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Lesson 9 — Average rate of change — the gateway to calculus

Δy/Δx as the central concept preceding the derivative. Geometric interpretation (slope of the secant) and physical interpretation (average velocity). The question that opens calculus: 'what if Δx becomes very small?'

Used in: 1.º ano EM · porta de entrada para Cálculo (Trim 5-6)

\text{ARC} = \frac{\Delta y}{\Delta x} = \frac{f(b) - f(a)}{b - a}

Average rate of change: how much y changed divided by how much x changed. Geometrically, it is the slope of the secant line passing through the points $(a,\, f(a))$ and $(b,\, f(b))$ on the graph of f.

Choose your door

Rigorous notation, full derivation, hypotheses

Definition and interpretation

"The ratio $[f(b) - f(a)]/(b - a)$ is called the average rate of change of $f$ on the interval $[a, b]$ ." — Active Calculus §1.3

Geometric interpretation

The ARC is the slope of the secant line to the graph of $f$ through the points $(a, f(a))$ and $(b, f(b))$ .

The secant line (gold) through the points (a, f(a)) and (b, f(b)). Its slope is exactly Δy / Δx, the average rate of change of f on [a, b].

Special cases

$f$ linear ( $f(x) = mx + n$ ): ARC is constant and equal to $m$ , regardless of the interval chosen.
$f$ quadratic: ARC varies with the interval; it equals $a(p + q) + b$ for $f(x) = ax^2 + bx + c$ on $[p, q]$ .
$f$ constant: ARC $= 0$ for any interval.

The question that opens calculus

What if $\Delta x \to 0$ ? The secant line "turns into" the tangent line, and the ARC converges to the instantaneous rate of change — which is exactly the derivative $f'(a)$ :

$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$

This is the topic of Trimesters 5 (limits) and 6 (derivatives). This lesson is the antechamber.

Worked examples

Example— 1· ARC of f(x) = x² on [1, 4] (direct application)

Problem: Calculate the average rate of change of $f(x) = x^2$ on the interval $[1, 4]$ .

Strategy: Apply the definition $\text{ARC} = (f(b) - f(a))/(b - a)$ directly with $a = 1$ , $b = 4$ .

Solution:

Compute $f(1) = 1^2 = 1$ and $f(4) = 4^2 = 16$ .
Apply the formula: $\text{ARC} = (16 - 1)/(4 - 1) = 15/3 = 5$ .
Answer: ARC = 5.

Verification: This is the slope of the secant line through $(1, 1)$ and $(4, 16)$ . Equation of the secant: $y - 1 = 5(x - 1)$ , or $y = 5x - 4$ . Substituting $x = 4$ : $y = 16$ , confirmed. Substituting $x = 1$ : $y = 1$ , confirmed.

Source. Active Calculus §1.3 — CC-BY-NC-SA license.

Example— 2· Average velocity in free fall (kinematics)

Problem: The height of a stone in free fall is $h(t) = 80 - 5t^2$ (m). Calculate the average velocity on the interval $[1, 3]$ s.

Strategy: Average velocity is the ARC of position. Since $h$ is in meters and $t$ in seconds, the ARC has units of m/s.

Solution:

Compute $h(1) = 80 - 5 = 75$ m and $h(3) = 80 - 45 = 35$ m.
Change in height: $\Delta h = 35 - 75 = -40$ m (negative: the stone fell).
Change in time: $\Delta t = 3 - 1 = 2$ s.
Average velocity: $v_m = \Delta h/\Delta t = -40/2 = -20$ m/s.

Verification: The derivative $h'(t) = -10t$ (calculus preview). At the midpoint $t = 2$ : $h'(2) = -20$ m/s. By the Mean Value Theorem, for quadratics the derivative at the midpoint equals the ARC — confirmed.

Source. OpenStax Calculus Volume 1 §2.1 — CC-BY-NC-SA license.

Example— 3· ARC on [a, a+h] and the limit that becomes the derivative (intermediate)

Problem: For $f(x) = x^2$ , find the ARC on the interval $[a, a+h]$ as a function of $a$ and $h$ . What happens when $h \to 0$ ?

Strategy: Apply the definition with endpoints $a$ and $a + h$ , then simplify algebraically.

