Question 142·Hard·Advanced Math

In $k b^x$, $g(0)=k$ and $g(1)=kb$, so $\frac{g(1)}{g(0)}=b$. Define $f$ and $g$ in Desmos, then type $\frac{g(1)}{g(0)}$. Desmos shows $5$, so the base is $5$. Two outputs suggest the base; the exponent work below confirms it for every $x$.

$f$ raises $5$ to $2$ times its input, minus $3$. Substitute $\frac{x}{2}+1$ for that whole input:

Replace the entire input before simplifying the exponent.

Multiply the $2$ by both terms inside its parentheses, so $2\left(\frac{x}{2}\right)=x$:

The division by $2$ in the new input cancels the multiplication by $2$ in the original exponent.

Combine $+2$ and $-3$ in the exponent:

Now increasing $x$ by $1$ increases the exponent by $1$, so the output is multiplied by $5$ every time. That confirms the Desmos ratio wasn't a coincidence.

In $k b^x$, the coefficient $k$ equals $g(0)$ because $b^0=1$. Add $g(0)$ in Desmos. It shows $0.04$; use its fraction button to see $\frac{1}{25}$. So $g(x)=\frac{1}{25}(5)^x$, with the outside factor included. Choice B.