Question

Show that 4ax ^{3}+3bx ^{2}+2cx=a+b+c has at least one root between 0 and 1.

Answer 1

\[f(x)=4ax^3+3bx^2+2cx-a-b-c\]

Answer 2

try
\[f(0)\] and
\[f(1)\]

Answer 3

f(0)=-a-b-c=-(a+b+c)
f(1)=4a+3b+2c-a-b-c=3a+2b+c

Answer 4

are a, b,and c positive?

Answer 5

if so we are done

Answer 6

oh actually this is not that straightforward is it?

Answer 7

because a+b+c is positive and -(a+b+c) is negative
and 3a+2b+c is positive
you know if a,b,c are all positive

Answer 8

but.. suppose not. can it be the case that for any a, b, c if
\[-(a+b+c)\] has one sign then
\[3a+2b+c\] has opposite sign?

Answer 9

maybe

Answer 10

lets see if i can find a counterexample

Answer 11

and i'm going to assume they can't all be zero

Answer 12

now i am wondering if this is a set up for mvt somehow...

Answer 13

any thoughts kwenisha?

Answer 14

i think i found a counterexample satellite

Answer 15

oh wait not yet lol

Answer 16

after a beer i cannot, but you have three variables so it seems likely that it is possible

Answer 17

i mean find three numbers a, b, c where
\[a+b+c>0\] and
\[3a+2b+c<0\]

Answer 18

how about a = -10 b = 9, c = 2

Answer 19

then
\[a+b+c>0\] and
\[3a+2b+c<0\] so both
\[-(a+b+c)<0\] and also
\[3a+2b+c<0\]

Answer 20

ok maybe we can try it with the mean value theorem

Answer 21

let
\[F(x)=ax^4+bx^3+cx^2\] now
\[F(1)=a+b+c\]
\[F(0)=0\] to there exists a number "r" in [0,1] with
\[F'(r)=\frac{F(1)-F(0)}{1-0}=a+b+c\] and that gives it to us because
\[\frac{F(1)-F(0)}{1-0}=a+b+c\] and
\[F'(x)=f(x)=4ax^3+3bx^2+2cx\]

Answer 22

so mean value theorem what the right way to go, and we know there is a number "r" in [0,1] that satisfies that equation