Chapter 1 – Functions and Rates of Change

1.1 Learning Objectives

By the end of this chapter, you should be able to:

• Explain what a function is in words, symbols, tables, and graphs.
• Distinguish independent and dependent variables in real situations.
• Evaluate and interpret functions given by formulas, tables, or graphs.
• Compute average rate of change on an interval and interpret it in context.

Recognize common function families (linear, quadratic, exponential, rational, piecewise) and what their graphs typically look like.

Connect average rate of change with the idea of a secant line on a graph.

1.2 Big Picture: Why Start Here?

Calculus is about change: how things move, grow, decay, speed up, slow down. To talk rigorously about change, we need:

A way to describe how one quantity depends on another → functions.

A way to measure “how fast” that dependence changes → rates of change.

You’ve seen functions before in algebra, but in calculus we treat them as objects we can analyze: we ask not just “what is f(3)?”, but “how does f behave near 3?”, “how fast is it changing there?”, “what patterns does its graph follow?”.

This chapter lays the language the rest of the course uses:

What is a function?

How do we read it from tables/graphs?

How do we talk about “change per unit” in a precise way?

1.3 Functions: Rules, Inputs, and Outputs

1.3.1 Intuitive idea

A function is a rule that assigns to each allowed input exactly one output.

Input → independent variable → usually written as xx or tt.

Output → dependent variable → usually written as yy or f(x)f(x).

Examples:

Temperature as a function of time: T(t)T(t).

Position along a line as a function of time: s(t)s(t).

Area of a circle as a function of radius: A(r)=πr2A(r)=πr2.

Key idea: for each input, there is exactly one output. If an “input” could give two different outputs, it’s not a function.

1.3.2 Domain and range

Every function has:

Domain: all inputs for which the rule makes sense.

Range: all outputs the function can actually produce.

Example:

f(x)=1xf(x)= x1​

Domain: all real numbers except 0 (you can’t divide by zero).

Range: all nonzero real numbers.

g(x)= xg(x)= x

​ (real-valued)

Domain: x≥0x≥0.

Range: g(x)≥0g(x)≥0.

On graphs:

Domain → shadow of the graph on the x-axis.

Range → shadow on the y-axis.

1.3.3 Ways functions show up

Formula – e.g. f(x)=2x2−3x+1f(x)=2x2−3x+1.

Table – discrete data (e.g. week vs height).

Graph – you read values visually.

Words/story – “cost is $4 plus $2 per mile” → translate to a rule.

A strong student can move between these forms.

1.4 Independent and Dependent Variables

The independent variable is what you choose, control, or what flows on its own (often time).

The dependent variable is what responds.

Examples:

“Height of a plant as a function of time”

Independent: time tt.

Dependent: height h(t)h(t).

“Fuel left as a function of distance driven”

Independent: distance xx.

Dependent: fuel F(x)F(x).

Sometimes the choice isn’t unique, but we pick whatever makes the story most natural.

1.5 Average Rate of Change

1.5.1 The formula

For a function f(x)f(x), the average rate of change of ff on [a,b][a,b] is

Average rate of change = f(b)−f(a)b−a.

Average rate of change = b−af(b)−f(a)​.

This is “change in output” divided by “change in input”.

If f(t)f(t) is position, average rate of change is average velocity.

If C(q)C(q) is cost, average rate of change is average cost per unit.

1.5.2 Units and interpretation

Units tell you what the rate actually means.

ff in meters, xx in seconds → rate in m/s.

ff in dollars, xx in years → rate in $/year.

Example: Volume V(t)V(t) in liters, time tt in minutes.

V(5)−V(2)5−2

5−2V(5)−V(2)​

has units liters/minute and measures how fast the volume changed on average between 2 and 5 minutes.

1.5.3 Geometric picture: secant lines

On the graph of y = f(x)y = f(x):

Point 1: (a,f(a))(a,f(a))

Point 2: (b,f(b))(b,f(b))

The line through them (a secant line) has slope

msecant = f(b)−f(a)b−a.

msecant​= b−af(b)−f(a)​.

So:

Slope of the secant line = average rate of change on [a,b].

Later, the derivative will be the slope of a tangent line, the limit of these secant slopes as b→ab→a.

1.6 Function Families: Patterns

1.6.1 Linear

f(x)= mx+bf(x)= mx+b

Straight line; slope mm is constant rate of change.

b = f(0)b = f(0) is starting value.

1.6.2 Quadratic

f(x)= ax2+bx+c, a≠0f(x)= ax2+bx+c, a=0

Parabola (up if a>0a>0, down if a<0a<0).

