Let-f-1-5-R-be-defined-by-f-x-6-x-1-Show-that-f-has-a-unique-fixed-point-and-find-it- Tinku Tara June 4, 2023 Limits 0 Comments FacebookTweetPin Question Number 125814 by Tanuidesire last updated on 14/Dec/20 Letf:[1,5]→Rbedefinedbyf(x)=6x+1.Showthatfhasauniquefixedpointandfindit. Answered by Olaf last updated on 14/Dec/20 fstrictlydecreasesand:f(1)=61+1=3f(5)=65+1=1f([1,5])=[1,3]⊆[1,5]and:f(x)=x⇔x2+x−6=0⇔x=−1±1−4(1)(−6)2=−3or2Onlyx=2isincludedin[1,5].⇒S={2}. Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: prove-that-the-function-f-x-1-e-x-2-2x-has-a-root-in-the-open-interval-0-1-recall-that-e-2-7-Next Next post: Question-60283 Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.
![Let f:[1,5]→R be defined by f(x)=(6/(x+1)). Show that f has a unique fixed point and find it.](https://www.tinkutara.com/question/Q125814.png)
![f strictly decreases and : f(1) = (6/(1+1)) = 3 f(5) = (6/(5+1)) = 1 f([1,5]) = [1,3] ⊆ [1,5] and : f(x) = x ⇔ x^2 +x−6 = 0 ⇔ x = ((−1±(√(1−4(1)(−6))))/2) = −3 or 2 Only x = 2 is included in [1,5]. ⇒ S = { 2 }.](https://www.tinkutara.com/question/Q125860.png)