show that the function has unique root in the interval[1,2]

x⁵-3x-1=0

FROM THE GIVEN EQN WE GET

x=(x⁵-1)/3

as it is given to find the soln in [1,2]

therefore 1<=x<=2

1<(x⁵-1)/3<2

on solving further

4<x⁵<7

now taking log both sides we get that x has a unique soln in [1,2]

hope u got it....

lmvt is a   better way..

let the function be F(x) . it is a increasing function in [1,2]



F(1) = -3

F(2) = 25



they are of opposite sign



so the function cuts the x - axis . so it has a unique solution

it doesnt  means it  has a unique root.. itmeans it  has odd no of roots...sorry!!

let f(x) = x⁵ - 3x -1

now f(1) = -3 =-ve

and f(2) = 25 = +ve

this means that either 1 or 3 or 5 roots lies in [1,2] ---------------------(I)

now, f ' (x) = 5x⁴ - 3 > 0 for all x belongs to [1,2]

this means f(x) is increasing in [1,2] -------------------------(II)

from (I) and (II)

there can only be one root lying in intervel[1,2]

hence proved .  :)

sorry

can someone explain lmvt..i don't get it

see this

http://www.goiit.com/chapters/tutorial/maths/application-of-derivatives-3.htm

hey guys what abt my method?

isnt that right?