AoPS Wiki talk:Problem of the Day/June 9, 2011

Problem

AoPSWiki:Problem of the Day/June 9, 2011

Solution

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 $\boxed{f(x)=4x+3}$ .

Another solution:

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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AoPS Wiki talk:Problem of the Day/June 9, 2011

Problem

Solution