Problem
(a) $f(x)=2 x+1$ \[ \begin{array}{l} g(x)=\frac{x-1}{2} \\ f(g(x))=\square \\ g(f(x))=\square \end{array} \] $f$ and $g$ are inverses of each other $f$ and $g$ are not inverses of each other
Solution
Step 1 ：Given that $f(x)=2x+1$ and $g(x)=\frac{x-1}{2}$, we can substitute $g(x)$ into $f(x)$ to find $f(g(x))$ and substitute $f(x)$ into $g(x)$ to find $g(f(x))$.
Step 2 ：First, let's find $f(g(x))$. Substituting $g(x)$ into $f(x)$, we get $f(g(x))=f\left(\frac{x-1}{2}\right)=2\left(\frac{x-1}{2}\right)+1=x-1+1=x$.
Step 3 ：Next, let's find $g(f(x))$. Substituting $f(x)$ into $g(x)$, we get $g(f(x))=g(2x+1)=\frac{2x+1-1}{2}=x$.
Step 4 ：Since $f(g(x))=x$ and $g(f(x))=x$, we can conclude that $f$ and $g$ are inverses of each other.
