3
d)
Now
that
we've
proven
there
is
at
least
one
root,
we
will
now
prove
there
is
not
at
least
two
roots.
We'll
do
this
using
contradiction.
Let's
(incorrectly)
assume
that
there
are
in
fact
two
different
roots,
and
use
that
to
blow
up
math.
The
logic
here
is
that
if
we
assume
something,
and
end
up
breaking
math
using
that
assumption,
then
our
assumption
should
be
wrong
and
we
can
correctly
conclude
its
opposite.
Let's
call
those
two
roots
a
and
b.
Mean
Value
Theorem
then
states
that
the
derivative
of
f(z)
should
equal
L8202
somewhere.
What
is
£8=1(2)
numerically
equal
to
in
this
case?
Assume
aLb
soon
vhet
fea)
=0
and
$ce)
=0

€0y
s
conv,
over
Lo,
0T}
ona
oF.
over
(ob).
So
by
MVT
dhee
exsts
¢
berwcen
0
ond
b,
Som
thee
fCy
=
FeB)
y
v
T(C)
boo

©
¢)
Calculate
f/(z)
and
find
its
range.
Does
your
answer
from
part
d)
fall
in
the
range?
In
other
words,
even
though
MVT
said
this
must
happen,
is
it
actually
true that
f'(z)
can
attain
the
value
found
in
part
d)?
'f'(A)
=
red
and
$TIZ2

¥0
=)
e
onsWeC
S0
pacy
d
does
ner
o\
ia
™e
fage
betovse
fR)X

ond
O
s
Vess.
Ffeardz
xyer
B
4
£00)9
1
1;':'
f)
If
you
did
parts
d)
and
e)
correctly,
then
congratulations,
you created
a
contradiction!
What
can
we
now
conclude?
Trece
orec
NOr
2
(oors
and
ynere
s
CAacrly
1
root.
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