Show that linear functions grow by equal differences and exponential functions by equal factors over equal intervals, and recognize constant-rate and constant-percent-rate situations

Learning goal: Distinguish between situations that can be modeled with linear functions and with exponential functions. Show that linear functions grow by equal differences over equal intervals and that exponential functions grow by equal factors over equal intervals; recognize situations in which one quantity changes at a constant rate per unit interval relative to another; and recognize situations in which a quantity grows or decays by a constant percent rate per unit interval relative to another.

Two ways to grow. Two hypothetical online clubs each start with 50 members. Club A gains 30 members every month. Club B's membership grows by 30% every month. The models are A(x) = 50 + 30x and B(x) = 50(1.3)ˣ, where x is the number of months. Club A has 50, 80, 110, 140 and 170 members in months 0 to 4: each month adds the same amount, 30. Club B's model gives 50, 65, 84.5, 109.85 and 142.805 (a model value; a real count would be a whole number): each month multiplies by the same number, 1.3. Subtract successive values of B and the differences are 15, 19.5, 25.35 and 32.955, not equal. Divide successive values of A and the ratios are 1.6, 1.375, 1.27 and 1.21, not equal. Equal differences belong to the linear function and equal factors to the exponential function. A table shows this for five rows; the next two paragraphs show it for every linear and every exponential function and for intervals of any length.

Showing that linear functions grow by equal differences over equal intervals. Let f(x) = mx + b be any linear function and let h be the length of an interval. Starting from any input x, the change in the function over the interval from x to x + h is

f(x + h) − f(x) = [m(x + h) + b] − [mx + b] = mx + mh + b − mx − b = mh.

The result, mh, does not contain x. So wherever the interval starts, an interval of length h changes a linear function by the same difference, mh. With h = 1 the difference is m: the slope is the constant rate of change per unit interval. For club A, m = 30, so every interval of 1 month adds 30 members and every interval of 5 months adds 150, whether it runs from month 0 to month 5 or from month 7 to month 12. "Grow" includes shrinking: when m is negative, the equal differences are equal decreases.

Showing that exponential functions grow by equal factors over equal intervals. Let g(x) = a·bˣ be any exponential function, with a ≠ 0, b > 0 and b ≠ 1. The factor by which the function changes over the interval from x to x + h is the ratio

g(x + h) ÷ g(x) = (a·bˣ⁺ʰ) ÷ (a·bˣ) = (a·bˣ·bʰ) ÷ (a·bˣ) = bʰ,

using the exponent property bˣ⁺ʰ = bˣ·bʰ. The result, bʰ, does not contain x. So wherever the interval starts, an interval of length h multiplies an exponential function by the same factor, bʰ. With h = 1 the factor is b. For club B, b = 1.3, so every month multiplies the membership by 1.3 and every interval of 2 months multiplies it by 1.3² = 1.69. When 0 < b < 1 the equal factors are less than 1 and the function decays.

The same algebra shows why an exponential function does not have equal differences: g(x + h) − g(x) = a·bˣ·bʰ − a·bˣ = g(x)·(bʰ − 1). The difference is a fixed multiple of the current value g(x), so it changes as the function changes. That sentence is the heart of the distinction: a linear function changes by a fixed amount, and an exponential function changes by a fixed fraction of what is already there.

The intervals must be equal. Both statements compare intervals of the same length. A table whose inputs jump by 1, then 2, then 3 cannot be tested by subtracting or dividing neighboring rows as they stand. Divide each change in y by its own change in x to test for a constant rate; to test for a constant factor, check whether each ratio equals one base raised to the length of its interval.

Recognizing a constant rate per unit interval. A situation can be modeled with a linear function when one quantity changes by the same amount for each unit of another quantity. The signal in words is a fixed amount attached to "per", "each" or "every": $12 per hour, 30 members every month, 2.5 cm of candle each hour, 6 liters drained each minute. The other quantity does not have to be time: a cost that rises $3.50 per kilogram changes at a constant rate relative to mass. The model is y = (starting value) + (rate) × x, and the rate is the slope.

Recognizing a constant percent rate per unit interval. A situation can be modeled with an exponential function when a quantity grows or decays by the same percent of its current value in each unit interval. Growing by a percent rate r, written as a decimal, means new value = old value + r × old value = old value × (1 + r), so each unit interval multiplies by the factor 1 + r; decaying by r multiplies by 1 − r. Membership that grows 30% a month has factor 1.3. A hypothetical laptop that loses 20% of its value each year has factor 0.8, so a $1,200 laptop is worth 1,200(0.8)ᵗ dollars after t years: $960, $768, $614.40. "Doubles every day" is a 100% increase per day, factor 2; a half-life is a 50% decrease per period, factor 0.5. Again the unit interval need not be time: light that loses 12% of its intensity for each meter of depth in a hypothetical lake decays exponentially with depth.