Rate of change itself changes linearly with xx.

1.6.3 Exponential

f(x)= Arxorf(x)= Aekxf(x)= Arxorf(x)= Aekx

Each unit of xx multiplies the output by a fixed factor.

Used for growth/decay, interest, etc.

1.6.4 Rational

f(x)= p(x)q(x)f(x)= q(x)p(x)​ with polynomials p,qp,q.

Can have vertical/horizontal asymptotes.

Behavior can change dramatically near where q(x)=0q(x)=0.

1.6.5 Piecewise

Different rules on different parts of the domain:

F(x)={2x+1x<0,x2x≥0.

F(x)={2x+1x2​x<0,x≥0.​

Useful for “if–then” pricing, taxes, and systems with regime changes.

1.7 Concept Check (Short)

In “height of a plant as a function of time,” identify independent and dependent variables.

For f(x)=7−0.5xf(x)=7−0.5x, what is the rate of change, and what does its sign mean?

For s(t)s(t) position in meters and tt in seconds, what are the units of s(8)−s(3)8−38−3s(8)−s(3)​?

If the graph of a function is a straight line, what can you say about its average rate of change over different intervals?

Give a real-world example that is likely modeled well by an exponential function.

1.8 Practice Problems (24 Problems)

A. Basic Function Evaluation and Interpretation

Problem 1.

Let f(x)=2x2−3x+1f(x)=2x2−3x+1. Compute:

a) f(0)f(0)

b) f(2)f(2)

c) f(−1)f(−1)

Problem 2.

Let g(t)=5e0.2tg(t)=5e0.2t represent a bacteria population (in millions) tt hours after an experiment begins.

a) What is the initial population?

b) Is this growth or decay?

c) Evaluate g(3)g(3) (leave as an exact expression).

Problem 3. (Table)

The table shows the height of a plant over time:

Week tt 0 1 2 3 4

Height h(t)h(t) (cm) 5 7 10 14 19

a) Find h(0)h(0) and h(4)h(4).

b) What is the average rate of change in height from week 1 to week 4?

Problem 4. (Domain)

Find the domain of each function:

a) f(x)=1x−3f(x)= x−31​

b) g(x)=5−2xg(x)=5−2x

​ (real-valued)

c) h(x)= ln⁡(x2−4)h(x)= ln(x2−4)

Problem 5. (Piecewise)

Define

F(x)={2x+1x<0,x2x≥0.

F(x)={2x+1x2​x<0,x≥0.​

a) Compute F(−2)F(−2), F(0)F(0), F(3)F(3).

b) Is FF linear, quadratic, neither, or piecewise?

B. Average Rate of Change – Numeric and Algebraic

Problem 6.

For f(x)= x2+4x+1f(x)= x2+4x+1, find the average rate of change from:

a) x =0x =0 to x =2x =2

b) x =1x =1 to x =4x =4

Problem 7. (General interval)

Let f(x)= x2f(x)= x2. Find the average rate of change from x = ax = a to x = a+hx = a+h, where h≠0h=0.

Problem 8. (Velocity from position)

A particle moves on a line, and its position is

s(t)=3t2−2t+1(meters),t in seconds.

s(t)=3t2−2t+1(meters),t in seconds.

a) Find the average velocity on the time interval [1,5][1,5].

b) Interpret your answer with units.

Problem 9. (From a graph)

Suppose you have a graph of a differentiable function y = f(x)y = f(x).

Estimate the average rate of change between x =1x =1 and x =4x =4 by reading approximate values f(1)f(1) and f(4)f(4) from the graph.

Problem 10. (Nonlinear units)

The radius of a spherical balloon (in centimeters) as a function of time (in seconds) is

r(t)=2+0.5t.

r(t)=2+0.5t.

a) Express the volume of the balloon V(t)V(t) in terms of tt (use V =43πr3V =34​πr3).

b) Compute the average rate of change of volume from t =0t =0 to t =2t =2.

C. Units, Context, and Interpretation

Problem 11.

A water tank contains W(t)W(t) liters of water at time tt hours. You know that:

W(5)−W(3)5−3=−250.

5−3W(5)−W(3)​=−250.

a) What are the units of this quantity?

b) What does the negative sign mean in context?

Problem 12.

The cost C(q)C(q) (in dollars) to produce qq widgets is approximately

C(q)=500+3q.

C(q)=500+3q.

a) Interpret the number 500.

b) Interpret the number 3.

c) Find the average cost per widget on the interval [0,100][0,100].

