MGU Mathematics Series — Calculus
Part III: Applications of Derivatives
Critical Points · Concavity · Inflection · Curve Sketching
Chapter 7 of 14
MGU Mathematics Series  ·  Calculus  ·  Library Textbook Edition

Chapter 7:
Graph Analysis
with Derivatives

Using the first and second derivative to reveal function behavior — where functions rise, fall, curve, peak, and bend — without point-by-point plotting

■ Math Level: Calculus I ■

■ 14 Sections  •  Part III of 5 ■

Follows Chapter 6: Implicit Differentiation and Related Rates  ·  Prepares students for Chapter 8: Optimization

Purpose and Overview

Derivatives do more than measure rates of change at a single point. When you know how a function's derivative behaves across an interval — where it is positive, where it is negative, where it equals zero — you know the shape of the function itself. This chapter teaches you to use derivatives as tools for reading the behavior of functions: where they rise and fall, where they reach peaks and valleys, how they curve, and where their concavity changes.

This kind of analysis turns algebraic information into geometric understanding. You will learn to construct sign charts for the first and second derivative, identify critical numbers and classify them as local maxima, local minima, or neither, determine where a function is concave up or concave down, find inflection points, and use all of this information together to sketch accurate graphs without relying on point-by-point plotting.

By the end of this chapter, you will understand that the derivative is not just a calculation — it is a language for describing function behavior.

Opening Question

Suppose you are analyzing the path of a roller coaster. You want to know where the track rises and where it falls, where the highest and lowest points are, where the ride curves upward like a bowl and where it curves downward like a hill, and where the curvature transitions from one type to the other. You are not asked to evaluate the height at every point. You are asked to understand the shape of the ride. Derivatives give you exactly this kind of structural information. The first derivative tells you whether the height is increasing or decreasing. The second derivative tells you whether the curvature is concave up or concave down. Their combined information tells you the story of the shape.

14Sections
9Objectives
10Key Terms
12+Worked Examples

Learning Objectives

  1. 1Use the sign of the first derivative to determine where a function is increasing and where it is decreasing.
  2. 2Find critical numbers and explain what they represent.
  3. 3Classify critical numbers as local maxima, local minima, or neither using the First Derivative Test.
  4. 4Use the sign of the second derivative to determine concavity on intervals.
  5. 5Find inflection points and explain what they represent.
  6. 6Apply the Second Derivative Test to classify critical numbers when applicable.
  7. 7Combine information from f, f′, and f″ to sketch accurate graphs of functions.
  8. 8Interpret derivative behavior in graphs, tables, and applied models.
  9. 9Recognize and avoid common mistakes in graph analysis involving derivatives.

Sections at a Glance

Sections 7.1 – 7.5
The First Derivative: Direction, Critical Numbers, and Local Extrema
Function Behavior · Increasing & Decreasing · Critical Numbers · First Derivative Test

Derivatives and Function Behavior

Before diving into formal rules and procedures, it is worth pausing to think about what derivatives actually reveal about a function's graph. Recall that the derivative f′(x) represents the instantaneous rate of change of f at the point x. If you think of the graph of f as a road you are driving along, then the derivative at each point tells you the slope of the road at that moment. When the slope is positive, you are going uphill. When the slope is negative, you are going downhill. When the slope is zero, you are momentarily on level ground — at a peak, a valley, or a flat section.

This connection between derivative sign and graph direction is one of the most useful insights in all of calculus. Instead of computing function values at dozens of points to understand a graph, you can compute the derivative and analyze its sign across intervals.

Core Relationships

If f′(x) > 0 on an interval, then f is increasing on that interval.

If f′(x) < 0 on an interval, then f is decreasing on that interval.

If f′(x) = 0 at a point, then f has a horizontal tangent there — a candidate for a local extremum.

The second derivative adds another layer of information. If f″(x) > 0 on an interval, the graph is concave up — it curves upward like the inside of a bowl. If f″(x) < 0 on an interval, the graph is concave down — it curves downward like the outside of a hill. Where concavity changes from one type to the other, the graph has an inflection point. The combination of first and second derivative information gives a complete portrait of a function's behavior without needing to evaluate the function at every point.

Graph Instruction

Draw a smooth curve that rises from left to right, reaches a peak, falls to a valley, then rises again. Mark the rising intervals with a "+" label for f′, the falling interval with a "–" label for f′, and place arrows on the curve showing the direction of travel. This is a simple visual version of the relationship between derivative sign and graph direction.

Increasing and Decreasing Intervals

A function f is increasing on an interval (a, b) if, for every pair of values x₁ and x₂ in that interval with x₁ < x₂, it follows that f(x₁) < f(x₂). In other words, as x moves to the right, f(x) moves upward. A function f is decreasing on (a, b) if f(x₁) > f(x₂) whenever x₁ < x₂ — as x moves right, f(x) moves down.

Increasing / Decreasing Test

Let f be differentiable on an open interval (a, b).

If f′(x) > 0 for all x in (a, b), then f is increasing on (a, b).

If f′(x) < 0 for all x in (a, b), then f is decreasing on (a, b).

