SAT Question f44b4125: Nonlinear Functions answer and explanation with Desmos solution

Choice B is correct. For the function $f$, since the base of the exponent, $1.25$, is greater than $1$, the value of ${\left(1.25\right)}^x$ increases as $x$ increases. Therefore, the value of $18{\left(1.25\right)}^x$ and the value of $18{\left(1.25\right)}^x+41$ also increase as $x$ increases. Since $f$ is therefore an increasing function where $x\ge 0$, the function $f$ has no maximum value. For the function $g$, since the base of the exponent, $0.73$, is less than $1$, the value of ${\left(0.73\right)}^x$ decreases as $x$ increases. Therefore, the value of $9{\left(0.73\right)}^x$ also decreases as $x$ increases. It follows that the maximum value of $g\left(x\right)$ for $x\ge 0$ occurs when $x=0$. Substituting $0$ for $x$ in the function $g$ yields $g\left(0\right)=9{\left(0.73\right)}^0$, which is equivalent to $g\left(0\right)=9\left(1\right)$, or $g\left(0\right)=9$. Therefore, the maximum value of $g\left(x\right)$ for $x\ge 0$ is $9$, which appears as a coefficient in equation II. So, of the two equations given, only II displays, as a constant or coefficient, the maximum value of the function it defines, where $x\ge 0$.