Last updated .
4.2 — Bike Lovers
You learned in 4.1 that a piecewise function uses different rules for different inputs. Today we use that idea to describe changing bicycle speeds, including how the pieces connect and how to describe their intervals precisely.
1. In class: A Solidify Understanding Task
Review: one function with several rules
A piecewise-defined function assigns one output to each input using whichever rule applies to that input. The domain restriction printed beside each rule tells you when to use it. For example,
For \(P(1)\), use the first rule and get \(4\). For \(P(3)\), use the second rule: \(8+2(3-2)=10\). For \(P(2)\), use the first rule because it includes equality at \(t=2\).
These rules agree where the pieces join: the first reaches \(8\) at \(t=2\), and the expression for the second would also give \(8\) there. We say the graph is continuous at the join: it has no jump or hole at that point.
Bike trips: the slope is speed
On a graph of cumulative distance traveled against time, the slope tells how fast distance is increasing. A line rising 16 miles in one hour has slope \(16\) miles per hour. A shallower line corresponds to slower riding.
In the first task, Michelle and Rashid take a four-hour ride represented by
Here \(x\) is time in hours and \(f(x)\) is total distance in miles. Notice how each new piece starts with the total distance already traveled: \(16\), then \(26\), then \(40\).
| Time interval | Speed (slope) | Distance at the end |
|---|---|---|
| 0–1 hours | 16 mi/h | \(f(1)=16\) |
| 1–2 hours | 10 mi/h | \(f(2)=10(1)+16=26\) |
| 2–3 hours | 14 mi/h | \(f(3)=14(1)+26=40\) |
| 3–4 hours | 12 mi/h | \(f(4)=12(1)+40=52\) |
Fastest: the first hour, at 16 miles per hour. Slowest: the second hour, at 10 miles per hour. Domain: \([0,4]\), because the ride lasts from hour 0 through hour 4. At 2 hours: \(f(2)=26\) miles. At 3 hours: \(f(3)=40\) miles. Entire ride: \(f(4)=52\) miles.
Average rate of change
For any function, the average rate of change over \([a,b]\) is the change in the output divided by the change in input:
From 1 to 3 hours, the riders go from 16 miles to 40 miles. Thus their average speed over the entire two-hour interval is
\[\frac{f(3)-f(1)}{3-1}=\frac{40-16}{2}=12\text{ mi/h}.\]That differs from either individual hourly speed because the interval spans two pieces.
Joining successive pieces: Rashid’s solo route
A convenient way to write a line segment with slope \(m\), starting at \((a,b)\), is \(m(x-a)+b\). When \(x=a\), the expression gives \(b\). This point-slope structure makes adjacent pieces easier to connect.
The solo-ride task provides a piecewise function in which time \(x\) is measured in minutes:
Compare the endpoint values:
| Time | Calculation | Cumulative distance |
|---|---|---|
| 20 minutes | \(\tfrac14(20)\) | 5 |
| 50 minutes | \(\tfrac15(50-20)+5\) | 11 |
| 92 minutes | \(\tfrac27(92-50)+11\) | 23 |
The final segment starts at time 92, after Rashid has traveled 23 distance units. Match the starting point \((a,b)=(92,23)\). The last rule is therefore \(\frac18(x-92)+23\), for \(92<x\le100\). At 100 minutes it reaches \(\frac18(8)+23=24\).
For \(g(30)\), use the second rule: \(\frac15(30-20)+5=7\). For \(g(64)\), use the third rule: \(\frac27(64-50)+11=15\). For \(g(10)\), use the first rule: \(\frac14(10)=2.5\). Each value represents total distance traveled after that many minutes.
The interval \([20,50]\) has average rate \((11-5)/(50-20)=\frac15\) distance units per minute. Among the solo route’s four segments, the greatest rate is \(\frac27\), on \((50,92]\).
To graph a continuous piecewise linear function, calculate its starting point and each joining point, plot them, and connect consecutive points with straight segments. Label the axes with quantities and units.
A piecewise function can contain a curve
Not every piece must be linear. The third task (problems 13–19) begins with a quadratic expression and then changes to two straight-line rules:
Here \(x\) represents minutes and \(h(x)\) represents kilometers traveled. Its first graph piece curves upward because \(x^2\) increases faster as \(x\) increases.
At \(x=10\), the first rule reaches \(h(10)=\frac14(10)^2=25\). The second expression begins from the number \(c\) when \(x=10\), so continuity requires \(c=25\). Check the next join: at 20, the second piece gives \(\frac12(20-10)+25=30\), matching the third piece’s starting value of 30.
The domain is \([0,30]\) minutes. The output rises from 0 km to \(h(30)=2(30-20)+30=50\) km, so the range is \([0,50]\) km. To evaluate \(h(8)\), use the quadratic rule: \(\frac14(8)^2=16\) km. To evaluate \(h(15)\), use the middle rule: \(\frac12(15-10)+25=27.5\) km.