How to Find Increasing and Decreasing Intervals

  1. Find f′(x).
  2. Set f′(x) = 0 and solve. Also find any x-values where f′(x) does not exist.
  3. These values divide the domain into intervals.
  4. Test one value of x from each interval by plugging it into f′(x).
  5. f′(x) > 0 → increasing on that interval. f′(x) < 0 → decreasing on that interval.
Sign Charts

Organizing the sign of f′ across intervals in a sign chart is the clearest way to track behavior across the whole domain. You will use sign charts repeatedly throughout this chapter and the next.

Example 7.2.1
Finding Increasing and Decreasing Intervals
  • ProblemFind the intervals on which f(x) = x³ − 3x² − 9x + 5 is increasing and decreasing.
  • ThinkI need the sign of f′(x) on each interval. I will find f′(x), set it to zero to find sign-change candidates, then test each interval.
  • Calculate f′(x) = 3x² − 6x − 9 Set f′(x) = 0: 3x² − 6x − 9 = 0 x² − 2x − 3 = 0 (x − 3)(x + 1) = 0 x = 3 or x = −1 These values divide the number line into three intervals: (−∞, −1), (−1, 3), and (3, ∞).

    Test a value from each interval in f′(x) = 3x² − 6x − 9: On (−∞, −1): test x = −2 f′(−2) = 3(4) − 6(−2) − 9 = 12 + 12 − 9 = 15 > 0 → increasing On (−1, 3): test x = 0 f′(0) = 0 − 0 − 9 = −9 < 0 → decreasing On (3, ∞): test x = 4 f′(4) = 3(16) − 6(4) − 9 = 48 − 24 − 9 = 15 > 0 → increasing
  • Sign Chart x: ... −1 ... 3 ... f′(x): + 0 − 0 + f(x): ↑ ↓ ↑
  • CheckThe derivative is a parabola opening upward (leading coefficient 3 > 0), so its sign should be positive outside its roots and negative between them. This matches the sign chart. ✓
  • InterpretThe function is increasing on (−∞, −1) and (3, ∞), and decreasing on (−1, 3).

Critical Numbers

Definition — Critical Number

A value x = c is a critical number of f if c is in the domain of f and either:

f′(c) = 0    — the tangent line is horizontal at x = c, or

f′(c) does not exist    — a corner, cusp, or vertical tangent at x = c.

Critical numbers are the candidates for local extrema — the peaks and valleys of the graph. The word candidate matters here: not every critical number is a local maximum or minimum. Some are inflection points with horizontal tangents, or corners, or cusps. But every local maximum and every local minimum of a differentiable function occurs at a critical number.

Tip — Critical Numbers vs. Critical Points

Critical numbers are x-values (inputs), not y-values. The critical point is the full coordinate pair (c, f(c)). Always compute both.

Example 7.3.1
Finding Critical Numbers
  • ProblemFind the critical numbers of g(x) = x⁴ − 4x³.
  • Calculate g′(x) = 4x³ − 12x² = 4x²(x − 3) Set g′(x) = 0: 4x²(x − 3) = 0 x = 0 or x = 3 Since g′(x) is a polynomial, it exists everywhere. There are no points where the derivative fails to exist.
  • InterpretThe critical numbers are x = 0 and x = 3. The function has horizontal tangents at these points. Whether each is a local max, local min, or neither will be determined by the First Derivative Test.

Local Maximums and Minimums

A function f has a local maximum at x = c if f(c) ≥ f(x) for all x in some open interval around c. It has a local minimum at x = c if f(c) ≤ f(x) for all x in some open interval around c. Together, local maxima and minima are called local extrema.

Local extrema are not the same as absolute extrema. A local maximum might not be the highest point on the entire graph — it is only the highest point near its location. Absolute extrema are studied more carefully in Chapter 8.

Near a local maximum, the function must first rise, reach the peak, then fall. This means f′ must transition from positive to negative at the critical number. Near a local minimum, the function must first fall, reach the valley, then rise — f′ transitions from negative to positive. This observation is the key idea behind the First Derivative Test.

First Derivative Test

Theorem — First Derivative Test

Suppose x = c is a critical number of a continuous function f.

If f′(x) changes from + to − at x = c, then f has a local maximum at c.

If f′(x) changes from − to + at x = c, then f has a local minimum at c.

If f′(x) does not change sign at x = c (remains positive or negative on both sides), then f has neither a local maximum nor a local minimum at c.

Example 7.5.1
Applying the First Derivative Test
  • ProblemFor f(x) = x³ − 3x² − 9x + 5, classify each critical number as a local maximum, local minimum, or neither.
  • Sign Chart x: ... −1 ... 3 ... f′(x): + 0 − 0 +
  • At x = −1f′ changes from + to −. By the First Derivative Test, f has a local maximum at x = −1. f(−1) = (−1)³ − 3(−1)² − 9(−1) + 5 = −1 − 3 + 9 + 5 = 10 Local maximum point: (−1, 10).
  • At x = 3f′ changes from − to +. By the First Derivative Test, f has a local minimum at x = 3. f(3) = 27 − 27 − 27 + 5 = −22 Local minimum point: (3, −22).
  • InterpretThe function rises to a local peak of 10 at x = −1, then falls to a local valley of −22 at x = 3, then rises again without bound.
Example 7.5.2
A Critical Number That Is Neither a Max Nor a Min
  • ProblemFind and classify the critical numbers of h(x) = x³.
  • Calculate h′(x) = 3x² Set h′(x) = 0: 3x² = 0 → x = 0
  • Sign Chart x: ... 0 ... h′(x): + 0 + h′(x) = 3x² ≥ 0 for all x. The sign does not change at x = 0.
  • InterpretBy the First Derivative Test, h has neither a local maximum nor a local minimum at x = 0. The function has a horizontal tangent at the origin but continues to increase through that point. The origin is an inflection point, not an extremum.
Student Warning

