Question #fa896

1 Answer
Nov 24, 2017

Using Bolzano's & Rolles Theorem w/ Monotony & proof of contradiction

Explanation:

f(x)=x^6+x^2-1 , f(x)=x6+x21,xin [0,1]subeR[0,1]R
ff is continuous in RR and so [0,1] [0,1]
also
f'(x)=6x^5+2x>0 , xin(0,1)
so f strictly uarr in [0,1]

  • f continuous in [0,1]

  • f(0)=-1<0

  • f(1)=1>0
    => f(0)f(1)<0

-So, according to Bolzano's Theorem there is at least one x_0 in (0,1) for which f(x_0)=0
x_0 is unique because f is strictly increasing

Supposed there is one more root r_1 with 0<x_0<r_1 (Without loss of generality) & f(r_1)=f(x_0)=0
- All the criteria for Rolle's theorem are being met at [x_0,r_1] for f

According to Rolle's theorem, there is one ξ in (x_0,r_1) with ξ >0 and f'(ξ)=0 <=> 6ξ^5+2ξ=0 <=> 2ξ(3ξ^4+1)=0 <=> 2ξ=0 <=> ξ=0 -------> CONTRADICTION!
Because 0<x_0<ξ<r_1

=> As a result f has exactly one positive root
(ξ is greek letter used to mark the second root, nothing important. )