Solution:

$f(a + h) = (a + h)^2 = a^2 + 2ah + h^2$ .
Numerator: $f(a + h) - f(a) = 2ah + h^2 = h(2a + h)$ .
Denominator: $(a + h) - a = h$ .
ARC $= h(2a + h)/h = 2a + h$ (valid for $h \neq 0$ ).
When $h \to 0$ : ARC $\to 2a$ .

Verification: For $a = 1$ and $h = 0{,}1$ : ARC $= 2{,}1$ . For $h = 0{,}01$ : ARC $= 2{,}01$ . The limit tends to $2$ . This is how Cauchy showed that $f'(x) = 2x$ for $f(x) = x^2$ .

Source. Active Calculus §1.3 — CC-BY-NC-SA license.

Example— 4· Marginal cost via ARC (economics, advanced)

Problem: A company has cost $C(q) = 0{,}5 q^2 + 30 q + 200$ (R$). Calculate the ARC of cost between $q = 100$ and $q = 101$ units. Interpret economically.

Strategy: ARC $= C(101) - C(100)$ . Since $\Delta q = 1$ , the ARC coincides with the cost of the marginal unit.

Solution:

$C(100) = 5{,}000 + 3{,}000 + 200 = 8{,}200$ R$.
$C(101) = 5{,}100{,}5 + 3{,}030 + 200 = 8{,}330{,}5$ R$.
ARC $= 8{,}330{,}5 - 8{,}200 = 130{,}5$ R$/unit.
Interpretation: producing the 101st unit costs approximately R$ 130.50 more.

Verification: The derivative $C'(q) = q + 30$ , so $C'(100) = 130$ R$/unit. The ARC (130.5) is a good estimate of the true marginal cost (130). The small difference (0.5) disappears as $\Delta q \to 0$ .

Source. Yoshiwara — Modeling, Functions, and Graphs ch. 5 — open license.

Example— 5· Estimating the derivative via successive ARCs (modeling)

Problem: For $f(x) = \sqrt{x}$ , calculate the ARC on $[4,\, 4{,}5]$ , $[4,\, 4{,}1]$ , and $[4,\, 4{,}01]$ . To what value do these ARCs converge?

Strategy: Compute each ARC numerically and observe convergence. As $\Delta x$ shrinks, the ARC approaches the derivative.

Solution:

$[4,\, 4{,}5]$ : $f(4) = 2$ , $f(4{,}5) \approx 2{,}1213$ . ARC $\approx 0{,}1213/0{,}5 \approx 0{,}2426$ .
$[4,\, 4{,}1]$ : $f(4{,}1) \approx 2{,}0248$ . ARC $\approx 0{,}0248/0{,}1 \approx 0{,}2485$ .
$[4,\, 4{,}01]$ : $f(4{,}01) \approx 2{,}002498$ . ARC $\approx 0{,}002498/0{,}01 \approx 0{,}2498$ .
The ARCs converge to $0{,}25 = 1/4$ . Therefore $f'(4) = 1/4$ .

Verification: The exact derivative is $f'(x) = 1/(2\sqrt{x})$ . At $x = 4$ : $f'(4) = 1/4$ . Confirmed.

Source. Active Calculus §1.5 — CC-BY-NC-SA license.

Exercise list

40 exercises · 10 with worked solution (25%)

