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Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 9, 2011"

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==Solution==
 
==Solution==
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===Solution 1===
 
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>.
 
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>.
  
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Another solution:
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===Solution 2===
  
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We know that f(x) is of the form ax+b, so we can start by plugging in x=1 which yields f(1)=7 and plugging in x=-1 gives f(0)=3, using the slope formula we can get f(x)=4x+3
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We know that<math> f(x)</math> is of the form <math>ax+b</math>, so we can start by plugging in <math>x=1</math> which yields <math>f(1)=7</math> and plugging in <math>x=-1</math> gives <math>f(0)=3</math>, using the slope formula we can get <math>\boxed{f(x)=4x+3}</math>

Latest revision as of 21:07, 9 June 2011

Problem

AoPSWiki:Problem of the Day/June 9, 2011

Solution

Solution 1

Let $g(x)=\frac{x+1}{2}$. Therefore, $f(g(x))=2x+5$. Now, let $g^{-1}(x)$ be the inverse function of $g(x)$, so that $f(g(g^{-1}(x)))=2(g^{-1}(x))+5$. However, the $g(g^{-1}(x))$ in the LHS cancels out by the definition of an inverse function. Therefore, we have $f(x)=2(g^{-1}(x))+5$. Now we must find $g^{-1}(x)$. Again by the definition of an inverse function, we have $g(g^{-1}(x))=x$, and $g(x)=\frac{x+1}{2}$, so $\frac{g^{-1}(x)+1}{2}=x$. Solving, we find that $g^{-1}(x)=2x-1$. Plugging this in to $f(x)=2(g^{-1}(x))+5$ yields $\boxed{f(x)=4x+3}$.

Solution 2

We know that$f(x)$ is of the form $ax+b$, so we can start by plugging in $x=1$ which yields $f(1)=7$ and plugging in $x=-1$ gives $f(0)=3$, using the slope formula we can get $\boxed{f(x)=4x+3}$

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