Problem 13. (Temperature)

The temperature in a room at time tt (hours) is modeled by

T(t)=20+4sin⁡(πt12)

T(t)=20+4sin(12πt​)

in degrees Celsius.

a) What is the range of possible temperatures?

b) Over 6 hours, from t =0t =0 to t =6t =6, is the average rate of change positive, negative, or zero? Explain qualitatively.

D. Function Identification and Classification

Problem 14.

For each of the following, identify the likely function type (linear, quadratic, exponential, or “other”) and justify briefly:

a) A population doubles every 5 years.

b) The trajectory of a thrown ball (height vs. time).

c) Total cost is roughly constant plus a fixed cost per item.

d) The intensity of light from a point source vs. distance (assume inverse-square law).

Problem 15. (Matching)

Match each formula with its best verbal description:

y =7−0.1xy =7−0.1x

y =100⋅(0.9)xy =100⋅(0.9)x

y =5xy = x5​

Descriptions:

A. Decreasing line; value drops by 0.1 each time x increases by 1.

B. Exponential decay by 10% per unit of x.

C. Inversely proportional to x.

Problem 16. (Piecewise story)

A taxi company charges:

A flat fee of $4 to start,

Plus $2 per mile for the first 5 miles,

And $1 per mile for every mile after that.

Define a piecewise function C(d)C(d) giving cost in terms of distance dd (miles), for d≥0d≥0.

E. More Challenging / Symbolic Problems

Problem 17. (General polynomial)

Let f(x)= ax2+bx+cf(x)= ax2+bx+c with a≠0a=0. Show that the average rate of change between x =−hx =−h and x = hx = h is bb. Interpret the case b =0b =0.

Problem 18. (Symmetric interval, concrete)

Consider f(x)= x2f(x)= x2.

a) Compute the average rate of change from x =−2x =−2 to x =2x =2.

b) Explain why this value is 0 by symmetry.

Problem 19. (Rational function domain and average rate)

Let

f(x)=2x+1x−4.

f(x)= x−42x+1​.

a) Find the domain of ff.

b) Compute the average rate of change from x =0x =0 to x =2x =2.

Problem 20. (Parameter dependence)

For k≠0k=0, let f(x)= kx2f(x)= kx2.

a) Find the average rate of change on [0,3][0,3] in terms of kk.

b) If kk doubles, how does this average rate of change change?

Problem 21. (Discrete data, non-constant rates)

A cyclist’s position along a track is measured every 10 seconds:

Time (s) 0 10 20 30 40

Position (m) 0 60 135 210 280

a) Find the average velocity on [0,10], [10,20], [20,30], [30,40].

b) Is the cyclist speeding up, slowing down, or neither? Justify from your values.

Problem 22. (Non-monotone behavior)

The function f(x)f(x) has the following values:

xx -3 -1 0 1 2 4

f(x)f(x) 5 1 0 2 2 1

a) Compute the average rate of change on [−3,−1][−3,−1], [−1,1][−1,1], [1,4][1,4].

b) On which interval does ff increase on average? On which does it decrease?

Problem 23. (Algebra practice with average rate)

If f(x)=1xf(x)= x1​, simplify the average rate of change on [a,b][a,b] (assume a≠0a=0, b≠0b=0, a≠ba= b):

f(b)−f(a)b−a.

b−af(b)−f(a)​.

Problem 24. (Conceptual)

True or False? Explain briefly.

a) If the average rate of change of a function on [a,b][a,b] is positive, then the function must be increasing everywhere on [a,b][a,b].

b) If a function is decreasing on [a,b][a,b], then its average rate of change on [a,b][a,b] is negative.

c) If the average rate of change on [a,b][a,b] is zero, then f(a)= f(b)f(a)= f(b).

1.9 Solutions

Concept Check

Independent: time; dependent: height of the plant.

Rate of change is −0.5. The function decreases by 0.5 units for each 1-unit increase in x.

Units: meters per second (m/s).

Average rate of change is the same on any interval; it’s the line’s slope.

Examples: population growth, radioactive decay, compound interest.

Solutions – Section A

Problem 1.

f(0)=2(0)2−3(0)+1=1,f(2)=2(2)2−3(2)+1=8−6+1=3,f(−1)=2(−1)2−3(−1)+1=2+3+1=6.

f(0)f(2)f(−1)​=2(0)2−3(0)+1=1,=2(2)2−3(2)+1=8−6+1=3,=2(−1)2−3(−1)+1=2+3+1=6.​

Problem 2.

a) g(0)=5e0=5g(0)=5e0=5 million.

b) Exponential growth (exponent coefficient 0.2 > 0).

c) g(3)=5e0.6g(3)=5e0.6.