Application 23Understanding 5Modeling 7Challenge 5

Ex. 9.1UnderstandingAnswer key
Can the average rate of change of a function be constant? Explain with an example.
Select the correct option
Yes — every linear function has constant average rate equal to its slope.No — the average rate always changes when the interval changes.Yes, but only for constant functions.No — only the instantaneous rate can be constant.
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 1
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For a linear function $f(x) = mx + b$ , the average rate on any interval $[a, c]$ is $\frac{(mc+b)-(ma+b)}{c-a} = \frac{m(c-a)}{c-a} = m$ , which is constant. For non-linear functions, the average rate varies with the interval.
Ex. 9.2Application
Find the average rate of change of $f(x) = 4x^2 - 7$ on the interval $[1,\, b]$ in simplest form.
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$4(b+1)$ $4(b-1)$ $8b$ $4b$
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 5
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We have $f(1) = 4-7 = -3$ and $f(b) = 4b^2-7$ . The average rate is $\frac{4b^2-7-(-3)}{b-1} = \frac{4b^2-4}{b-1} = \frac{4(b-1)(b+1)}{b-1} = 4(b+1)$ .
Show step-by-step (with the why)
1. Compute $f(1) = 4(1)^2 - 7 = -3$ .
2. Write $f(b) = 4b^2 - 7$ .
3. Form the quotient $\frac{4b^2-7-(-3)}{b-1} = \frac{4b^2-4}{b-1}$ .
4. Factor: $4(b^2-1) = 4(b-1)(b+1)$ ; cancel $b-1$ .
Ex. 9.3ApplicationAnswer key
Find the average rate of change of $g(x) = 2x^2 - 9$ on the interval $[4,\, b]$ in simplest form.
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$2(b+4)$ $2(b-4)$ $4b$ $2b+8$
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 6
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We have $g(4) = 32 - 9 = 23$ and $g(b) = 2b^2-9$ . The average rate is $\frac{2b^2-9-23}{b-4} = \frac{2(b^2-16)}{b-4} = \frac{2(b-4)(b+4)}{b-4} = 2(b+4)$ .
Ex. 9.4Application
Find the average rate of change of $p(x) = 3x + 4$ on the interval $[2,\; 2+h]$ in simplest form.
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$3$ $3h$ $3 + h$ $6 + 3h$
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 7
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We have $p(2) = 10$ and $p(2+h) = 3(2+h)+4 = 10+3h$ . The average rate is $\frac{10+3h-10}{h} = \frac{3h}{h} = 3$ . The function is linear, so the rate is always $3$, regardless of $h$.
Ex. 9.5ApplicationAnswer key
Find the average rate of change of $k(x) = 4x - 2$ on the interval $[3,\; 3+h]$ in simplest form.
Select the correct option
$4$ $4h$ $4 + h$ $8 + 4h$
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OpenStax College Algebra 2e · §3.3 · ex. 8
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We have $k(3) = 12-2 = 10$ and $k(3+h) = 4(3+h)-2 = 10+4h$ . The average rate is $\frac{10+4h-10}{h} = 4$ . Since $k$ is linear, the rate is always $4$.
Ex. 9.6Application
Find the average rate of change of $f(x) = 2x^2 + 1$ on the interval $[x,\; x+h]$ in simplest form.
Select the correct option
$4x + 2h$ $4x$ $2x + h$ $4x + h$
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 9
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We have $f(x+h) = 2(x+h)^2+1 = 2x^2+4xh+2h^2+1$ . The numerator is $4xh+2h^2$ . Dividing by $h$: $4x + 2h$ .
Show step-by-step (with the why)
1. Expand $f(x+h) = 2(x+h)^2+1$ .
2. Subtract $f(x) = 2x^2+1$ ; the constant terms cancel.
3. Divide $4xh + 2h^2$ by $h$.
Ex. 9.7Application
Find the average rate of change of $g(x) = 3x^2 - 2$ on the interval $[x,\; x+h]$ in simplest form.
Select the correct option
$6x + 3h$ $6x$ $3x + h$ $6x + h$
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 10
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We have $g(x+h) = 3(x+h)^2-2 = 3x^2+6xh+3h^2-2$ . Subtracting $g(x)$ : numerator $6xh+3h^2$ . Dividing by $h$: $6x+3h$ .
Ex. 9.8Application
Find the average rate of change of $a(t) = \dfrac{1}{t+4}$ on the interval $[9,\; 9+h]$ in simplest form.
Select the correct option
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 11
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We have $a(9) = \frac{1}{13}$ and $a(9+h) = \frac{1}{13+h}$ . The average rate is $\frac{\frac{1}{13+h}-\frac{1}{13}}{h} = \frac{13-(13+h)}{13h(13+h)} = \frac{-1}{13(13+h)}$ . When $h \to 0$ , the limit is $-\frac{1}{169}$ .