Do not assume that every critical number is a local maximum or minimum. Example 7.5.2 shows this is wrong. A sign chart is needed to determine the correct classification every time.

Sections 7.6 – 7.8
The Second Derivative: Concavity, Inflection Points, and the Second Derivative Test
Concavity · Inflection Points · Second Derivative Test

Concavity

The first derivative tells you whether a function is increasing or decreasing. The second derivative tells you something deeper: how the rate of change itself is changing. This is called concavity.

Concavity Definitions

The graph of f is concave up on an interval if f″(x) > 0 for all x on that interval. The graph curves upward like the inside of a bowl. Tangent lines lie below the curve.

The graph of f is concave down on an interval if f″(x) < 0 for all x on that interval. The graph curves downward like the outside of a hill. Tangent lines lie above the curve.

All Four Combinations Are Possible

A function can be increasing and concave up; increasing and concave down; decreasing and concave up; or decreasing and concave down. Concavity is about how the slope is changing, not about whether the function is rising or falling.

Direction of fConcavityMeaningExample
IncreasingConcave UpSlope positive and growing (accelerating upward)Right half of y = x²
IncreasingConcave DownSlope positive but shrinking (decelerating upward)y = √x for x > 0
DecreasingConcave UpSlope negative but becoming less negative (decelerating downward)Left half of y = x²
DecreasingConcave DownSlope negative and growing more negative (accelerating downward)y = −ex
Example 7.6.1
Finding Concavity Intervals
  • ProblemFind the intervals on which f(x) = x³ − 3x² − 9x + 5 is concave up and concave down.
  • Calculate f′(x) = 3x² − 6x − 9 f″(x) = 6x − 6 Set f″(x) = 0: 6x − 6 = 0 → x = 1
  • Test Intervals On (−∞, 1): test x = 0 → f″(0) = −6 < 0 → concave down On (1, ∞): test x = 2 → f″(2) = 6 > 0 → concave up
  • InterpretThe function is concave down to the left of x = 1 and concave up to the right of x = 1. The concavity changes at x = 1 — this is an inflection point candidate.

Inflection Points

Definition — Inflection Point

A point (c, f(c)) on the graph of f is an inflection point if f is continuous at c and the concavity changes — from concave up to concave down, or vice versa — at c.

How to Find Inflection Points

  1. Find f″(x).
  2. Find values where f″(x) = 0 or f″(x) does not exist. These are candidates only.
  3. Check whether the concavity actually changes sign at each candidate value.
  4. If it does, compute f(c) to find the full inflection point (c, f(c)).
Candidate Warning

A value where f″(c) = 0 is only a candidate for an inflection point. If the concavity does not actually change there, it is not an inflection point. Always verify the sign change.

Example 7.7.1
Finding Inflection Points
  • ProblemFind the inflection points of f(x) = x³ − 3x² − 9x + 5.
  • ThinkFrom Section 7.6: f″(x) = 6x − 6, and the concavity changes at x = 1.
  • Confirm Sign Change On (−∞, 1): f″(x) < 0 (concave down) On (1, ∞): f″(x) > 0 (concave up) ✓ Concavity changes at x = 1.
  • Compute f(1) = 1 − 3 − 9 + 5 = −6
  • InterpretThe inflection point is (1, −6). This is the point where the graph transitions from curving downward (like a hill) to curving upward (like a bowl).
Example 7.7.2
A Candidate That Is Not an Inflection Point
  • ProblemFind the inflection points of p(x) = x⁴.
  • Calculate p′(x) = 4x³ p″(x) = 12x² Set p″(x) = 0: 12x² = 0 → x = 0
  • Check Sign ChangeSince p″(x) = 12x² ≥ 0 for all x, the second derivative does not change sign at x = 0. It is non-negative on both sides.
  • InterpretAlthough p″(0) = 0, there is no inflection point at x = 0. The graph of x⁴ is concave up everywhere. Always confirm a sign change before declaring an inflection point.

Second Derivative Test

Theorem — Second Derivative Test

Suppose f′(c) = 0 and f″(c) exists.

If f″(c) > 0, then f has a local minimum at c. (Concave up at that point — bottom of a bowl.)

If f″(c) < 0, then f has a local maximum at c. (Concave down at that point — top of a hill.)

If f″(c) = 0, the test is inconclusive. Use the First Derivative Test instead.

