PrecalculusFunctionsMathematical Models and Applications

Mathematical Models and Function Applications

Chapter 2 — Precalculus (Stewart)

Functions become powerful when they model the real world. This guide covers the complete toolkit: representing functions four ways, reading graphs, building linear, quadratic, exponential, and power models, and applying the full mathematical modeling process to problems from physics, economics, and biology.

Chapter 2 Practice Problems

Functions and modeling — 80+ questions covering all sections

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The Four Ways to Represent a Function

Every function can be described four ways. The best modeling problems require you to move fluently between all four representations.

1. Verbal

A description in plain language. Example: the cost of a taxi ride is $2.50 plus $1.80 per mile.

2. Algebraic

A formula. The taxi example becomes C(x) = 2.50 + 1.80x, where x is miles traveled.

3. Visual

A graph on coordinate axes. The taxi function is a line with slope 1.80 and y-intercept 2.50.

4. Numerical

A table of input-output pairs. Miles 0, 1, 2, 3 give costs $2.50, $4.30, $6.10, $7.90.

Modeling tip

Real problems usually start verbal. Your job is to translate to algebraic, then use the formula to answer questions. The graph and table help you check your algebra and develop intuition for the function's behavior.

Vertical Line Test, Domain, and Range

Vertical Line Test

A curve in the plane is the graph of a function if and only if no vertical line intersects the curve more than once. Each x-value must map to exactly one y-value.

Passes the test (is a function)

y = x\u00b2, y = sin x, y = |x|, any parabola opening up or down

Fails the test (not a function)

x\u00b2 + y\u00b2 = r\u00b2 (circle), x = y\u00b2 (sideways parabola)

Domain and Range from Equations

Polynomial (any degree)

All real numbers: (−∞, ∞)

No denominators, no radicals

Rational: p(x)/q(x)

All x where q(x) ≠ 0

Division by zero is undefined

Even root: √(g(x))

All x where g(x) ≥ 0

Square root of negative undefined in reals

Logarithm: log(g(x))

All x where g(x) > 0

Log of non-positive undefined

Even root in denominator

All x where g(x) > 0 (strict)

Both undefined at 0 AND division by zero

Domain and Range from Graphs

Read domain by projecting the graph onto the x-axis: all x-values covered. Read range by projecting onto the y-axis: all y-values reached.

Graph: parabola y = \u2212x\u00b2 + 4, vertex at (0, 4), crosses x-axis at x = \u00b12

Domain: (\u2212\u221e, \u221e) \u2014 parabola extends left and right forever

Range: (\u2212\u221e, 4] \u2014 graph goes down to \u2212\u221e but tops out at y = 4

Increasing, Decreasing, and Local Extrema

Increasing on (a, b)

f is increasing on (a, b) if f(x\u2081) < f(x\u2082) whenever a < x\u2081 < x\u2082 < b. Moving right, the graph goes up.

Decreasing on (a, b)

f is decreasing on (a, b) if f(x\u2081) > f(x\u2082) whenever a < x\u2081 < x\u2082 < b. Moving right, the graph goes down.

Local Maximum and Local Minimum

Local Maximum at c

f(c) is a local max if f(c) \u2265 f(x) for all x near c. The graph has a peak at (c, f(c)). f switches from increasing to decreasing at c.

Local Minimum at c

f(c) is a local min if f(c) \u2264 f(x) for all x near c. The graph has a valley at (c, f(c)). f switches from decreasing to increasing at c.

Why this matters for modeling

In applied problems, local maxima represent peak profit, maximum height, or optimal quantity. Local minima represent minimum cost, lowest temperature, or break-even boundary. Identifying where a function switches from increasing to decreasing is the core task of optimization.

Average Rate of Change and the Difference Quotient

Definition

The average rate of change of f from x = a to x = b is:

[f(b) \u2212 f(a)] / (b \u2212 a)

This is the slope of the secant line connecting (a, f(a)) to (b, f(b)).

