Once you know where the maximum occurs (based on whether the function increases or decreases), plug that $x$-value into each function to get its maximum value.

After you find the maximum value of each function, look back at the equation as written. Does that number appear directly as a standalone constant or as a coefficient multiplying something, without doing any extra algebra?

For function I, notice that the number is inside the exponent on $4$. For function II, notice there is a number in front of $4^{-x}$. Which presentation more clearly shows the maximum value as a coefficient?

In Desmos, enter the two equations on separate lines:

• y = 4^(-(x+1))
• y = (1/4)*4^(-x) You will see two decreasing exponential curves.

To see only $x \ge 0$, you can restrict each function, for example:

• y = 4^(-(x+1)) {x>=0}
• y = (1/4)*4^(-x) {x>=0} Then, click each graph near $x=0$ to see the coordinate of the point where $x=0$. The $y$-value there is the maximum for that function on $x \ge 0$.

Note the $y$-value you see at $x=0$ for each graph. Then look back at each equation and check whether that $y$-value is written directly as a standalone number or coefficient in the formula. Decide which function(s) meet this condition.

The question is not only about finding the maximum value of each function; it is asking whether that maximum value appears directly as a constant or coefficient in the equation as written.

So for each function:

• First: Find its maximum value when $x \ge 0$.
• Second: Check if that number is actually written somewhere in the formula as a constant (like a standalone number) or as a coefficient (a number multiplying something).

Both functions are exponential decay functions because they involve $4$ raised to a negative exponent.

• As $x$ increases, $4^{-x}$ gets smaller.
• For $x \ge 0$, this means each function is decreasing as $x$ increases.

If a function is decreasing on $x \ge 0$, its largest value in that domain happens at the smallest $x$, which is $x=0$.

So for both functions, the maximum value occurs at $x=0$.{

Now plug in $x=0$ to each function.

For I: $f(x)=4^{-(x+1)}$.

• $f(0)=4^{-(0+1)}=4^{-1}=1/4$.

For II: $g(x)=\dfrac14\,4^{-x}$.

• $g(0)=\dfrac14\,4^{0}=\dfrac14\cdot 1=1/4$.

So both functions have the same maximum value $1/4$ on $x \ge 0$.{

Now compare the maximum value $1/4$ with the way each equation is written.

• Function I: $f(x)=4^{-(x+1)}$.

  • The constants you see are $4$, $-1$ (from the negative sign), and $+1$ in the exponent.
  • The number $1/4$ is not written as a separate constant or coefficient; it only appears after simplifying $4^{-1}$ when we plug in $x=0$.
  • So the equation for I does not display the maximum value as a constant or coefficient.

• Function II: $g(x)=\dfrac14\,4^{-x}$.

  • The coefficient in front is $1/4$.
  • The maximum value we found is also $1/4$.
  • Here, the maximum value is exactly the coefficient in the equation.

Therefore, only function II displays its maximum value as a constant or coefficient. So the correct answer is II only.