Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

03453660 Group TitleBest ResponseYou've already chosen the best response.0
dw:1350309182854:dw
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
\[f(x)=(x+1)^{2}2\]
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.2
we have f(f(x))  f(x) =0 or (x^2 + 2x1)^2 + (x^2 + 2x 1)  1 =0 its a quadratic in x^2 + 2x1 it'll give 2 real vales of that,, then we'll solve for x 4 roots shall be obtained..
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
is this correct ? f(f(x))=f(x) f(x)=f^1 (f(x)) f(x) = x to get x^2 +2x 1 = x so solving this : x^2+x1=0
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
Can't see the wood for the trees... The equation is f(x)^2 + 2f(x) 1 = f(x) > f(x)^2 + f(x) 1 = 0 as shubhamsrg said...
 one year ago

harsh314 Group TitleBest ResponseYou've already chosen the best response.0
if we do it like this \[f(x)=(x+1)^{2}2 \implies f(f(x))=(f(x)+1)^{2}2\] using it in the given equation we have \[f(x)=(f(x)+1)^{2}2\] putting the value of f(x) \[(x+1)^{2}2=((x+1)^{2}2+1)^{2}2\] solving that we get \[x+1=(x+1)^{2}1\] and then solve again
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.