Show step-by-step (with the why)
1. Compute $a(9) = \frac{1}{9+4} = \frac{1}{13}$ .
2. Compute $a(9+h) = \frac{1}{13+h}$ .
3. Form the quotient and use the difference of fractions in the numerator.
Ex. 9.9Application
Find the average rate of change of $b(x) = \dfrac{1}{x+3}$ on the interval $[1,\; 1+h]$ in simplest form.
Select the correct option
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 12
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We have $b(1) = \frac{1}{4}$ and $b(1+h) = \frac{1}{4+h}$ . The average rate is $\frac{\frac{1}{4+h}-\frac{1}{4}}{h} = \frac{4-(4+h)}{4h(4+h)} = \frac{-1}{4(4+h)}$ . When $h \to 0$ , the limit is $-\frac{1}{16}$ .
Ex. 9.10ApplicationAnswer key
Find the average rate of change of $j(x) = 3x^3$ on the interval $[1,\; 1+h]$ in simplest form.
Select the correct option
$9 + 9h + 3h^2$ $9$ $9h + 3h^2$ $3 + 3h + h^2$
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 13
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We have $j(1) = 3$ and $j(1+h) = 3(1+h)^3 = 3(1+3h+3h^2+h^3)$ . The numerator is $3(3h+3h^2+h^3)$ . Dividing by $h$: $3(3+3h+h^2) = 9+9h+3h^2$ . When $h \to 0$ , the limit is $9$.
Show step-by-step (with the why)
1. Expand $(1+h)^3 = 1+3h+3h^2+h^3$ .
2. Multiply by $3$ and subtract $j(1) = 3$ .
3. Divide by $h$; simplify.
Ex. 9.11Application
Find the average rate of change of $r(t) = 4t^3$ on the interval $[2,\; 2+h]$ in simplest form.
Select the correct option
$48 + 24h + 4h^2$ $48$ $24 + 12h + 4h^2$ $12 + 6h + h^2$
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 14
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We have $r(2) = 32$ and $r(2+h) = 4(2+h)^3 = 4(8+12h+6h^2+h^3)$ . The numerator is $4(12h+6h^2+h^3)$ . Dividing by $h$: $4(12+6h+h^2) = 48+24h+4h^2$ . When $h \to 0$ , the limit is $48$, which is the derivative of $4t^3$ at $t=2$.
Ex. 9.12Application
Compute $\dfrac{f(x+h)-f(x)}{h}$ for $f(x) = 2x^2 - 3x$ in simplest form.
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$4x + 2h - 3$ $4x - 3$ $2x + h$ $4x + 2h$
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 15
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We have $f(x+h) = 2(x+h)^2 - 3(x+h) = 2x^2+4xh+2h^2-3x-3h$ . Subtracting $f(x) = 2x^2-3x$ : numerator $4xh+2h^2-3h$ . Dividing by $h$: $4x+2h-3$ . When $h\to 0$ : $4x-3$ , which is the derivative.
Show step-by-step (with the why)
1. Expand $f(x+h) = 2(x+h)^2-3(x+h)$ .
2. Subtract $f(x)$ .
3. Divide by $h$ and simplify.
Ex. 9.13Application
Calculate the average rate of change of $f(x) = x^2$ on the interval $[1,5]$ .
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$6$ $5$ $4$ $24$
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 28
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We have $f(1) = 1$ and $f(5) = 25$ . The average rate is $\frac{25-1}{5-1} = \frac{24}{4} = 6$ .
Ex. 9.14Application
Calculate the average rate of change of $h(x) = 5 - 2x^2$ on the interval $[-2,4]$ .
Select the correct option
$-4$ $4$ $-12$ $-2$
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 29
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We have $h(-2) = 5-2(4) = -3$ and $h(4) = 5-2(16) = -27$ . The average rate is $\frac{-27-(-3)}{4-(-2)} = \frac{-24}{6} = -4$ .
Ex. 9.15ApplicationAnswer key
Calculate the average rate of change of $q(x) = x^3$ on the interval $[-4,2]$ .
Select the correct option
$12$ $-12$ $6$ $-6$
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 30
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We have $q(-4) = -64$ and $q(2) = 8$ . The average rate is $\frac{8-(-64)}{2-(-4)} = \frac{72}{6} = 12$ .
Ex. 9.16Application
Calculate the average rate of change of $g(x) = 3x^3 - 1$ on the interval $[-3,3]$ .
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$27$ $-27$ $54$ $9$
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 31
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We have $g(-3) = 3(-27)-1 = -82$ and $g(3) = 3(27)-1 = 80$ . The average rate is $\frac{80-(-82)}{3-(-3)} = \frac{162}{6} = 27$ .
Ex. 9.17Application