Even on the curved segment, calculate average rate by using the two endpoint values:
| Interval | Endpoint calculation | Average rate (km/min) |
|---|---|---|
| \([0,5]\) | \((6.25-0)/5\) | 1.25 |
| \([5,10]\) | \((25-6.25)/5\) | 3.75 |
| \([10,15]\) | \((27.5-25)/5\) | 0.5 |
| \([25,30]\) | \((50-40)/5\) | 2 |
The greatest rate is 3.75 km/min on \([5,10]\). Notice that the rate on a curved piece depends on the particular interval selected; there is no single constant slope for that piece.
2. Ready homework: solving absolute value equations (1–12)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
The absolute value \(|u|\) measures the distance of \(u\) from zero. A distance cannot be negative. That gives the basic rule:
When a coefficient or constant lies outside the absolute-value bars, first isolate the absolute-value expression, then make the two cases. The bars act as grouping symbols: do not remove them before isolating.
Solve \(|x-7|=4\).
Case 1: \(x-7=4\), so \(x=11\). Case 2: \(x-7=-4\), so \(x=3\). Check: both values are four units from 7.
Solve \(3|2x-1|-4=11\).
\[3|2x-1|=15\quad\Longrightarrow\quad |2x-1|=5.\]Case 1: \(2x-1=5\Rightarrow x=3\). Case 2: \(2x-1=-5\Rightarrow x=-2\). Substituting either value into the original equation gives 11.
Solve \(-2|x+6|+1=-9\). Subtract 1, then divide by \(-2\): \(|x+6|=5\). Therefore \(x+6=5\) or \(x+6=-5\), giving \(x=-1\) or \(x=-11\).
In contrast, \(-4|x+2|=12\) becomes \(|x+2|=-3\). There is no solution, because an absolute value cannot equal \(-3\).
Check your work — Ready 1–12 answer key
- \(x=\pm7\)
- \(x=3,9\)
- \(w=-15,7\)
- \(m=\pm7\)
- \(d=\pm5\)
- \(x=-2,\frac{16}{3}\)
- \(m=-16,10\)
- \(m=\pm16\)
- \(x=-3,1\)
- \(c=-5,-1\)
- \(p=-1,4\)
- No solution: \(|m|\ge0\) cannot equal \(-3\).
3. Set homework: domain, range, intervals, and grading steps (13–17)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
Read domain and range from a graph
The domain is the set of horizontal input coordinates used by a graph. The range is the set of vertical output coordinates. Interval notation uses square brackets for included endpoints and parentheses for excluded endpoints:
| Inequality | Interval notation | Meaning |
|---|---|---|
| \(-2\le x\le5\) | \([-2,5]\) | Both endpoints included |
| \(-2<x\le5\) | \((-2,5]\) | Left excluded, right included |
| \(-2\le x<5\) | \([-2,5)\) | Left included, right excluded |
For a connected graph, identify the leftmost and rightmost inputs for the domain and the lowest and highest outputs for the range. For a graph containing gaps, write the union of separate intervals as needed.
Suppose a connected graph runs from the filled point \((-3,4)\) down to \((0,-1)\), up to \((4,2)\), and finishes at the filled point \((6,1)\). Its domain is \([-3,6]\); its range is \([-1,4]\). The graph’s changing direction does not change this method.
Write a domain for each piece
To separate a graph into three adjacent pieces with change points at \(x=0\) and \(x=4\), one possible partition is:
Together, these intervals cover \([-3,6]\), and they do not assign a shared boundary to more than one rule. You could choose the ownership of a joining point differently if the formulas agree and the graph is unchanged; always match the graph’s open and filled circles when those are shown.
A graph exists for \(-5\le x<-1\) and \(2\le x\le6\), with no points between \(-1\) and 2. Its domain is \([-5,-1)\cup[2,6]\), not \([-5,6]\). The union symbol \(\cup\) combines the two permitted input intervals.
Step functions: letter grades
A step function is piecewise constant: its output stays at one level over an interval, then jumps to another level. Letter grades are one example: every test score in a particular percentage band receives the same letter.
Under this example grading scale for scores \(0\le x\le100\), let \(G(x)\) be the letter grade:
\[G(x)=\begin{cases}\mathrm{F},&0\le x<60\\\mathrm{D},&60\le x<70\\\mathrm{C},&70\le x<80\\\mathrm{B},&80\le x<90\\\mathrm{A},&90\le x\le100.\end{cases}\]Thus \(G(79)=\mathrm C\), but \(G(80)=\mathrm B\). The graph has five horizontal pieces and jumps at 60, 70, 80, and 90. This scale is an illustrative convention; use the intervals shown in the packet’s graph to complete its own blanks.