Example 7.8.1
Applying the Second Derivative Test
  • ProblemFind and classify the critical numbers of f(x) = 2x³ − 3x² − 12x + 4.
  • Calculate f′ f′(x) = 6x² − 6x − 12 = 6(x² − x − 2) = 6(x − 2)(x + 1) Critical numbers: x = 2 and x = −1
  • Calculate f″ f″(x) = 12x − 6 f″(2) = 12(2) − 6 = 18 > 0 → local minimum at x = 2 f″(−1) = 12(−1) − 6 = −18 < 0 → local maximum at x = −1
  • Compute Values f(2) = 16 − 12 − 24 + 4 = −16 f(−1) = −2 − 3 + 12 + 4 = 11
  • InterpretThe function has a local maximum of 11 at x = −1 and a local minimum of −16 at x = 2.
  • CheckThe leading coefficient is positive and the degree is odd (cubic), so the function rises to the right and falls to the left. A local max before a local min is consistent with this end behavior. ✓
When to Use Each Test

Use the First Derivative Test when f″(c) = 0 at a critical number, when f″ is difficult to compute, or when you are already constructing a sign chart for f′. Use the Second Derivative Test when f″ is easy to compute and evaluate, and you want a quick classification at an isolated critical number.

Sections 7.9 – 7.12
Curve Sketching, Combined Analysis, and Applied Models
Curve Sketching · f, f′, f″ Together · Technology · Function Behavior in Models

Curve Sketching with Derivatives

With all the pieces in place — domain, intercepts, increasing/decreasing intervals, critical points and their classifications, concavity intervals, and inflection points — you now have enough information to sketch a fairly accurate graph of a function without plotting dozens of points.

Systematic Curve Sketching — Eight Steps

1. Domain — Identify the domain. Are there restrictions from square roots, denominators, logarithms?

2. Intercepts — Find f(0) for the y-intercept. Solve f(x) = 0 for x-intercepts if practical.

3. Asymptotes — Vertical asymptotes where denominators are zero; horizontal asymptotes using limits at infinity; end behavior for polynomials.

4. First derivative — Find f′(x). Build a sign chart. Identify increasing/decreasing intervals and critical numbers.

5. Local extrema — Classify critical numbers. Compute f-values at each critical number.

6. Second derivative — Find f″(x). Build a sign chart. Identify concavity intervals.

7. Inflection points — Confirm sign changes in f″. Compute f-values at each inflection point.

8. Sketch — Plot intercepts, extrema, and inflection points. Connect with correct direction and concavity. Add asymptotes.

Example 7.9.1
Full Curve Sketching
  • ProblemSketch the graph of f(x) = x³ − 3x² − 9x + 5.
  • Step 1: DomainAll real numbers. No restrictions.
  • Step 2: Interceptsy-intercept: f(0) = 5. Point: (0, 5). The x-intercepts of this cubic exist but are irrational and not essential for the sketch.
  • Step 3: AsymptotesPolynomial — no asymptotes. End behavior: as x → ∞, f(x) → ∞; as x → −∞, f(x) → −∞ (positive leading coefficient, odd degree).
  • Steps 4–5: f′ and Extrema f′(x) = 3(x − 3)(x + 1) → critical numbers: x = −1, x = 3 Sign chart: x: (−∞, −1) −1 (−1, 3) 3 (3, ∞) f′: + 0 − 0 + f: ↑ ↓ ↑ Local maximum at x = −1: f(−1) = 10 → point (−1, 10) Local minimum at x = 3: f(3) = −22 → point (3, −22)
  • Steps 6–7: f″ and Inflection f″(x) = 6x − 6 → candidate: x = 1 Sign chart: x: (−∞, 1) 1 (1, ∞) f″: − 0 + concave ↓ concave ↑ Concavity changes at x = 1. f(1) = 1 − 3 − 9 + 5 = −6 → inflection point (1, −6)
  • Step 8: Sketch
    Graph Instruction

    Plot these key points: local max at (−1, 10), y-intercept at (0, 5), inflection point at (1, −6), and local min at (3, −22). Coming from the left, the curve rises steeply (concave down as it approaches the local max), reaches the peak at (−1, 10), curves through the inflection at (1, −6) where it transitions to concave up, then falls to the minimum at (3, −22), and rises steeply thereafter. The overall shape is the typical S-shaped cubic.

  • InterpretThis is a cubic polynomial with a local max followed by a local min. The function rises from −∞ on the left, peaks at (−1, 10), passes through the inflection at (1, −6), reaches its valley at (3, −22), and rises to +∞ on the right.

Interpreting f, f′, and f″ Together

One of the richest skills in calculus is reading information about f, f′, and f″ simultaneously — from separate graphs, from a table, or by reasoning through the relationships. The following table summarizes the connections between all three.

Condition on fCondition on f′Condition on f″
f is increasingf′(x) > 0—
f is decreasingf′(x) < 0—
f has a local maximumf′(c) = 0, sign changes + → −f″(c) < 0 (if SDT applies)
f has a local minimumf′(c) = 0, sign changes − → +f″(c) > 0 (if SDT applies)
f is concave upf′ is increasingf″(x) > 0
f is concave downf′ is decreasingf″(x) < 0
f has an inflection pointf′ has a local extremum at cf″(c) = 0 and sign changes
Three-Panel Graph Instruction

Draw three curves vertically aligned. The top curve is a smooth cubic f. The middle curve is its derivative f′ — a parabola opening upward with roots matching the critical numbers of f. The bottom curve is f″ — a line. Mark clearly where f′ = 0 corresponds to extrema on f, and where f″ = 0 corresponds to the inflection point of f and the vertex of f′. This three-panel diagram is one of the most instructive exercises in first-year calculus.

