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 April 24th, 2014, 03:19 PM #1 Senior Member   Joined: Nov 2013 Posts: 434 Thanks: 8 f(x) f(x^2 +1)=x^4 +2x^2+1 find domain and rang of f
 April 24th, 2014, 04:46 PM #2 Global Moderator   Joined: May 2007 Posts: 6,852 Thanks: 743 It looks like f(x) = x^2. I am not sure what you want.
 April 24th, 2014, 06:11 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,697 Thanks: 2681 Math Focus: Mainly analysis and algebra Well the domain would be $[1,+\infty)$ I suppose. The range would be the corresponding values of $f(x^2 + 1)$.
 April 24th, 2014, 11:20 PM #4 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra The function is $f(g(x))=x^4+2x^2+1$ where $g(x)=x^2+1$ and $f(x)=x^2$ so the domain and range are both $[1,\,\infty)$.
 April 24th, 2014, 11:44 PM #5 Newbie   Joined: Apr 2014 From: USA Posts: 8 Thanks: 2 I think the domain is (-infinity. +infinity) because x can take any value and range is [1, +infinity)
April 24th, 2014, 11:49 PM   #6
Senior Member

Joined: Apr 2014
From: Greater London, England, UK

Posts: 320
Thanks: 156

Math Focus: Abstract algebra
Quote:
 Originally Posted by rickkkyrich I think the domain is (-infinity. +infinity) because x can take any value
Nope. $(-\infty,\,\infty)$ is the domain of $g(x)$. The domain of $f(g(x))$ is the range of $g(x)$, namely $[1,\,\infty)$.

 April 25th, 2014, 12:03 AM #7 Newbie   Joined: Apr 2014 From: USA Posts: 8 Thanks: 2 Yes you are right. Thanks for correcting Thanks from Olinguito

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