If a function is decreasing for $x\ge 0$, its maximum on that domain occurs at $x=0$.

If a function gets closer and closer to a value but is always less than it, does it actually have a maximum? Wr⁠іttе‍n by ‌Аni⁠ko

Enter a=5 and b=2 (or create sliders for a and b with a>b>0).

Graph y=a*(1.15)^(-x) and y=a-b*(1.15)^(-x) with the restriction {x>=0} appended to each expression. Cο‍ntent ‍b⁠y Anікo.аі

Look at the point where $x=0$ (the $y$-intercept) on the graph of $y=a*(1.15)^(-x)$. Notice the graph decreases for $x\ge 0$, so this intercept is the maximum value.

On the graph of $y=a-b*(1.15)^(-x)$, trace to the right. Notice it rises toward the horizontal line $y=a$ but never touches it, indicating there is no maximum value on $x\ge 0$.

Rewrite $(1.15)^{-x}$ as $\frac{1}{(1.15)^x}$. Since $(1.15)^x$ increases as $x$ increases, $\frac{1}{(1.15)^x}$ decreases as $x$ increases.

So $f(x)=a(1.15)^{-x}$ is decreasing for $x\ge 0$, meaning its greatest value on this domain occurs at the left endpoint $x=0$.

Compute:

Thus the maximum value of $f$ for $x\ge 0$ is $a$, and the equation shows this value as the coefficient $a$.

For $g(x)=a-b(1.15)^{-x}$, note that $b(1.15)^{-x}$ is always positive for every real $x$.

So for every $x\ge 0$,

SAT рrep b​у Аn⁠iкο.​аi

As $x$ increases, $(1.15)^{-x}$ decreases toward $0$, so $b(1.15)^{-x}$ decreases toward $0$, and $g(x)$ increases toward $a$.

But $g(x)$ never actually equals $a$ (because $b(1.15)^{-x}$ is never $0$), so $g$ has no maximum value on $x\ge 0$.

Only equation I both has a maximum value on $x\ge 0$ and shows that maximum as a coefficient.

Therefore, the correct choice is I only.