Example 7.10.1
Reading Derivative Information from a Graph
  • ProblemA function f is increasing on (−∞, 2), has a local maximum at x = 2, is decreasing on (2, 5), has a local minimum at x = 5, and is increasing on (5, ∞). The curve is concave down on (−∞, 3.5) and concave up on (3.5, ∞). Describe the behavior of f′ and f″.
  • For f′ f′(x) > 0 on (−∞, 2) — f is increasing f′(2) = 0 — local maximum f′(x) < 0 on (2, 5) — f is decreasing f′(5) = 0 — local minimum f′(x) > 0 on (5, ∞) — f is increasing again
  • For f″ f″(x) < 0 on (−∞, 3.5) — concave down f″(3.5) = 0 — inflection point candidate f″(x) > 0 on (3.5, ∞) — concave up Also: since f transitions from concave down to concave up at x = 3.5, f′ has a local minimum at x = 3.5.
  • InterpretThe graph of f′ is positive, then negative, then positive, with zeros at x = 2 and x = 5 and a minimum at x = 3.5. The graph of f″ is negative on the left and positive on the right, crossing zero at x = 3.5.

Graph Analysis with Technology

Graphing calculators and computer algebra systems can confirm your algebraic analysis and help you visualize functions that are too complex to sketch by hand. However, technology is a verification tool, not a replacement for understanding.

Using Technology Effectively

Setting a good window. Before you trust a graph, make sure the window shows all relevant features. Include the domain of interest, the critical points, the inflection points, any asymptotes, and enough end behavior to see the overall shape. A default window from −10 to 10 may miss important features entirely.

Zooming to verify local behavior. Near a critical point, zoom in to see whether the function actually reaches a local max or min there. Near a potential inflection point, zoom in to confirm the curvature change.

Comparing your sketch to the technology graph. Before graphing on a calculator, produce your own sketch using the derivative analysis from Sections 7.2 through 7.9. Then compare. If they agree, your analysis is confirmed. If they disagree, find the error in your algebra or your reasoning.

Technology Tip

If the calculator graph shows a local maximum where your analysis predicts a local minimum, go back and check the sign chart. Either the derivative was computed incorrectly, or the test values were substituted wrong. Technology confirms your algebra; it does not replace the need for correct algebra.

Example 7.11.1
Technology Verification for f(x) = sin x − x
  • ProblemDoes f(x) = sin x − x have any local extrema on (0, 4π)?
  • Calculate f′(x) = cos x − 1 Now, cos x − 1 ≤ 0 for all x, and cos x − 1 = 0 only when cos x = 1, i.e., x = 0, 2π, 4π, …
  • Sign AnalysisOn (0, 4π), f′(x) ≤ 0 everywhere, with f′(x) = 0 only at x = 2π. The sign of f′ does not change — it is zero at one isolated point and negative everywhere else.
  • InterpretBy the First Derivative Test, f has no local extrema on (0, 4π). The graph of f(x) = sin x − x is always decreasing with a brief horizontal tangent at multiples of 2π. A graphing tool confirms this conclusion.

Function Behavior in Models

Graph analysis with derivatives is not purely abstract. The same ideas describe meaningful behavior of real quantities across many fields.

In motion. If s(t) is the position of an object and s′(t) = v(t) is its velocity, then v(t) > 0 means the object moves in the positive direction, v(t) < 0 means it moves in the negative direction, and v(t) = 0 is a critical number where the object momentarily stops. The acceleration a(t) = v′(t) plays the role of f″ — where a(t) > 0, the velocity is increasing.

In economics. If P(x) is the profit from producing x units, P′(x) > 0 means profit is increasing as production increases, and P′(x) = 0 at a critical number — possibly the production level that maximizes profit. The condition P″(x) < 0 at a critical number confirms a local maximum profit by the Second Derivative Test.

In biology. If N(t) is a population at time t, N′(t) > 0 means the population is growing. An inflection point in N(t) corresponds to when the rate of growth is at its maximum — a critical moment in epidemiology and ecology. After the inflection point, the population still grows, but more slowly.

In data science and machine learning. Optimization algorithms that train machine learning models search for local minima in a loss function. Gradient descent uses information analogous to f′(x) — the gradient — to move in the direction of steeper descent. Understanding concavity helps practitioners understand whether their algorithms are converging.

