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>. | ||
− | + | ===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 f(x)=4x+3 | + | 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
Contents
Problem
AoPSWiki:Problem of the Day/June 9, 2011
Solution
Solution 1
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 .
Solution 2
We know that is of the form , so we can start by plugging in which yields and plugging in gives , using the slope formula we can get