Formula:[f(b) − f(a)] / (b − a)
Geometric meaning:Slope of the secant line from (a, f(a)) to (b, f(b))
Difference quotient:[f(x+h) − f(x)] / h — as h→0 this becomes the derivative

Worked Example: Average Rate of Change

Find the average rate of change of f(x) = x\u00b2 \u2212 3x from x = 1 to x = 4.

f(1) = 1 \u2212 3 = \u22122

f(4) = 16 \u2212 12 = 4

Average rate of change = (4 \u2212 (\u22122)) / (4 \u2212 1) = 6/3 = 2

Interpretation: on average, f increases 2 units for every 1 unit increase in x over this interval.

The Difference Quotient

The difference quotient is a special form using h for the interval width:

[f(x + h) \u2212 f(x)] / h

Find the difference quotient for f(x) = 2x\u00b2 \u2212 5x.

f(x + h) = 2(x + h)\u00b2 \u2212 5(x + h)

= 2(x\u00b2 + 2xh + h\u00b2) \u2212 5x \u2212 5h

= 2x\u00b2 + 4xh + 2h\u00b2 \u2212 5x \u2212 5h

f(x + h) \u2212 f(x) = 4xh + 2h\u00b2 \u2212 5h

[f(x + h) \u2212 f(x)] / h = (4xh + 2h\u00b2 \u2212 5h) / h

= 4x + 2h \u2212 5

As h approaches 0, this approaches 4x \u2212 5, which is the derivative f'(x). The difference quotient bridges precalculus and calculus.

Piecewise Functions in Modeling

Many real-world situations follow different rules over different input ranges. Piecewise functions capture this naturally and are among the most practical function types in applied mathematics.

Income Tax Brackets

Tax rate differs by income level. Each bracket is a separate linear function.

T(x) = 0.10x if x ≤ 9000; 900 + 0.12(x−9000) if 9000 < x ≤ 41000; ...

Shipping Cost

Flat rate up to 1 lb, then per-ounce charge above that.

C(w) = 4.00 if w ≤ 1; 4.00 + 0.25(w−1) if w > 1

Electric Utility Rate

Lower rate for first 500 kWh, higher rate beyond that.

E(x) = 0.09x if x ≤ 500; 45 + 0.12(x−500) if x > 500

How to Evaluate a Piecewise Function

For the absolute value function f(x) = |x|, which equals x if x \u2265 0 and \u2212x if x < 0:

f(3) = 3 (since 3 \u2265 0, use first piece)

f(\u22125) = \u2212(\u22125) = 5 (since \u22125 < 0, use second piece)

f(0) = 0 (boundary: use whichever piece includes 0)

Common mistake

Always check which piece applies before substituting. The condition is about the input value x, not the output.

The Four Core Mathematical Models

These four model families handle the vast majority of precalculus applications. Learning to identify which model fits a situation is the central modeling skill.

Linear Model

f(x) = mx + b

Constant rate of change

Use when: Equal changes in input produce equal changes in output

  • Cost per unit
  • Distance at constant speed
  • Simple interest

Quadratic Model

f(x) = ax² + bx + c

Parabola with single max or min

Use when: Projectile motion, area optimization, revenue maximization

  • Ball height vs. time
  • Profit = Revenue − Cost
  • Fenced area

Exponential Model

f(x) = C · aˣ

Constant percentage change

Use when: Population growth, radioactive decay, compound interest

  • P(t) = P₀e^(kt)
  • A = P(1 + r/n)^(nt)
  • Cooling curves

Power Model

f(x) = k · xⁿ

Direct or inverse variation

Use when: Physical laws, scaling relationships, allometric growth

  • Kepler: T² = k · a³
  • Free fall: d = ½ gt²
  • Weight vs. height

Linear Models

A linear model has the form f(x) = mx + b. The slope m is the rate of change per unit input. The y-intercept b is the starting value when x = 0.

Worked Example: Cost/Revenue/Profit Model

A company makes widgets. Fixed cost is $800 per month. Variable cost is $3.50 per widget. Each widget sells for $7.

