Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 9, 2011"
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==Solution== | ==Solution== | ||
| − | {{ | + | Let <math> g(x)=\frac{x+1}{2} </math>. Therefore, <math> f(g(x))=2x+5 </math>. Now, let <math> g^{-1}(x) </math> be the [[inverse function]] of <math> g(x) </math>, so that <math> f(g(g^{-1}(x)))=2(g^{-1}(x))+5 </math>. However, the <math> g(g^{-1}(x)) </math> in the [[LHS]] cancels out by the definition of an inverse function. Therefore, we have <math> f(x)=2(g^{-1}(x))+5 </math>. Now we must find <math> g^{-1}(x) </math>. Again by the definition of an inverse function, we have <math> g(g^{-1}(x))=x </math>, and <math> g(x)=\frac{x+1}{2} </math>, so <math> \frac{g^{-1}(x)+1}{2}=x </math>. Solving, we find that <math> g^{-1}(x)=2x-1 </math>. Plugging this in to <math> f(x)=2(g^{-1}(x))+5 </math> yields <math> \boxed{f(x)=4x+3} </math>. |
Revision as of 20:43, 8 June 2011
Problem
AoPSWiki:Problem of the Day/June 9, 2011
Solution
Let 
. Therefore, 
. Now, let 
be the inverse function of 
, so that 
. However, the 
in the LHS cancels out by the definition of an inverse function. Therefore, we have 
. Now we must find . Again by the definition of an inverse function, we have 
, and , so 
. Solving, we find that 
. Plugging this in to yields 
.