Problem 3.

a) h(0)=5h(0)=5 cm, h(4)=19h(4)=19 cm.

b) From week 1 to 4:

h(4)−h(1)4−1=19−73=123=4 cm/week.

4−1h(4)−h(1)​=319−7​=312​=4 cm/week.

Problem 4.

a) x−3≠0⇒x≠3x−3=0⇒x=3. Domain: all real xx except 3.

b) Inside square root nonnegative:

5−2x≥0⇒−2x≥−5⇒x≤52.

5−2x≥0⇒−2x≥−5⇒x≤25​.

Domain: (−∞,5/2](−∞,5/2].

c) For ln, argument > 0:

x2−4>0⇒x2>4⇒x<−2 or x>2.

x2−4>0⇒x2>4⇒x<−2 or x>2.

Domain: (−∞,−2)∪(2,∞)(−∞,−2)∪(2,∞).

Problem 5.

a)

F(−2)=2(−2)+1=−4+1=−3F(−2)=2(−2)+1=−4+1=−3

F(0)=02=0F(0)=02=0

F(3)=32=9F(3)=32=9

b) FF is piecewise: linear for x<0x<0 and quadratic for x≥0x≥0.

Solutions – Section B

Problem 6.

Given f(x)= x2+4x+1f(x)= x2+4x+1.

a) On [0,2][0,2]:

f(0)=1f(0)=1

f(2)=4+8+1=13f(2)=4+8+1=13

Average rate:

13−12−0=122=6.

2−013−1​=212​=6.

b) On [1,4][1,4]:

f(1)=1+4+1=6f(1)=1+4+1=6

f(4)=16+16+1=33f(4)=16+16+1=33

Average rate:

33−64−1=273=9.

4−133−6​=327​=9.

Problem 7.

For f(x)= x2f(x)= x2 on [a,a+h][a,a+h]:

f(a)= a2f(a)= a2

f(a+h)=(a+h)2= a2+2ah+h2f(a+h)=(a+h)2= a2+2ah+h2

Average rate:

f(a+h)−f(a)(a+h)−a =(a2+2ah+h2)−a2h =2ah+h2h =2a+h.

(a+h)−af(a+h)−f(a)​= h(a2+2ah+h2)−a2​= h2ah+h2​=2a+h.

Problem 8.

s(t)=3t2−2t+1.

s(t)=3t2−2t+1.

a) On [1,5][1,5]:

s(1)=3−2+1=2s(1)=3−2+1=2

s(5)=3(25)−2(5)+1=75−10+1=66s(5)=3(25)−2(5)+1=75−10+1=66

Average velocity:

66−25−1=644=16.

5−166−2​=464​=16.

b) The particle’s average velocity from 1 to 5 seconds is 16 m/s.

Problem 9.

If from the graph f(1)≈2f(1)≈2 and f(4)≈10f(4)≈10, then

Avg rate≈10−24−1=83≈2.67 units of f per unit of x.

Avg rate≈4−110−2​=38​≈2.67 units of f per unit of x.

(Use your actual graphed values.)

Problem 10.

a) r(t)=2+0.5tr(t)=2+0.5t. Volume:

V(t)=43πr(t)3=43π(2+0.5t)3.

V(t)=34​πr(t)3=34​π(2+0.5t)3.

b) From t =0t =0 to t =2t =2:

V(0)=43π(2)3=323πV(0)=34​π(2)3=332​π

V(2)=43π(3)3=43π(27)=36πV(2)=34​π(3)3=34​π(27)=36π

Average rate:

36π−323π2=(108−323)π2=763π2=766π=383π cm3/s.

236π−332​π​=2(3108−32​)π​=2376​π​=676​π=338​π cm3/s.

Solutions – Section C

Problem 11.

a) Units: liters per hour (L/hr).

b) The water amount is decreasing; on average 250 liters per hour are leaving the tank between 3 and 5 hours.

Problem 12.

a) 500 is a fixed cost (overhead) even with 0 widgets.

b) 3 is the variable cost per widget (each widget adds $3).

c) At 100 widgets:

C(100)=500+3⋅100=800 dollars.

C(100)=500+3⋅100=800 dollars.

Average cost per widget on [0,100]:

800100=8 $/widget.

100800​=8 $/widget.

Problem 13.

a) Since sin⁡sin ranges from −1 to 1:

T(t)∈[20−4, 20+4]=[16,24] °C.

T(t)∈[20−4, 20+4]=[16,24] °C.

b) Between t =0t =0 and t =6t =6, the sine argument goes from 0 to π/2π/2, so sin increases from 0 to 1 and T increases from 20°C toward 24°C. Average rate of change is positive.