C(x) = 800 + 3.50x (total cost: x widgets)

R(x) = 7x (revenue)

P(x) = R(x) \u2212 C(x) = 7x \u2212 800 \u2212 3.50x = 3.50x \u2212 800

Break-even: P(x) = 0 when 3.50x = 800, so x = 800/3.50 \u2248 229 widgets.

Slope of P is $3.50/widget — each additional widget adds $3.50 of profit.

Worked Example: Simple Interest

$5,000 invested at 4% simple annual interest. What is the account value after t years?

A(t) = P + Prt = 5000 + 5000(0.04)t

A(t) = 5000 + 200t

A(10) = 5000 + 2000 = $7,000

This is a linear model: constant rate of $200/year, starting at $5,000.

Linear model key facts

  • Slope = rate of change (units of output per unit of input)
  • y-intercept = initial value or fixed cost
  • Equal increases in x always produce equal increases in y
  • Graph is a non-vertical, non-horizontal straight line

Quadratic Models

A quadratic model f(x) = ax\u00b2 + bx + c produces a parabola. It's the right model when a quantity increases then decreases (or vice versa), with a single maximum or minimum.

Vertex Formula (the critical tool)

x-coordinate of vertex:

x = \u2212b / (2a)

y-coordinate of vertex:

y = f(\u2212b / 2a)

If a > 0: parabola opens up, vertex is a minimum. If a < 0: opens down, vertex is a maximum. In models, this is where max profit, max height, or min cost occurs.

Worked Example: Projectile Motion

A ball is thrown upward from a height of 6 feet with initial velocity 40 ft/s. Height model: h(t) = \u221216t\u00b2 + 40t + 6.

Find maximum height:

t = \u2212b/(2a) = \u221240/(2 \u00b7 \u221216) = \u221240/(\u221232) = 1.25 seconds

h(1.25) = \u221216(1.25)\u00b2 + 40(1.25) + 6

= \u221216(1.5625) + 50 + 6

= \u221225 + 50 + 6

Maximum height = 31 feet at t = 1.25 sec

Worked Example: Revenue Optimization

A concert venue has 1000 seats. At $40 per ticket, all seats sell. For each $5 price increase, 50 fewer tickets sell. Find the ticket price that maximizes revenue.

Let n = number of $5 increases

Price = 40 + 5n

Tickets sold = 1000 \u2212 50n

R(n) = (40 + 5n)(1000 \u2212 50n)

= 40000 \u2212 2000n + 5000n \u2212 250n\u00b2

= \u2212250n\u00b2 + 3000n + 40000

Vertex: n = \u22123000/(2 \u00b7 \u2212250) = 3000/500 = 6

Optimal price = 40 + 5(6) = $70. Revenue = 70 \u00b7 700 = $49,000

Exponential Models

An exponential model has the form f(t) = C \u00b7 a^t or f(t) = Ce^(kt). It applies when a quantity grows or decays by a constant percentage per time period.

Exponential Growth

P(t) = P\u2080 \u00b7 e^(kt), k > 0

Population growth, compound interest, viral spread. k is the continuous growth rate.

Exponential Decay

Q(t) = Q\u2080 \u00b7 e^(\u2212kt), k > 0

Radioactive decay, drug concentration, Newton's law of cooling. Half-life = ln(2)/k.

Worked Example: Compound Interest

$10,000 invested at 6% annual interest, compounded monthly. Value after 20 years?

A = P(1 + r/n)^(nt)

A = 10000(1 + 0.06/12)^(12\u00b720)

A = 10000(1.005)^(240)

A \u2248 $33,102

Continuously compounded: A = Pe^(rt) = 10000e^(0.06\u00b720) = 10000e^(1.2) \u2248 $33,201. The difference is small but continuous compounding is the theoretical maximum.

Worked Example: Population Growth

A bacterial colony starts with 500 cells. It doubles every 3 hours. How many cells after 12 hours?