MGU Library Connection
  • MGU Physics chapters on kinematics — applications of position, velocity, and acceleration graphs
  • MGU Economics chapters on marginal analysis — derivative sign changes in profit and cost functions
  • MGU Data Science Foundations — the role of derivatives in gradient descent
  • MGU Differential Equations — logistic and exponential growth models
Example 7.12.1
Population Growth and the Inflection Point
  • ProblemA population is modeled by the logistic equation N(t) = 1000 / (1 + 9e−0.5t), where t is years. Find the inflection point and explain its meaning.
  • ThinkFor a logistic model, the inflection point occurs at half the carrying capacity. The carrying capacity (horizontal asymptote as t → ∞) is 1000, so the inflection occurs when N = 500.
  • Solve Set N(t) = 500: 1000 / (1 + 9e^{−0.5t}) = 500 1 + 9e^{−0.5t} = 2 9e^{−0.5t} = 1 e^{−0.5t} = 1/9 −0.5t = ln(1/9) = −ln 9 t = 2 ln 9 ≈ 4.39 years
  • InterpretThe inflection point occurs at approximately t ≈ 4.39 years, when the population reaches 500. This is the moment when the population growth rate is at its peak — the population is growing fastest at this moment. Before this point, the growth rate is accelerating; after this point, the growth rate decelerates as the population approaches the carrying capacity of 1000.
Sections 7.13 – 7.14
Common Mistakes and the Road Ahead
Graph Analysis Mistakes · Preparing for Optimization

Common Graph Analysis Mistakes

Mistake 1 — Confusing increasing with concave up
"If a function is increasing, it must be concave up."

Increasing/decreasing depends on f′. Concavity depends on f″. A function can be increasing and concave down (slopes are positive but getting smaller). All four combinations — increasing/concave up, increasing/concave down, decreasing/concave up, decreasing/concave down — are possible and all occur in real functions.

Mistake 2 — Assuming every critical number is a local extremum
"f′(c) = 0, therefore c is a local max or min."

Example 7.5.2 showed that f(x) = x³ has a critical number at x = 0 that is neither a local max nor a local min. Always check for a sign change in f′ before classifying.

Mistake 3 — Declaring an inflection point when f″(c) = 0 without checking
"f″(c) = 0, therefore (c, f(c)) is an inflection point."

Example 7.7.2 showed that f(x) = x⁴ has f″(0) = 0 but no inflection point at x = 0 — the concavity does not change. Always check whether f″ actually changes sign at the candidate.

Mistake 4 — Misusing the inconclusive Second Derivative Test
"The Second Derivative Test is inconclusive at x = c, therefore there is no extremum there."

Inconclusive means the test cannot decide — not that the answer is "neither." When f″(c) = 0, switch to the First Derivative Test to make a correct classification.

Mistake 5 — Reporting only x-coordinates of critical points and inflection points
"The local maximum occurs at x = −1." (No y-value given.)

Identifying the x-value is only half the work. The actual point on the graph requires both the x-coordinate and f(x). Always compute f(c) at critical numbers and inflection point candidates.

Mistake 6 — Choosing test values at interval boundaries
Testing the critical number itself, or an endpoint, as a test value in the sign chart.

Each test value must be strictly inside its interval, not at its boundary. Choosing a critical number as a test value gives f′(c) = 0, which does not reveal the sign on the interval.

Mistake 7 — Ignoring domain restrictions
Performing sign chart analysis on an interval outside the function's domain.

If f(x) = √(x − 2), the domain is [2, ∞). Critical number analysis and sign charts must stay within the domain. Values outside the domain are irrelevant.

Mistake 8 — Conflating local and absolute extrema
"The local maximum at x = c is the highest value the function reaches."

A local maximum is the highest point locally — near c. An absolute maximum is the highest point on the entire domain. A local maximum can be well below an absolute maximum elsewhere. Chapter 8 treats absolute extrema fully.

Mistake 9 — Reading the graph of f′ as if it were the graph of f
"The graph of f′ has a local max at x = 3, so f has a local max at x = 3."

A local max of f′ is an inflection point of f, not a local max of f. When a graph of f′ is given, translate carefully: peaks and valleys of f′ become inflection points of f; zeros of f′ become critical numbers of f.

Mistake 10 — Forgetting that f can be increasing while its derivative has a local minimum
"f′ has a minimum at x = c, so f must be decreasing near x = c."

A local minimum of f′ at c means f′(c) is the smallest slope value nearby — but f′(c) may still be positive, meaning f is still increasing through that inflection point. The sign of f′ determines direction; the derivative of f′ determines concavity.

Preparing for Optimization

Chapter 7 established all the derivative tools needed to understand function behavior: increasing and decreasing intervals, critical points, local extrema, concavity, and inflection points. Chapter 8 (Optimization and Applied Derivative Models) will use these tools to solve applied maximum and minimum problems — finding the largest area, the shortest distance, the maximum profit, the minimum cost, the most efficient design.

The transition from Chapter 7 to Chapter 8 is from analyzing function behavior to using that behavior to answer real questions about the best or worst value of a quantity. To prepare, keep these connections in mind:

Critical numbers are where extrema can occur. In Chapter 8, you will also need to check the endpoints of a closed interval (the Closed Interval Method). The Second Derivative Test will help confirm whether a critical number gives a maximum or minimum in physical models. Units will return with full force — when the function being optimized represents cost in dollars, area in square meters, or volume in cubic centimeters, every critical number has a physical meaning that must be interpreted.

Bridge to Chapter 8

The derivative techniques of Chapters 3 through 7 are the foundation for everything in applied mathematics that involves maximizing or minimizing a quantity. The optimization problems in Chapter 8 connect directly to engineering design, economic profit analysis, physics energy minimization, machine learning loss minimization, and scientific computing.

