In I, the base $0.83$ is between $0$ and $1$. In II, the base $1.07$ is greater than $1$, but the coefficient is negative. How do these features affect whether each function increases or decreases as $x$ gets larger?

For each function, plug in $x=0$ and then a few larger $x$-values (like $1,2,5$). Does the function approach some value from above or below, or does it keep going down without bound? Is there an actual smallest value that the function reaches for $x \ge 0$?

If you believe a function has a minimum, ask: is that minimum exactly equal to one of the numbers you see in the equation (like the final $+6$ or $+11$), or is it only approached but never reached?

Enter y = 14(0.83)^x + 6 in Desmos and focus on $x \ge 0$. Then either:

• Use the table feature (click on the equation and select the table icon) to see $y$-values for increasing $x$ (for example, $x=0,1,2,5,10,20$), or
• Pan/zoom the graph to see what $y$-value the curve gets close to as $x$ becomes large. Compare the $y$-values you observe to the constants and coefficients (like $14$, $0.83$, and $6$) and decide whether there is an actual smallest $y$-value and whether it matches any of those numbers.

Enter y = -9(1.07)^x + 11 and again focus on $x \ge 0$. Use a table or zoom out to larger $x$-values (like $x=0,1,2,5,10,20$). Watch how the $y$-values change as $x$ increases and see whether they settle at some smallest value or keep decreasing. Then compare any candidate smallest value with the constants and coefficients in the equation ($-9$, $1.07$, $11$) to decide if one of them represents a minimum.

The problem asks whether each equation shows (as one of its numbers: a constant or a coefficient) the minimum value of that function for $x \ge 0$.

For each function, you must:

• Decide whether it has a minimum value on $x \ge 0$.
• If it does, check whether that minimum equals one of the numbers in the equation (for example, $14$, $0.83$, or $6$ in I; $-9$, $1.07$, or $11$ in II).

Look at the structure:

• $14(0.83)^x$ is always positive for $x \ge 0$ because $14>0$ and $(0.83)^x>0$.
• The base $0.83$ is between $0$ and $1$, so $(0.83)^x$ decreases as $x$ increases.

So as $x$ gets larger:

• $(0.83)^x$ gets closer and closer to $0$.
• $14(0.83)^x$ gets closer and closer to $0$.
• $r(x)=14(0.83)^x+6$ gets closer and closer to $6$ from above.

Check a couple of values:

• $r(0)=14(0.83)^0+6=14(1)+6=20$.
• $r(1)=14(0.83)+6 \approx 11.62+6=17.62$.
• For large $x$, $r(x)$ is a bit more than $6$.

Because $14(0.83)^x>0$ for all $x \ge 0$, you always have

So $r(x)$ never actually equals $6$; it just approaches $6$. That means $6$ is a horizontal asymptote, not an attained minimum value.

On the domain $x \ge 0$, $r(x)$ keeps decreasing but stays strictly above $6$. Therefore:

• There is no smallest value that $r(x)$ actually reaches.
• The function has no minimum on $x \ge 0$.

Since there is no minimum value at all, equation I cannot display a minimum value as a constant or coefficient.

Now consider $s(x)$:

• $(1.07)^x$ is always positive and increases as $x$ increases because the base $1.07>1$.
• The coefficient $-9$ is negative, so $-9(1.07)^x$ is always negative and becomes more negative as $x$ increases.

As $x$ gets larger:

• $(1.07)^x \to \infty$.
• $-9(1.07)^x \to -\infty$.
• So $s(x)=-9(1.07)^x+11 \to -\infty$.

Check a couple of values:

• $s(0)=-9(1.07)^0+11=-9+11=2$.
• $s(1)=-9(1.07)+11\approx -9.63+11=1.37$.

The function is decreasing and the $y$-values go lower and lower without bound; there is no lowest value it settles at.

Because $s(x)$ decreases without bound as $x$ increases, it has no minimum value on $x \ge 0$.

If a function has no minimum, no constant or coefficient in its equation can equal that (nonexistent) minimum. This is true for both functions I and II.

Therefore, neither I nor II displays, as a constant or coefficient, the minimum value of the function it defines on $x \ge 0$.