Calculate the average rate of change of $y = \dfrac{1}{x}$ on the interval $[1,3]$ .
Select the correct option
$-\dfrac{1}{3}$ $\dfrac{1}{3}$ $-\dfrac{2}{3}$ $-3$
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 32
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We have $y(1) = 1$ and $y(3) = \frac{1}{3}$ . The average rate is $\frac{\frac{1}{3}-1}{3-1} = \frac{-\frac{2}{3}}{2} = -\frac{1}{3}$ .
Ex. 9.18Understanding
If a function $f$ is increasing on $(a,b)$ and decreasing on $(b,c)$ , what can be said about the local extremum of $f$ on $(a,c)$ ?
Select the correct option
If $f$ increases on $(a,b)$ and decreases on $(b,c)$ , then $f$ has a local maximum at $x=b$ .If $f$ increases on $(a,b)$ and decreases on $(b,c)$ , then $f$ has a local minimum at $x=b$ .It is not possible to determine whether there is an extremum at $x=b$ . $f$ has an inflection point at $x=b$ , not an extremum.
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 2
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When a function changes from increasing to decreasing at a point, it reaches a peak — that is, a local maximum. If the change were from decreasing to increasing, there would be a local minimum.
Ex. 9.19Modeling
At the start of a trip, the odometer read $21\,395$ miles. At the end, $13{,}5$ hours later, it read $22\,125$ miles. What was the average speed of the trip?
Select the correct option
Approximately $54{,}07$ miles per hour.Approximately $730$ miles per hour.Approximately $27$ miles per hour.Approximately $108$ miles per hour.
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 44
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The distance traveled was $22125 - 21395 = 730$ miles. The time was $13{,}5$ hours. The average speed was $\frac{730}{13{,}5} \approx 54{,}07$ miles per hour.
Show step-by-step (with the why)
1. Compute the distance: $22125 - 21395 = 730$ miles.
2. Divide by the duration: $730 \div 13{,}5 \approx 54{,}07$ mph.
Ex. 9.20Modeling
Near the surface of the Moon, the distance a falling object travels is given by $d(t) = 2{,}6667\,t^2$ feet, where $t$ is in seconds. What is the distance traveled at $t=2$ s, and what is the average speed between $t=1$ s and $t=2$ s?
Select the correct option
$d(2) = 10{,}6668$ feet; average rate on $[1,2]$ is $\approx 8$ ft/s. $d(2) = 10{,}6668$ feet; average rate on $[1,2]$ is $\approx 5{,}3$ ft/s. $d(2) = 2{,}6667$ feet; average rate on $[1,2]$ is $\approx 2{,}7$ ft/s. $d(2) = 5{,}3334$ feet; average rate on $[1,2]$ is $\approx 4$ ft/s.
Select an option first
Solve online OpenStax College Algebra 2e · §3.3 · ex. 46
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We have $d(t) = 2{,}6667t^2$ . So $d(1) = 2{,}6667$ and $d(2) = 2{,}6667 \times 4 = 10{,}6668$ feet. The average rate is $\frac{10{,}6668 - 2{,}6667}{2-1} \approx 8{,}0001 \approx 8$ ft/s.
Ex. 9.21ModelingAnswer key
A ball is thrown upward with initial velocity of $36$ ft/s. Its height is $y = 36t - 12t^2$ feet. What is the average velocity on the interval $[1,2]$ ?
Select the correct option
$0$ m/s $36$ m/s $12$ m/s $-12$ m/s
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Solve online Active Calculus · §1.1 · ex. 2
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We have $y(1) = 36 - 12 = 24$ m and $y(2) = 72 - 48 = 24$ m. The average velocity on $[1,2]$ is $\frac{24-24}{2-1} = 0$ m/s. The object rose and fell symmetrically over that interval.
Show step-by-step (with the why)
1. Compute $y(1) = 36(1) - 12(1)^2 = 24$ feet.
2. Compute $y(2) = 36(2) - 12(2)^2 = 72-48 = 24$ feet.
3. The average rate is $(24-24)/(2-1) = 0$ .
Ex. 9.22Modeling
For $f(x) = 25 - x^2$ , what is the average rate of change on the interval $[0,3]$ ?
Select the correct option
$-3$ $3$ $-8$ $0$
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Solve online Active Calculus · §1.1 · ex. 3
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We have $f(3) = 25-9 = 16$ , $f(0) = 25$ . ARC on $[0,3]$ : $\frac{16-25}{3} = -3$ . We have $f(5) = 0$ . ARC on $[3,5]$ : $\frac{0-16}{2} = -8$ . ARC on $[0,5]$ : $\frac{0-25}{5} = -5$ . The average rate on $[0,3]$ is $-3$.
Ex. 9.23Modeling
According to the US census, the population of Grand Rapids, Michigan, was $181\,843$ in 1980 and $197\,800$ in 2000. What was the average annual growth rate over that period?