Solutions – Section D

Problem 14.

a) Population doubling every 5 years → exponential growth.

b) Height vs time for a thrown ball → quadratic (projectile motion).

c) Total cost = fixed + per-item → linear.

d) Intensity ~ 1/(distance²) → other, specifically inverse-square (rational).

Problem 15.

1 → A

2 → B

3 → C

(1 is a line with slope −0.1, 2 is exponential decay with factor 0.9, 3 is inverse proportionality.)

Problem 16.

For d≥0d≥0:

If 0≤d≤50≤d≤5:

C(d)=4+2d.

C(d)=4+2d.

If d>5d>5: first 5 miles cost 4+2⋅5=144+2⋅5=14; remaining d−5d−5 miles cost $1 each:

C(d)=14+(d−5)= d+9.

C(d)=14+(d−5)= d+9.

So

C(d)={4+2d0≤d≤5,d+9d>5.

C(d)={4+2dd+9​0≤d≤5,d>5.​

Solutions – Section E

Problem 17.

Average rate from x =−hx =−h to x = hx = h:

f(h)= ah2+bh+cf(h)= ah2+bh+c

f(−h)= ah2−bh+cf(−h)= ah2−bh+c

Difference:

f(h)−f(−h)=(ah2+bh+c)−(ah2−bh+c)=2bh.

f(h)−f(−h)=(ah2+bh+c)−(ah2−bh+c)=2bh.

Average rate:

f(h)−f(−h)h−(−h)=2bh2h = b.

h−(−h)f(h)−f(−h)​=2h2bh​= b.

If b =0b =0, the average rate on any symmetric interval is 0; the parabola is symmetric about the y-axis (no linear tilt).

Problem 18.

a) For f(x)= x2f(x)= x2 on [−2,2][−2,2]:

f(−2)=4f(−2)=4, f(2)=4f(2)=4

Average rate:

4−42−(−2)=04=0.

2−(−2)4−4​=40​=0.

b) By symmetry: the graph is symmetric about the y-axis, so the “upward” change from −2 to 0 cancels the “downward” change from 0 to 2, yielding zero net change.

Problem 19.

a) Domain: denominator ≠ 0 → x≠4x=4.

b) On [0,2]:

f(0)=1−4=−14f(0)=−41​=−41​

f(2)=5−2=−52f(2)=−25​=−25​

Average rate:

f(2)−f(0)2−0=−52+142=−104+142=−942=−98.

2−0f(2)−f(0)​=2−25​+41​​=2−410​+41​​=2−49​​=−89​.

Problem 20.

a) f(x)= kx2f(x)= kx2 on [0,3]:

f(0)=0f(0)=0, f(3)=9kf(3)=9k

Average rate:

9k−03−0=3k.

3−09k−0​=3k.

b) If kk doubles, average rate also doubles (it’s proportional to kk).

Problem 21.

Positions: 0, 60, 135, 210, 280.

[0,10]: 60−010=6 m/s1060−0​=6 m/s

[10,20]: 135−6010=7.5 m/s10135−60​=7.5 m/s

[20,30]: 210−13510=7.5 m/s10210−135​=7.5 m/s

[30,40]: 280−21010=7 m/s10280−210​=7 m/s

b) Speed: 6 → 7.5 → 7.5 → 7. So cyclist speeds up at first, holds roughly constant, then slows slightly.

Problem 22.

Average rates:

[−3,−1][−3,−1]: 1−5−1−(−3)=−42=−2−1−(−3)1−5​=2−4​=−2

[−1,1][−1,1]: 2−11−(−1)=121−(−1)2−1​=21​

[1,4][1,4]: 1−24−1=−134−11−2​=3−1​

So it decreases on average on [−3,−1][−3,−1] and [1,4][1,4], increases on [−1,1][−1,1].

Problem 23.

For f(x)=1/xf(x)=1/x:

f(b)−f(a)b−a =1b−1ab−a = a−babb−a = a−bab(b−a)=−1ab.

b−af(b)−f(a)​= b−ab1​−a1​​= b−aaba−b​​= ab(b−a)a−b​=−ab1​.

(Used a−b =−(b−a)a−b =−(b−a).)

Problem 24.

a) False. Average rate positive only means f(b)>f(a)f(b)>f(a) overall; the function might dip down in between.

b) True (for a strictly decreasing function). If it’s decreasing on [a,b], then f(b)

c) True. If average rate is 0:

f(b)−f(a)b−a =0⇒f(b)−f(a)=0⇒f(a)= f(b).

b−af(b)−f(a)​