Chapter Summary

Key Terms Review

increasing
f is increasing on (a, b) if f(x₁) < f(x₂) whenever x₁ < x₂. Guaranteed when f′(x) > 0 on the interval.
decreasing
f is decreasing on (a, b) if f(x₁) > f(x₂) whenever x₁ < x₂. Guaranteed when f′(x) < 0 on the interval.
critical number
A value x = c in the domain of f where f′(c) = 0 or f′(c) does not exist. Every local extremum occurs at a critical number.
local maximum
f(c) ≥ f(x) for all x near c. The function is higher at c than at all nearby points.
local minimum
f(c) ≤ f(x) for all x near c. The function is lower at c than at all nearby points.
First Derivative Test
Classifies critical numbers by the sign change of f′. Positive-to-negative → local max. Negative-to-positive → local min. No change → neither.
concave up
f″(x) > 0 on the interval. The graph curves upward like a bowl. Tangent lines lie below the curve.
concave down
f″(x) < 0 on the interval. The graph curves downward like a hill. Tangent lines lie above the curve.
inflection point
A point where f is continuous and concavity changes. Requires a sign change in f″, not just f″(c) = 0.
Second Derivative Test
If f′(c) = 0: f″(c) > 0 gives local min; f″(c) < 0 gives local max; f″(c) = 0 is inconclusive.
sign chart
A visual record of where f′ or f″ is positive, negative, or zero on each interval between critical or inflection candidates.

Practice Problems

Part AConcept Check
  1. In your own words, explain why f′(c) = 0 is necessary but not sufficient for f to have a local maximum or minimum at x = c.
  2. A student claims: "If a function is increasing on (−∞, ∞), then it has no local extrema." Is this student correct? Explain why.
  3. Suppose f″(x) > 0 on (1, 5) and f″(x) < 0 on (5, 9). What can you say about the behavior of f near x = 5?
  4. A graph of f′ shows a local minimum at x = 3. What does this tell you about the graph of f?
  5. Explain the difference between a local maximum and an absolute maximum. Can a function have a local maximum that is not an absolute maximum? Give an example.
  6. If f″(c) = 0, does f necessarily have an inflection point at x = c? Explain and give a counterexample.
Part BDerivative Sign Chart Practice

For each function, find the critical numbers, build a sign chart for f′(x), and determine increasing and decreasing intervals.

  1. f(x) = 2x³ − 6x + 1
  2. f(x) = x⁴ − 8x² + 3
  3. f(x) = x / (x² + 1)
  4. f(x) = x e−x
  5. f(x) = x2/3
  6. f(x) = ln(x² + 1)
Part CCritical Point Practice — First Derivative Test

Find critical numbers and classify each as local maximum, local minimum, or neither using the First Derivative Test.

  1. f(x) = x³ − 12x + 4
  2. g(x) = 2x⁴ − 4x² + 1
  3. h(x) = (1/3)x³ − x² − 3x + 5
  4. f(x) = x⁵ − 5x⁴
  5. f(x) = |x² − 4|
Part DConcavity Practice

Find the intervals of concavity and any inflection points.

  1. f(x) = x³ − 6x² + 9x + 1
  2. f(x) = x⁴ − 12x²
  3. f(x) = e−x²
  4. f(x) = x + sin x on (−π, π)
  5. f(x) = 1 / (x² + 1)
Part ESecond Derivative Test

Find critical numbers and classify them using the SDT where possible. If inconclusive, use the First Derivative Test.

  1. f(x) = x³ − 9x² + 15x − 2
  2. f(x) = 4x⁴ − 8x² + 3
  3. f(x) = x⁴ − 4x³
  4. f(x) = cos x − x on (0, 2π)
Part FCurve Sketching

Use the full curve sketching strategy to sketch each function. Label all critical points, inflection points, and any asymptotes.

  1. f(x) = x³ − 3x²
  2. f(x) = x⁴ − 4x³ + 6
  3. f(x) = x² / (x² + 1)
  4. f(x) = xe−x for x ≥ 0

Applications

  1. Profit analysis. A company's daily profit in dollars is modeled by P(x) = −2x³ + 24x² − 60x − 50 where x is units produced (in thousands). (a) Find the critical numbers of P. (b) Determine where profit is increasing and decreasing. (c) Find the local maximum and interpret its meaning. (d) Find any inflection points and explain their significance.
  2. Motion. A particle's position (in meters) at time t (in seconds) is s(t) = t⁴ − 8t³ + 18t² + 5 for t ≥ 0. (a) Find the velocity v(t) and acceleration a(t). (b) When is the particle moving in the positive direction? (c) When is the particle accelerating? Decelerating? (d) Find inflection points of s(t) and interpret their meaning in terms of motion.
  3. Drug concentration. A drug's concentration in the bloodstream at time t hours is C(t) = te−0.5t for t ≥ 0. (a) When is concentration increasing and when is it decreasing? (b) Find the maximum concentration and when it occurs. (c) Find the inflection point of C(t) and explain what it represents.
  4. Average cost. A firm's average cost per unit is A(x) = x² − 20x + 150 where x is units produced. (a) Find the production level that minimizes average cost. (b) What is the minimum average cost? (c) Use the Second Derivative Test to verify this is a minimum.

