Problem (Algebra):
Suppose that the function $g$ satisfies $g(x)=2 x-4$ for all real numbers $x$ and that $g^{-1}$ is the inverse function of $g$. Suppose that the function $f$ satisfies $g\left(f\left(g^{-1}(x)\right)\right)=2 x^{2}+16 x+26$ for all real numbers $x$. What is the value of $f(\pi)$ ?

Solution:
Since the function $g$ is linear and has positive slope, then it is one-to-one and so invertible. This means that $g^{-1}(g(a))=a$ for every real number $a$ and $g\left(g^{-1}(b)\right)=b$ for every real number $b$.

Therefore, $g\left(f\left(g^{-1}(g(a))\right)\right)=g(f(a))$ for every real number $a$.

This means that

$$
\begin{aligned}
g(f(a)) & =g\left(f\left(g^{-1}(g(a))\right)\right) \\
& =2(g(a))^{2}+16 g(a)+26 \\
& =2(2 a-4)^{2}+16(2 a-4)+26 \\
& =2\left(4 a^{2}-16 a+16\right)+32 a-64+26 \\
& =8 a^{2}-6
\end{aligned}
$$

Furthermore, if $b=f(a)$, then $g^{-1}(g(f(a)))=g^{-1}(g(b))=b=f(a)$.

Therefore,

$$
f(a)=g^{-1}(g(f(a)))=g^{-1}\left(8 a^{2}-6\right)
$$

Since $g(x)=2 x-4$, then $y=2 g^{-1}(y)-4$ and so $g^{-1}(y)=\frac{1}{2} y+2$.

Therefore,

$$
f(a)=\frac{1}{2}\left(8 a^{2}-6\right)+2=4 a^{2}-1
$$

and so $f(\pi)=4 \pi^{2}-1$.

Since the function $g$ is linear and has positive slope, then it is one-to-one and so invertible. To find a formula for $g^{-1}(y)$, we start with the equation $g(x)=2 x-4$, convert to $y=2 g^{-1}(y)-4$ and then solve for $g^{-1}(y)$ to obtain $2 g^{-1}(y)=y+4$ and so $g^{-1}(y)=\frac{y+4}{2}$. We are given that $g\left(f\left(g^{-1}(x)\right)\right)=2 x^{2}+16 x+26$.

We can apply the function $g^{-1}$ to both sides to obtain successively:

$$
\begin{aligned}
f\left(g^{-1}(x)\right) & =g^{-1}\left(2 x^{2}+16 x+26\right) \\
f\left(g^{-1}(x)\right) & \left.=\frac{\left(2 x^{2}+16 x+26\right)+4}{2} \quad \text { (knowing a formula for } g^{-1}\right) \\
f\left(g^{-1}(x)\right) & =x^{2}+8 x+15 \\
f\left(\frac{x+4}{2}\right) & \left.=x^{2}+8 x+15 \quad \text { (knowing a formula for } g^{-1}\right) \\
f\left(\frac{x+4}{2}\right) & =x^{2}+8 x+16-1 \\
f\left(\frac{x+4}{2}\right) & =(x+4)^{2}-1
\end{aligned}
$$

We want to determine the value of $f(\pi)$.

Thus, we can replace $\frac{x+4}{2}$ with $\pi$, which is equivalent to replacing $x+4$ with $2 \pi$.

Thus, $f(\pi)=(2 \pi)^{2}-1=4 \pi^{2}-1$.

Final answer: $4 \pi^{2}-1$