The misconception: a percent in the problem means exponential. What matters is what the percent is taken of. A hypothetical account pays simple interest of 5% of the original $1,000 deposit each year. That is $50 every year, a constant amount, so the balance is 1,000 + 50t: linear. A second account pays 5% of the current balance each year, so each year multiplies the balance by 1.05 and the balance is 1,000(1.05)ᵗ: $1,050, $1,102.50, $1,157.63: exponential. After 10 years the first holds $1,500 and the second $1,628.89. A percent of a fixed amount is a constant rate; a percent of the current amount is a constant percent rate. A related error is to call anything that grows quickly exponential. The values 10, 40, 90, 160, 250 at x = 1, 2, 3, 4, 5 grow quickly, but their differences are 30, 50, 70, 90 and their ratios are 4, 2.25, 1.78, 1.56: neither is constant, so the pattern is neither linear nor exponential (it is 10x², a quadratic function).

Worked check of both statements on longer intervals: For A(x) = 50 + 30x the general result says every interval of length 5 adds mh = 30 × 5 = 150. Check two such intervals: A(7) − A(2) = 260 − 110 = 150 and A(12) − A(7) = 410 − 260 = 150. For B(x) = 50(1.3)ˣ the general result says every interval of length 2 multiplies by bʰ = 1.3² = 1.69. Check two: B(2) ÷ B(0) = 84.5 ÷ 50 = 1.69 and B(4) ÷ B(2) = 142.805 ÷ 84.5 = 1.69. The numerical checks agree with the algebra, but only the algebra shows the statements for every starting point and every interval length.

Worked tables with unequal intervals: Table 1 has inputs 0, 2, 5, 9 and outputs 7, 13, 22, 34. The neighboring differences 6, 9, 12 are not equal, but the intervals are 2, 3 and 4 units long: 6 ÷ 2 = 3, 9 ÷ 3 = 3 and 12 ÷ 4 = 3. The rate is 3 per unit throughout, so the table fits the linear function y = 7 + 3x; check: 7 + 3(9) = 34. Table 2 has inputs 0, 1, 3, 6 and outputs 5, 10, 40, 320. The rates of change are 5 ÷ 1 = 5, 30 ÷ 2 = 15 and 280 ÷ 3 = 93.3, not constant, so it is not linear. The ratios are 10 ÷ 5 = 2 over 1 unit, 40 ÷ 10 = 4 = 2² over 2 units and 320 ÷ 40 = 8 = 2³ over 3 units: each is 2 raised to the length of its interval, so the factor per unit interval is 2 throughout and the table fits the exponential function y = 5(2)ˣ; check: 5 × 2⁶ = 320.

Worked recognition of situations: Each situation is hypothetical. (a) A tank holding 240 liters drains 6 liters each minute. A fixed amount per minute is a constant rate per unit interval: linear, V = 240 − 6t. (b) A hypothetical car bought for $18,000 loses 15% of its value each year. A fixed percent of the current value per year is a constant percent rate: exponential decay with factor 0.85, V = 18,000(0.85)ᵗ. (c) A plan costs $25 a month plus $0.10 for each text message. The cost changes by a constant $0.10 per message, relative to the number of messages rather than time: linear, C = 25 + 0.10n. (d) A culture of bacteria doubles every 3 hours. Equal intervals of 3 hours give equal factors of 2: exponential. With t in hours the model is N = N₀·2^(t ÷ 3), and the factor for one hour is 2^(1 ÷ 3) = 1.26, a growth rate of about 26% per hour. (e) Light intensity in a lake falls 12% for each meter of depth: exponential decay relative to depth, I = I₀(0.88)ᵈ, so at 5 m the intensity is 0.88⁵ = 0.528 of its surface value, about 52.8%.

Worked limit of two readings: A quantity is 100 at x = 0 and 120 at x = 1. A linear function through both readings is 100 + 20x; an exponential function through both is 100(1.2)ˣ. Two readings cannot distinguish the situations, because a single interval has one difference and one ratio and nothing to compare them with. A third reading at x = 2 decides: the linear function predicts 140 and the exponential function predicts 144. A third reading, or a description of how the change happens, is needed to tell these two candidates apart; with measured data, more readings still are wanted before either form is trusted.

Represent in tables and graphs: Make a table for months 0 to 6 with four columns: A(x) = 50 + 30x, the difference from the previous month, B(x) = 50(1.3)ˣ rounded to one decimal, and the ratio to the previous month. Graph both functions on one set of axes with labeled scales. On the graph of A, draw two "steps" of width 2 months at different places and label the rise of each; on the graph of B, do the same and label each rise and also the ratio of the heights at the two ends of the step. Write one sentence saying what stayed the same for A and what stayed the same for B.

Independent performance: (1) Write a general argument, in your own words and symbols, showing that f(x) = mx + b changes by the same difference over every interval of length h, and that g(x) = a·bˣ changes by the same factor over every interval of length h; state the exponent property you used and the conditions on a and b. (2) Write four short situations of your own: two in which one quantity changes at a constant rate per unit interval relative to another, and two in which a quantity grows or decays by a constant percent rate per unit interval; at least one of the four must use a unit interval that is not a unit of time, and at least one must be a decay. (3) For each situation write the function, make a table of four equally spaced values and show the equal differences or equal factors in the table. (4) Write one situation that mentions a percent but is linear, and explain why. Success means the argument works for any starting point x and not only for chosen numbers, each situation is classified by what stays constant (an amount or a percent of the current value), every function reproduces its table, and the percent-but-linear example states clearly what the percent is taken of.