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Approximately $798$ people per year.Approximately $1,600$ people per year.Approximately $399$ people per year.Approximately $15,957$ people per year.
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Solve online Active Calculus · §1.1 · ex. 11
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The population grew from $181\,843$ in 1980 to $197\,800$ in 2000, a growth of $15\,957$ people over $20$ years. The average rate is $\frac{15957}{20} \approx 798$ people per year.
Show step-by-step (with the why)
1. Compute the change: $197800 - 181843 = 15957$ people.
2. Divide by the interval: $15957 \div 20 = 797{,}85 \approx 798$ people/year.
Ex. 9.24Challenge
For $f(x) = \dfrac{1}{x}$ , find the number $c$ such that the average rate of change of $f$ on the interval $(1, c)$ equals $-\dfrac{1}{4}$ .
Select the correct option
$c = 4$ $c = 2$ $c = -4$ $c = 5$
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 42
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We have $f(x) = \frac{1}{x}$ . The average rate on $(1,c)$ is $\frac{\frac{1}{c}-1}{c-1} = \frac{1-c}{c(c-1)} = \frac{-(c-1)}{c(c-1)} = -\frac{1}{c}$ . Setting equal to $-\frac{1}{4}$ : $c = 4$ .
Show step-by-step (with the why)
1. Write the average rate: $\frac{f(c)-f(1)}{c-1} = \frac{\frac{1}{c}-1}{c-1}$ .
2. Simplify: $\frac{1-c}{c(c-1)} = -\frac{1}{c}$ .
3. Solve $-\frac{1}{c} = -\frac{1}{4}$ .
Ex. 9.25Challenge
For $f(x) = \dfrac{1}{x}$ , find the number $b$ such that the average rate of change of $f$ on the interval $(2, b)$ equals $-\dfrac{1}{10}$ .
Select the correct option
$b = 5$ $b = 3$ $b = 10$ $b = -5$
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 43
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The average rate of $f(x) = \frac{1}{x}$ on $(2,b)$ is $\frac{\frac{1}{b}-\frac{1}{2}}{b-2} = \frac{2-b}{2b(b-2)} = -\frac{1}{2b}$ . Setting equal to $-\frac{1}{10}$ : $2b = 10$ , so $b = 5$ .
Ex. 9.26Understanding
The temperature $H$ (in °C) of a cup of coffee placed on the kitchen counter is given by $H = f(t)$ , where $t$ is in minutes. What is the sign of $f'(t)$ ?
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Negative, because the coffee cools over time.Positive, because the temperature increases over time.Zero, because the temperature is constant.It is not possible to determine the sign without more data.
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Solve online Active Calculus · §1.1 · ex. 1
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Coffee placed on the counter cools progressively. Therefore, the temperature $H = f(t)$ is decreasing in $t$, that is, $f'(t) < 0$ .
Ex. 9.27Understanding
The cost $C$ (in dollars) of producing $g$ gallons of ice cream is given by $C = f(g)$ . In the expression $f(125) = 325$ , what are the units of $125$ ?
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Gallons of ice cream produced.Dollars per gallon.Gallons per dollar.Dollars squared.
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Solve online Active Calculus · §1.1 · ex. 2
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In the expression $f(125) = 325$ , the argument $125$ is the input of the function $C = f(g)$, where $g$ is the quantity produced in gallons. Therefore $125$ has units of gallons.
Ex. 9.28ApplicationAnswer key
The table below gives values of $f(x)$ : $x = 0, 2, 4, 6, 8, 10, 12$ and $f(x) = -16, -13, -12, -13, -17, -21, -25$ . What is the average rate of change of $f$ on the interval $[0,4]$ ?
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$1{,}0$ °C/cm² $-1{,}0$ °C/cm² $0{,}5$ °C/cm² $2{,}0$ °C/cm²
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Solve online Active Calculus · §1.1 · ex. 5
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The requested quotient is the average rate of change of $f$ on $[0,4]$: $\frac{f(4)-f(0)}{4-0}$ . From the table: $f(0) = -16$ and $f(4) = -12$ . So $\frac{-12-(-16)}{4} = \frac{4}{4} = 1{,}0$ .
Ex. 9.29Application
Using the table ( $x = 0, 2, 4, 6, 8, 10, 12$ and $f(x) = -16, -13, -12, -13, -17, -21, -25$ ), estimate $f'(2)$ by central difference.