P(t) = 500 \u00b7 2^(t/3)

P(12) = 500 \u00b7 2^(12/3) = 500 \u00b7 2^4 = 500 \u00b7 16

P(12) = 8,000 cells

Worked Example: Radioactive Decay (Half-life)

Carbon-14 has a half-life of 5730 years. What fraction remains after 2000 years?

Q(t) = Q\u2080 \u00b7 (1/2)^(t/5730)

Q(2000)/Q\u2080 = (0.5)^(2000/5730)

= (0.5)^(0.3491) \u2248 0.785

About 78.5% remains after 2000 years

Power Functions and Variation

A power function has the form f(x) = k \u00b7 x^n. The exponent n can be any real number, including fractions and negatives. Power functions model physical laws and scaling relationships throughout science and engineering.

Common Power Function Shapes

nNameShapeExample
1LinearStraight lined = vt
2QuadraticParabola (up)A = πr²
3CubicS-curveV = s³
1/2Square rootIncreasing, concave downv = √(2gh)
−1ReciprocalHyperbolaP = k/V
−2Inverse squareSteep hyperbolaF = k/r²

Direct Variation

y = kx

y varies directly as x. Doubling x doubles y.

Distance d = rt: distance varies directly with time at constant speed r.

Inverse Variation

y = k/x

y varies inversely as x. Doubling x halves y.

Boyle’s Law: PV = k, so P = k/V. Pressure varies inversely with volume.

Joint Variation

y = kxz

y varies jointly as x and z.

Area of triangle: A = ½ bh varies jointly with base and height.

Combined Variation

y = kxⁿ/zᵐ

Mix of direct and inverse variation.

Newton’s gravity: F = Gm₁m₂/r² varies jointly as masses, inversely as r².

Worked Example: Kepler's Third Law

The orbital period T of a planet varies directly as the 3/2 power of its mean distance a from the Sun: T = k \u00b7 a^(3/2). Earth's period is 1 year at distance 1 AU. Find Mars's period if Mars is 1.524 AU from the Sun.

For Earth: 1 = k \u00b7 1^(3/2) = k, so k = 1

T(Mars) = (1.524)^(3/2)

= (1.524)^1 \u00b7 (1.524)^(1/2)

= 1.524 \u00b7 1.2345 \u2248 1.881

Mars year \u2248 1.88 Earth years (actual: 1.881)

Combining Functions for Models

Real models are often built by combining simpler functions. The four arithmetic combinations and composition are the primary tools.

Arithmetic Combinations

(f + g)(x) = f(x) + g(x)

Total cost = materials cost + labor cost

(f − g)(x) = f(x) − g(x)

Profit = Revenue − Cost

(f · g)(x) = f(x) · g(x)

Revenue = Price × Quantity

(f/g)(x) = f(x)/g(x)

Average cost = Total cost / Units produced

Composition in Modeling

When output of one process becomes input to another, use composition.

A factory's daily output depends on the number of workers w: Q(w) = 50w \u2212 w\u00b2. Profit per unit depends on quantity: P(q) = 8 \u2212 0.01q. Total profit as function of workers:

(P \u2218 Q)(w) = P(Q(w)) = 8 \u2212 0.01(50w \u2212 w\u00b2) = 8 \u2212 0.5w + 0.01w\u00b2

Torricelli's Law: Water Draining from a Tank

This is the signature modeling example in Stewart Chapter 2. It shows how a physical law becomes a mathematical function through the modeling process.

The Setup

Water drains from a circular tank (radius 2 ft, initial height 9 ft) through a small hole at the bottom. Physics gives us: the rate of flow is proportional to the square root of the water height.

The Model

h(t) = (3 \u2212 0.127t)\u00b2

where h is in feet, t is in minutes

Initial height:h(0) = (3 \u2212 0)\u00b2 = 9 ft \u2714
Tank empties when:3 \u2212 0.127t = 0 \u2192 t \u2248 23.6 min
Height after 10 min:h(10) = (3 \u2212 1.27)\u00b2 = (1.73)\u00b2 \u2248 2.99 ft

What makes this a power/quadratic model?