Graph Interpretation

  1. The graph of f′(x) crosses the x-axis at x = −2, x = 1, and x = 4. It is positive on (−∞, −2), negative on (−2, 1), positive on (1, 4), and negative on (4, ∞). (a) Where is f increasing? Where is it decreasing? (b) Classify each critical number as a local max, local min, or neither. (c) If the graph of f′ has local maxima at x = −0.5 and x = 3, what does this tell you about inflection points of f?
  2. A table of values of f′(x) is given below. (a) On what intervals is f increasing? (b) Classify x = 2 and x = 6 as local max, local min, or neither. (c) Where is f′ increasing? What does this tell you about concavity of f?
    xf′(x)
    0−3
    1−1
    20
    32
    44
    53
    60
    7−2

Error Analysis

Each problem shows a student's work. Identify the error and write the correct solution.

Error Analysis Problem 1
"For f(x) = x³, f′(x) = 3x². Since f′(0) = 0, there is a local extremum at x = 0. I compute f″(x) = 6x and f″(0) = 0. Since the second derivative is zero, the point is neither a max nor a min."

Identify the logical error. The conclusion may be correct, but the reasoning is flawed. What should the student have done instead of citing the Second Derivative Test result?

Error Analysis Problem 2
"For f(x) = x⁴, I find f″(x) = 12x² and f″(0) = 0. Since the second derivative is zero at x = 0, there is an inflection point at x = 0."

Identify the error and state the correct conclusion. What step did this student skip?

Error Analysis Problem 3
"A student builds a sign chart for f(x) = 3x⁵ − 5x³. They find f′(x) = 15x⁴ − 15x² = 15x²(x² − 1) and identify critical numbers at x = 0, 1, −1. They compute f″(0) = 0 and conclude the Second Derivative Test is inconclusive. Since the test is inconclusive, they skip x = 0 and move on."

What is wrong with this reasoning? What should the student do?

Error Analysis Problem 4
"A student analyzing f uses a graph of f′. They see that f′ reaches a local maximum at x = 4, so they report that f has a local maximum at x = 4."

Explain the error. A local maximum of f′ at x = 4 corresponds to what feature on the graph of f?

Chapter 7 Checkpoint

Chapter 7 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 7
Part A — Concepts
  1. What does it mean for a function to be increasing on an interval? State the derivative condition that guarantees this.
  2. Define a critical number. Give an example of a function with a critical number that is not a local extremum.
  3. State the First Derivative Test in your own words.
  4. What is the difference between concave up and concave down? How is concavity connected to the second derivative?
  5. State the condition for an inflection point. Why must you check for a sign change in f″?
Part B — Skills

For f(x) = x³ − 6x² + 9x + 2:

  1. Find f′(x) and identify all critical numbers.
  2. Build a sign chart for f′(x) and determine where f is increasing and decreasing.
  3. Use the First Derivative Test to classify each critical number as a local max, local min, or neither.
  4. Find f″(x) and determine the intervals of concavity.
  5. Find any inflection points.
  6. Use your information from problems 6–10 to sketch the graph of f. Label the key points.
Part C — Application
  1. A position function is given by s(t) = 2t³ − 9t² + 12t + 1 for t ≥ 0 (meters, t in seconds). (a) Find the velocity v(t) and acceleration a(t). (b) When is the object at rest? (c) When is the object moving in the positive direction? (d) When is the velocity increasing? Interpret in terms of motion.
  2. Challenge. A function f has these properties: f′(x) > 0 on (−∞, −1) ∪ (3, ∞); f′(x) < 0 on (−1, 3); f″(x) > 0 on (1, ∞); f″(x) < 0 on (−∞, 1); f(−1) = 4, f(1) = 0, f(3) = −5. (a) Identify all local maxima and local minima. (b) Identify any inflection points. (c) Sketch a graph consistent with these conditions.
Looking Ahead
Chapter 8: Optimization and Applied Derivative Models

Chapter 7 gave you the tools to understand function behavior: where functions rise, fall, peak, and curve. Chapter 8 will use these tools to answer a more pointed question — what is the best value? Finding the largest area, the shortest distance, the maximum profit, the minimum cost, the most efficient design — all of these are optimization problems. You will apply the critical number machinery from this chapter to real-world objective functions, learn the Closed Interval Method for finding absolute extrema on bounded domains, and practice reading, representing, solving, checking, and explaining every answer in context with correct units.

MGU Library Connection
  • Derivative Rules Formula Sheet — Appendix F of this textbook
  • Optimization Strategy Guide — Appendix I of this textbook
  • MGU Calculus Chapter 6 — Implicit Differentiation and Related Rates
  • MGU Physics chapters on kinematics — position, velocity, and acceleration graphs in applied motion
  • MGU Economics chapters on marginal analysis — profit maximization and cost minimization
  • MGU Data Science Foundations — gradient descent and loss function optimization
  • MGU Dictionary entries: critical number, inflection point, concavity, First Derivative Test, Second Derivative Test