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$1$ $-1$ $0{,}5$ $-0{,}5$
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Solve online Active Calculus · §1.1 · ex. 5
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From the table: $f(0) = -16$ and $f(4) = -12$ . The estimate of $f'(2)$ by central difference is $\frac{f(4)-f(0)}{4-0} = \frac{-12-(-16)}{4} = \frac{4}{4} = 1$ .
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1. Use the central difference: $f'(2) \approx \frac{f(4)-f(0)}{4-0}$ .
2. Substitute $f(4) = -12$ and $f(0) = -16$ .
3. Compute $\frac{4}{4} = 1$ .
Ex. 9.30Modeling
The velocity of a vertically launched ball is given by $v(t) = 16 - 32t$ (in ft/s). At what instant does the ball reach maximum height (zero velocity)?
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$t = 0{,}5$ s $t = 1$ s $t = 2$ s $t = 0{,}25$ s
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Solve online Active Calculus · §1.1 · ex. 8
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The velocity is $v(t) = 16 - 32t$ . To find when the ball stops, solve $16 - 32t = 0$ , giving $t = \frac{16}{32} = 0{,}5$ s.
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1. The ball stops when $v(t) = 0$.
2. Solve $16 - 32t = 0$ .
3. Isolate $t$: $t = 0{,}5$ s.
Ex. 9.31Modeling
The value $V$ (in dollars) of a car depends on the mileage $m$ driven: $V = h(m)$ . If $h(40\,000) = 15\,500$ and $h(55\,000) = 13\,200$ , what is the average rate of change of the car's value with respect to mileage on that interval?
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Approximately $-153{,}33$ dollars per mile.Approximately $153{,}33$ dollars per mile.Approximately $-2300$ dollars per mile.Approximately $-76{,}67$ dollars per mile.
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Solve online Active Calculus · §1.1 · ex. 9
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We have $h(40000) = 15500$ and $h(55000) = 13200$ . The average rate is $\frac{13200 - 15500}{55000 - 40000} = \frac{-2300}{15000} \approx -0{,}1533$ dollars per mile, i.e., approximately $-153{,}33$ cents per thousand miles.
Ex. 9.32ApplicationAnswer key
Consider $f(x) = -7x - 4$ . What is the average rate of change of $f$ between the points $(-1, f(-1))$ and $(2, f(2))$ ?
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$-5$ $5$ $-7$ $-3$
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Solve online Active Calculus · §1.1 · ex. 7
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We have $f(-1) = -7(-1)-4 = 3$ and $f(2) = -7(2)-4 = -18$ . ARC on $[-1,2]$ : $\frac{-18-3}{2-(-1)} = \frac{-21}{3} = -7$ . Since $f$ is linear with slope $-7$, the ARC between any two points is always $-7$.
Ex. 9.33Understanding
What is the precise relationship between the average rate of change and the derivative of a function?
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The derivative $f'(a)$ is the limit of the average rate of change as the interval shrinks to a point.The derivative $f'(a)$ is the arithmetic mean of $f$ over an interval.The average rate and the derivative are always equal.The derivative $f'(a)$ exists only for linear functions.
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Solve online Active Calculus · §1.1 · ex. 5
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By definition, $f'(a) = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$ . The quotient inside the limit is exactly the average rate of change on the interval $[a, a+h]$ . When $h \to 0$ , the secant line becomes the tangent line.
Ex. 9.34ApplicationAnswer key
Compute $\dfrac{f(x+h)-f(x)}{h}$ for $f(x) = 5 - 2x^2$ in simplest form.
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$-4x - 2h$ $-4x$ $4x + 2h$ $-2x - h$
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Solve online Active Calculus · §1.1 · ex. 3
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We have $f(x) = 5 - 2x^2$ . $f(x+h) = 5 - 2(x+h)^2 = 5-2x^2-4xh-2h^2$ . Numerator: $-4xh-2h^2$ . Dividing by $h$: $-4x-2h$ . When $h \to 0$ : $-4x$ .
Ex. 9.35Application
The function $s(t) = 4t^2$ describes the position of an object in meters. What is the average rate of change between $t=1$ s and $t=3$ s?
