Question 502·Hard·Math

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The functions $r$ and $s$ are defined for $x \ge 0$ by the following equations.

I. $r(x)=14(0.83)^x+6$

II. $s(x)=-9(1.07)^x+11$

Which of the following equations displays, as a constant or coefficient, the minimum value of the function it defines, where $x \ge 0$ ?

A shifted exponential function has a changing power term plus a fixed number. For a minimum on $x \ge 0$ , graph only the allowed inputs in Desmos, then ask whether the curve actually reaches a lowest output. A level the curve approaches need not be a minimum, and a negative coefficient can make a growing power pull the graph down forever.

Hints

Hint 1
An exponential base is the number raised to $x$ . Each step to the right multiplies the power part by that base. Which power part shrinks as $x$ grows, and which grows?
Hint 2
A minimum must be an output the function actually takes. In the first equation, the power stays positive even while it shrinks. Can the output ever equal the added constant?
Hint 3
The domain is the set of allowed inputs. Here it has no right endpoint. As $x$ keeps growing, what does the minus sign in the second equation do to its outputs?

Step-by-step

Graph the allowed inputs, then check for a lowest output
Step 1
See how the first function behaves
The domain is the set of allowed inputs. Type $r(x)=14(0.83)^x+6\{x\ge0\}$ ; the curly braces tell Desmos to show only allowed inputs. The graph falls as $x$ grows and levels off near the added constant, $6$ .
Step 2
Check whether the first function reaches a minimum
The amount above $6$ is $14(0.83)^x$ , which is positive for every allowed $x$ . Each increase of $1$ in $x$ multiplies that amount by $0.83$ , making the output smaller but never equal to $6$ . The line $y=6$ is a horizontal asymptote, a level the graph approaches but never reaches. Getting closer to a level is not the same as having a minimum.
Step 3
See how the second function behaves
Below the first line, type $s(x)=-9(1.07)^x+11\{x\ge0\}$ . Desmos shows this graph falling farther down as $x$ grows, rather than leveling off. The negative coefficient in front of the power is the cue to check whether it can keep falling.
Step 4
Check whether the second function has a bottom
The base $1.07$ is greater than $1$ , so $9(1.07)^x$ grows without bound as $x$ grows. Because that growing term is subtracted from $11$ , $s(x)$ falls without bound and has no minimum. Neither equation displays a minimum value as a constant or coefficient. Choice D.

Question 502·Hard·Math

Strategy

Hints

Step-by-step

Lessons that teach this

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Strategy

Hints

Step-by-step

Lessons that teach this