[Algebra]
Problem: If $f(x)=2 x+1$ and $g(f(x))=4 x^{2}+1$, determine an expression for $g(x)$.

Solution: We use the fact that $g(x)=g\left(f\left(f^{-1}(x)\right)\right)$.

Since $f(x)=2 x+1$, then to determine $f^{-1}(x)$ we solve $x=2 y+1$ for $y$ to get $2 y=x-1$ or $y=\frac{1}{2}(x-1)$. Thus, $f^{-1}(x)=\frac{1}{2}(x-1)$.

Since $g(f(x))=4 x^{2}+1$, then

$$
\begin{aligned}
g(x) & =g\left(f\left(f^{-1}(x)\right)\right) \\
& =g\left(f\left(\frac{1}{2}(x-1)\right)\right) \\
& =4\left(\frac{1}{2}(x-1)\right)^{2}+1 \\
& =4 \cdot \frac{1}{4}(x-1)^{2}+1 \\
& =(x-1)^{2}+1 \\
& =x^{2}-2 x+2
\end{aligned}
$$
We use the expressions for $f(x)$ and $g(f(x))$ to construct $g(x)$.

Since $f(x)$ is linear and $g(f(x))$ is quadratic, then it is likely that $g(x)$ is also quadratic.

Since $f(x)=2 x+1$, then $(f(x))^{2}=4 x^{2}+4 x+1$.

Since $g(f(x))$ has no term involving $x$, then we subtract $2 f(x)$ (to remove the $4 x$ term) to get

$$
(f(x))^{2}-2 f(x)=\left(4 x^{2}+4 x+1\right)-2(2 x+1)=4 x^{2}-1
$$

To get $g(f(x))$ from this, we add 2 to get $4 x^{2}+1$.

Therefore, $g(f(x))=(f(x))^{2}-2 f(x)+2$, and so an expression for $g(x)$ is $x^{2}-2 x+2$.
We use the expressions for $f(x)$ and $g(f(x))$ to construct $g(x)$.

Since $f(x)$ is linear and $g(f(x))$ is quadratic, then it is likely that $g(x)$ is also quadratic.

Suppose that $g(x)=a x^{2}+b x+c$ for some real numbers $a, b, c$.

Then

$$
\begin{aligned}
g(f(x)) & =g(2 x+1) \\
& =a(2 x+1)^{2}+b(2 x+1)+c \\
& =a\left(4 x^{2}+4 x+1\right)+b(2 x+1)+c \\
& =4 a x^{2}+(4 a+2 b) x+(a+b+c)
\end{aligned}
$$

Since we are told that $g(f(x))=4 x^{2}+1$, then we can compare coefficients to deduce that $4 a=4$ and $4 a+2 b=0$ and $a+b+c=1$.

From the first equation, $a=1$.

From the second equation, $b=-2 a=-2$.

From the third equation, $c=1-a-b=2$.

Therefore, an expression for $g(x)$ is $x^{2}-2 x+2$.

Final answer: $g(x)=x^2-2x+2$