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$16$ $8$ $-16$ $32$
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 28
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We have $s(1) = 4$ and $s(3) = 4 \times 9 = 36$ . The average rate is $\frac{36-4}{3-1} = \frac{32}{2} = 16$ .
Ex. 9.36ApplicationAnswer key
A tank has $V(t) = 100 - 5t$ liters after $t$ minutes. What is the average rate of change of the volume between $t=2$ and $t=6$ minutes?
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$-5$ $5$ $-15$ $-10$
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 7
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Since $V(t) = 100 - 5t$ is linear with slope $-5$, the average rate of change is always $-5$ liters per minute, regardless of the interval.
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1. Compute $V(6) = 100 - 30 = 70$ L.
2. Compute $V(2) = 100 - 10 = 90$ L.
3. ARC: $\frac{70 - 90}{6-2} = \frac{-20}{4} = -5$ .
Ex. 9.37Application
A cyclist travels $s(t) = 2t^2 + 3t$ km in $t$ hours. What is the average speed between $t=1$ h and $t=4$ h?
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$13$ km/h $-13$ km/h $6{,}5$ km/h $39$ km/h
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 9
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We have $s(4) = 2 \times 16 + 3 \times 4 = 32 + 12 = 44$ km and $s(1) = 2 + 3 = 5$ km. The average speed is $\frac{44-5}{4-1} = \frac{39}{3} = 13$ km/h.
Show step-by-step (with the why)
1. Evaluate $s(4) = 2(16) + 3(4) = 44$ .
2. Evaluate $s(1) = 2 + 3 = 5$ .
3. Divide the difference by the interval of $3$ hours.
Ex. 9.38Challenge
For $s(t) = t^3$ , find the average rate of change between $t=1$ and $t = 1 + \Delta t$ in simplest form.
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$3 + 3\Delta t + (\Delta t)^2$ $3$ $3 + (\Delta t)^2$ $1 + \Delta t$
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Solve online Active Calculus · §1.1 · ex. 2
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We have $s(1+\Delta t) = (1+\Delta t)^3 = 1 + 3\Delta t + 3(\Delta t)^2 + (\Delta t)^3$ and $s(1) = 1$ . The numerator is $3\Delta t + 3(\Delta t)^2 + (\Delta t)^3$ . Dividing by $\Delta t$ : $3 + 3\Delta t + (\Delta t)^2$ . When $\Delta t \to 0$ , the limit is $3$, which is the derivative of $t^3$ at $t=1$.
Show step-by-step (with the why)
1. Expand $(1+\Delta t)^3$ .
2. Subtract $s(1) = 1$.
3. Divide by $\Delta t$ and simplify.
Ex. 9.39Challenge
What is the average rate of change of $f(x) = \sin x$ on the interval $[0, \pi]$ ?
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$0$ — the average rate of $\sin x$ on $[0,\pi]$ is zero. $\dfrac{2}{\pi}$ $1$ $-1$
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 32
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The average rate of change of $\sin x$ on $[0, \pi]$ is $\frac{\sin \pi - \sin 0}{\pi - 0} = \frac{0 - 0}{\pi} = 0$ . The sine rises and falls symmetrically, starting and ending at zero.
Ex. 9.40Challenge
The centripetal force is $F(r) = \dfrac{mv^2}{r}$ . What is the average rate of change of $F$ when $r$ varies from $1$ m to $2$ m, with $m$ and $v$ constant?
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$\dfrac{-mv^2}{2}$ $\dfrac{mv^2}{2}$ $-mv^2$ $\dfrac{mv^2}{4}$
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Solve online OpenStax College Algebra 2e · §3.3 · ex. 30
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We have $F(r) = \frac{mv^2}{r}$ . Then $F(1) = mv^2$ and $F(2) = \frac{mv^2}{2}$ . The average rate is $\frac{F(2)-F(1)}{2-1} = \frac{mv^2}{2} - mv^2 = -\frac{mv^2}{2}$ .

Sources

Active Calculus 2.0 — Matt Boelkins · 2024 · EN · CC-BY-NC-SA · §1.1, §1.3, §1.5 (ARC as motivation for the derivative). Primary source for this lesson.
Calculus Volume 1 — OpenStax · 2016 · EN · CC-BY-NC-SA · §2.1 (preview of calculus) and §3.1 (defining the derivative) and §4.4 (Mean Value Theorem).
Modeling, Functions, and Graphs — Katherine Yoshiwara · 2020 · EN · open · ch. 5 (ARC in economic and biological modeling).

This lesson is the gateway to Calculus — the ARC will appear again in Lessons 41–50 (Trimesters 5–6) under the name "derivative."