynomial
has
fix
at most
B
n
roots in IR
proofi Almost the
EXI Suppose f
and
show
fas
same
g is
as above
cont and
C
1
on
IR with
f Cx glad the HR
gas
fix gtx
fix gtx
h f g and prove by
Hint set ahssxdeffhae
pnofiie
diff
I
x
O
XO
G
G
contradiction
FI.aoysos.tn fog egcy
sohnohwdthaftxis gkxb.JO rallxhEfRebyo assumption
NEY s.snPw0sewfh1
there exist
true
not
g so sit
1is
fade gag
So
by
hkcj
MVT
there exists
fcgs scgi
o
Y
hcsty.no
tcg gig
SE
C C Co y
a
Cfcos gcoD
SO
So this contradict to A So 1 is true
For proving G this is similar to phony 1
EXI
pHooIf
thx
Show
for
X I
c
y
I by using MVT
X
sestet f ftp.xtnaxtxttn.dsouseuetheneedsanteosthmouk
as in EXI
o
for X 1
ki
fax
t
l CA
0 So
MIT
y
by
what
is f G
1
ex
Inx
consider
there existsfixt
C E Cl y
a
sit Cx ie what is the
behaviour
and what is the
of f C n1 1tl
ytl
Now
curve
l
x
o
change
of fixftp.lnb
Ty
a
CE
this contradict to CA
Tx7
ta
show
fit
a
Cby sit
So
thx
HEX if
X
CD
Infact this solution isIneyastery1
there exists
for
o
b I so
ftc
X I
X
o
for
0
and
X
I
D
by using MIT
EX8
p
Show
3
0
2Byshowfantheis rooaftisoncnqeabdyofftrhd.ec oh7u5rth3
5 so thus
IIT there exists
mdmvfp.co 7 0 0 5
E Co 1
by
fcdt 0
sit
have 2
N8wonLYforshgEEthhiseEEIif unique Suppose we
rootsfor fax say X and Xz Assure Xick Now
thereexists a
by MI
f
Also foot
f
O
x
positive
fH
220
roots So Xi Xz
f
Cc
10
21
0
C
0
So
0
fan
So
has exactly 1
D
be 2nd diff
f
so
0
on
4 then there exists
proofi By MIL
CA
o
Vx EIR
13
x
root
This antoudict to CA
real root
f
t
of flex is 4 0 only
5 and flax 20 FX EIR we have
for all X E O So we only have
f
Suppose
sit
C C Cx Xz
a
and f4x7 BX 121
and fkx o
Ex9
has exactly 1 root in IR
5 0
IDistshowe sthhoweisfhx
HextttzxhEhfarsstabtyleasvtt.ttroot
a
pe
X 17
a
If f Coto fCD
CE Co 2
Sit
O
c
there exists
co 23
a
X eco 1
KE Cl
2
2
Hint
Afpfg met fhineek
flask
then apply
me
CE Ex Xz
C
f
Co
at
tattoo
7
2
z
fo th
t
again the exists
2
Sit
2
I
tkkh.IE
nEIa
o
D
Ext
suppose
If
f
g be cont
fca fcb
O
ga
and diff
on ca b
show thee exist
fcc tf'Cd
c e
Ca b
on
ca b
s1
o
Sox
h
eBx xpCgcxs expcgcxs glex
sqfmhcxhffixexpcgcxy.FI
tfCx
then apply MV
Now by MVI there exists
h
MBEKI
a
C C Cgb
tcbexpcgcb
O
f'G exp gas
fccsexpcg.co
flat fcc g'Cc
Sit
O
b
xpgbD
a
96
0
D