Check your work — Set 13–19 answer key
- a. Domain \([-2,7]\). b. Range \([-3,6]\).
- a. Domain \([-3,6]\). b. Range \([-2,2]\).
- a. \(-2\le x<0\). b. \(0\le x<4\). c. \(4\le x\le6\). d. \([-2,6]\). At a shared endpoint either adjacent formula may include it when their values agree.
- a. \(-5\le x<-2\). b. \(-2\le x<1\). c. \(1\le x\le3\). d. \([-5,3]\).
- One conventional percentage grading rule: \(f(x)=\begin{cases}A,&90\le x\le100\\B,&80\le x<90\\C,&70\le x<80\\D,&60\le x<70\\F,&0\le x<60\end{cases}\). Plot not shown; label the horizontal pieces F, D, C, B, A in ascending order. The packet gives no numerical scale or cutoffs, so these cutoffs are an assumption.
- \(f(x)=\begin{cases}1-2x,&0\le x\le3\\\frac14(x-4)-1,&4\le x\le8\end{cases}\)
- \(f(x)=\begin{cases}4x-5,&1\le x\le3\\1-\frac17(x-\frac72),&\frac72\le x\le7\end{cases}\)
Go homework: quadratic transformations (20–22)
Study the worked examples first, then try the problems yourself. An answer key at the bottom of this section is provided to check your work.
Review the earlier function instruction and the quadratic transformation methods in Math II 2.1.
Check your work — Go 20–22 answer key
- \(g(x)=-2(x+3)^2-5\). Vertex \((-3,-5)\), opens downward. Plot not shown.
- \(g(x)=3(x-1)^2+4\). Vertex \((1,4)\), opens upward. Plot not shown.
- \(g(x)=-\frac12(x+6)^2+3\). Vertex \((-6,3)\), opens downward. Plot not shown. This uses the usual final vertical shift of 3; if “up 3” is performed first literally, then reflected and scaled, the constant is \(-\frac32\).
4. Additional learning resources
These optional resources provide extra practice, with the closest sections noted so you do not have to study ahead.
| Specific resource | Best use | Scope / What to skip |
|---|---|---|
| Professor Leonard — Intermediate Algebra Lecture 8.3: Study of Piecewise Functions and Basic Translations of Graphs | Extended, classroom-style explanation of piecewise rules and graphing. | Some extra material. Concentrate on the piecewise-function discussion. Graph translations extend beyond today’s focus; no verified timestamps available. |
| Khan Academy — Solve absolute value equations (practice) | Practice for Ready 1–12, including two-solution and no-solution situations. | Close match. Exercises focus on the relevant absolute-value algebra. |
| Khan Academy — Intro to absolute value equations and graphs | Video refresher on splitting an absolute-value equation into cases. | Some extra material. Watch the equation-solving explanation; graphical absolute-value function examples are optional background. |
5. Teacher’s review (teacher/parent reference)
Main task, problems 13–19: The formula \(h(x)=\frac14x^2\) produces a curved segment with unusually large numeric cycling distances in the given units (25 km after 10 minutes and 50 km after 30 minutes). It is internally consistent but not realistic for a human cyclist; explicitly distinguish the exercise’s mathematical model from practical cycling speeds.
Ready instructions, problems 1–12: The statement “You will have two answers” is inconsistent with problem 12, which intentionally has no solution. Suggested revision: “Find all real solutions; some equations may have no solution.”
Set 17: The printed grading-step graph does not sufficiently label the horizontal scale to uniquely determine exact score cutoffs. Provide the intended percentage thresholds in the text or label the graph’s x-axis.
6. Coverage of packet Section 4.2
| Packet assignment | Skills taught | Where in lesson |
|---|---|---|
| Main task 1–6 | Evaluate four pieces; compare speeds, domain, distance, graph | In class |
| Main task 7–12 | Rates, matching endpoints, missing parameters, evaluation, graphing | In class |
| Main task 13–19 | Quadratic and linear pieces, continuity, rate comparisons, domain/range | In class |
| Ready 1–12 | Absolute-value equations including impossible cases | Section 2 |
| Set 13–14 | Read domain and range from piecewise graphs in interval notation | Section 3 |
| Set 15–16 | Assign input intervals to segments; endpoint inclusion | Section 3 |
| Set 17 | Noncontinuous step functions and piecewise grade rules | Section 3 |
Curriculum: Mathematics Vision Project, Secondary Math II, Module 4 (2017), Section 4.2, “Bike Lovers,” pp. 5–9. This original supplement follows the daily scope of the packet and does not teach Section 4.3’s absolute-value functions as piecewise graphs. MathJax mathematics requires an internet connection when the HTML is opened.
Last updated .