SAT Question 01668cd6: Nonlinear Functions answer and explanation with Desmos solution

Choice B is correct. Functions $f$ and $g$ are both exponential functions with a base of $0.40$. Since $0.40$ is less than $1$, functions $f$ and $g$ are both decreasing exponential functions. This means that $f\left(x\right)$ and $g\left(x\right)$ decrease as $x$ increases. Since $f\left(x\right)$ and $g\left(x\right)$ decrease as $x$ increases, the maximum value of each function occurs at the least value of $x$ for which the function is defined. It's given that functions $f$ and $g$ are defined for $x\ge 0$. Therefore, the maximum value of each function occurs at $x=0$. Substituting $0$ for $x$ in the equation defining $f$ yields $f\left(0\right)=33{\left(0.4\right)}^{0+3}$, which is equivalent to $f\left(0\right)=33{\left(0.4\right)}^3$, or $f\left(0\right)=2.112$. Therefore, the maximum value of $f$ is $2.112$. Since the equation $f\left(x\right)=33{\left(0.4\right)}^{x+3}$ doesn't display the value $2.112$, the equation defining $f$ doesn't display the maximum value of $f$. Substituting $0$ for $x$ in the equation defining $g$ yields $g\left(0\right)=33\left(0.16\right){\left(0.4\right)}^{0-2}$, which can be rewritten as $g\left(0\right)=33\left(0.16\right)\left(\frac{1}{0.4^2}\right)$, or $g\left(0\right)=33\left(0.16\right)\left(\frac{1}{0.16}\right)$, which is equivalent to $g\left(0\right)=33$. Therefore, the maximum value of $g$ is $33$. Since the equation $g\left(x\right)=33\left(0.16\right){\left(0.4\right)}^{x-2}$ displays the value $33$, the equation defining $g$ displays the maximum value of $g$. Thus, only equation II displays, as a constant or coefficient, the maximum value of the function it defines.