Question

Question #fa896

Answer 1

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

Explanation:

`f(x)=x^6+x^2-1 ,`x`in` `[0,1]subeR`
`f` is continuous in `R` and so `[0,1]`
also
`f'(x)=6x^5+2x>0 ,`x`in` `(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. )