The height function h(t) = (3 \u2212 0.127t)\u00b2 is quadratic in t. But its derivation comes from Torricelli's law that the outflow velocity is proportional to \u221ah, a power function relationship. This is how physics-based models work: the underlying law is one type, but after solving the differential equation, you get a different function form.

The Mathematical Modeling Process

Stewart describes four stages that apply to every modeling problem. Practice moving through all four stages — many students stop after setting up the equation.

1

Identify Variables

Decide what you are modeling and what inputs affect it. Define symbols clearly — quantity, units, and range. Draw a diagram if geometry is involved.

2

Set Up the Model

Express the relationship as a function. Use geometry (area, volume, distance), physics (F = ma, Hooke’s law), or data patterns. Eliminate extra variables until you have one function of one variable.

3

Analyze the Model

Find domain, range, maximum, minimum, zeros, or intercepts. Use algebra to solve for the quantity of interest. In calculus this stage uses derivatives, but in precalculus you use vertex formulas, factoring, or technology.

4

Interpret Results

Translate mathematical answers back into real-world meaning. Check units, check reasonableness, and state a conclusion. x = 6 is meaningless; &apos;the optimal price is $70 per ticket&apos; is an answer.

Full Example: Fencing an Enclosure

A farmer has 1200 ft of fence to enclose a rectangular field adjacent to a barn (one side needs no fence). Find the dimensions that maximize area.

Step 1: Variables

Let x = width (two sides needed), L = length (one side needed)

Step 2: Model

Constraint: 2x + L = 1200, so L = 1200 \u2212 2x

Objective: A(x) = x \u00b7 L = x(1200 \u2212 2x) = 1200x \u2212 2x\u00b2

Domain: 0 < x < 600

Step 3: Analyze

A(x) = \u22122x\u00b2 + 1200x, a = \u22122 < 0 (opens down, max exists)

x = \u22121200/(2 \u00b7 \u22122) = \u22121200/(\u22124) = 300

L = 1200 \u2212 2(300) = 600

A(300) = 300 \u00b7 600 = 180,000 sq ft

Step 4: Interpret

The maximum area is 180,000 sq ft, achieved with a 300 ft \u00d7 600 ft rectangle. The long side (600 ft) runs along the barn. Using all the fence is optimal.

Falling Objects and Free Fall Models

Objects in free fall (ignoring air resistance) follow a quadratic model derived from physics. This appears constantly in precalculus and is the bridge to calculus mechanics.

The Free Fall Model

In feet (US customary)

s(t) = s\u2080 + v\u2080t \u2212 16t\u00b2

s\u2080 = initial height (ft)

v\u2080 = initial velocity (ft/s)

g = 32 ft/s\u00b2 \u2192 coefficient 16

In meters (SI)

s(t) = s\u2080 + v\u2080t \u2212 4.9t\u00b2

s\u2080 = initial height (m)

v\u2080 = initial velocity (m/s)

g = 9.8 m/s\u00b2 \u2192 coefficient 4.9

Worked Example: Stone Dropped from a Cliff

A stone is dropped from rest from a 400 ft cliff. When does it hit the ground? How fast is it falling when it hits?

s(t) = 400 \u2212 16t\u00b2 (dropped: v\u2080 = 0)

Hit ground when s(t) = 0:

400 \u2212 16t\u00b2 = 0 \u2192 t\u00b2 = 25 \u2192 t = 5 seconds

Stone hits after 5 seconds

The average speed over the 5-second fall: 400 ft / 5 s = 80 ft/s. At impact, the instantaneous speed (from calculus) is 160 ft/s — twice the average, which is a general result for constant-acceleration motion.

Practice Function Models and Applications

NailTheTest Chapter 2 has 80+ practice problems covering all model types — linear, quadratic, exponential, power, and piecewise. Work through real applications with instant feedback.