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# AoPS Wiki talk: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 $\boxed{f(x)=4x+3}$