Derivatives and Curve Sketching
Combine domains, limits and derivative sign charts to explain the shape of a graph and recover information from changing rates.
Explain the shape before plotting
A graphing window can hide an asymptote or make a nearly flat curve look constant. Derivatives let us justify where a graph rises, falls or bends. Begin with its domain and limits, then organize the signs of the derivatives into a coherent sketch.
Critical numbers and sign-chart boundaries
Critical number
A critical number c belongs to the domain of f and satisfies \(f'(c)=0\) or has no finite derivative. An excluded input is not a critical number of f.
Partition number
For a sign chart of an expression, a partition number is a zero or a discontinuity of that expression. State which expression is being studied: f, f′ or f′′. Its partition numbers divide the number line into intervals for sign testing.
A domain exclusion is not a critical number
For \(f(x)=x^2/(1-x^2)\), \(f'(x)=2x/(1-x^2)^2\). Identify the domain and critical numbers.
Show worked solution
The domain excludes ±1. The derivative is zero at x=0. Although the derivative is undefined at ±1, these inputs are not in the function domain. Thus 0 is the only critical number; −1,0,1 partition the derivative sign chart.
Increasing and decreasing intervals
Derivative sign test
If f is differentiable on an interval and f′ is positive throughout it, f is increasing there. If f′ is negative throughout it, f is decreasing there. Apply the test separately on intervals within the domain.
Read the sign, not the size
Use \(f'(x)=2x/(1-x^2)^2\) to determine monotonicity.
Show worked solution
The denominator is positive wherever defined. The sign is therefore the sign of x: f decreases on (−∞,−1) and (−1,0), and increases on (0,1) and (1,∞). Do not combine across the excluded inputs.
Positive height versus positive slope
A graph lies below the x-axis but has f′>0 throughout an interval. Is it increasing or decreasing?
Hint
Height and slope are different quantities.
Show worked solution
It is increasing. A negative function value says the point lies below the axis; a positive derivative says the output rises as x increases.
Local extrema and the first derivative test
Local extremum
A local maximum at c has f(c) at least as large as all nearby domain values; a local minimum has f(c) at most as large. These comparisons concern a neighborhood, not the entire domain.
First derivative test
At a critical number where f is continuous, a change in f′ from positive to negative gives a local maximum; negative to positive gives a local minimum. The same sign on both sides gives neither.
A stationary point need not turn
Compare \(f(x)=x^3\) and \(g(x)=x^2\) at zero.
Show worked solution
For x³, f′=3x² is positive on both sides, so zero is not a local extremum. For x², g′=2x changes from negative to positive, so zero is a local minimum.
Classify a sign change
Near c=4, f′ is positive to the left and negative to the right, and f is continuous at 4. Classify f(4).
Hint
Follow how the graph moves into and out of the point.
Show worked solution
The function rises into the point and falls afterward, so f(4) is a local maximum. Its absolute status requires information about the rest of the domain.
Second derivatives describe changing slopes
Second derivative
\(f''(x)\) is the derivative of \(f'(x)\), where it exists.
Concavity
For a differentiable function on an interval, concavity upward means its slopes are nondecreasing; concavity downward means they are nonincreasing. In the twice-differentiable case, f′′>0 guarantees upward concavity and f′′<0 guarantees downward concavity.
A falling function can bend upward: its negative slopes may be becoming less negative. Separate the question “Does it rise?” from “Are its slopes increasing?”
Decreasing and concave upward
Analyze \(f(x)=e^{-x}\).
Show worked solution
f′=−e⁻ˣ<0 and f′′=e⁻ˣ>0 everywhere. The function decreases while its slopes increase toward zero.
Read a derivative graph
If a graph of f′ lies below the axis and rises throughout an interval, describe f.
Hint
The height of f′ controls monotonicity; its trend controls concavity.
Show worked solution
The function decreases because f′<0 and is concave upward because f′ is increasing.
Read slopes and bending
Inflection requires a change in concavity
Inflection point
An inflection point is a point on a continuous curve at which the concavity changes. A zero or undefined second derivative is only a candidate.
Why f′′=0 is not enough
Compare \(f(x)=x^4\) and \(g(x)=x^3\) at zero.
Show worked solution
For x⁴, f′′=12x² is positive on both sides; there is no inflection. For x³, g′′=6x changes from negative to positive, so (0,0) is an inflection point.
A bend across an asymptote?
For \(f(x)=1/x\), the concavity changes across zero. Is zero an inflection point?
Hint
An inflection point must be on the continuous curve.
Show worked solution
No. The function is undefined and discontinuous at zero, so there is no point on the graph there.
A non-finite derivative can occur at an inflection
Classify the origin on \(f(x)=\sqrt[3]x\).
Show worked solution
The real cube root is continuous at zero. For x≠0, f″(x)=−2/(9x^(5/3)), which is positive on the negative side and negative on the positive side. Hence the origin is an inflection point, although its tangent is vertical and f′(0) is not finite.
A zero second derivative is only a candidate
The second derivative test and its limits
Second derivative test
Suppose f′(c)=0 and the second derivative exists near c. If f′′(c)>0, c gives a local minimum; if f′′(c)<0, a local maximum. If f′′(c)=0, this test is inconclusive.
Inconclusive means investigate
At zero, compare \(x^4\), \(-x^4\) and \(x^3\).
Show worked solution
All have first and second derivative zero at zero. Nevertheless, x⁴ has a minimum, −x⁴ a maximum, and x³ neither. A first-derivative sign chart resolves all three.
Build a complete rational-function sketch
Put every piece of evidence together
Analyze \(f(x)=x^2/(1-x^2)\), using \(f'=2x/(1-x^2)^2\) and \(f''=(6x^2+2)/(1-x^2)^3\).
Show worked solution
The domain excludes ±1; the only intercept is (0,0). The function is even. There are vertical asymptotes x=±1 and horizontal asymptote y=−1. Since f=−1+1/(1−x²), its outer branches lie below −1 and its middle branch is nonnegative.
| Interval | f′ | f′′ | Behavior |
|---|---|---|---|
| (−∞,−1) | − | − | decreasing, concave down |
| (−1,0) | − | + | decreasing, concave up |
| (0,1) | + | + | increasing, concave up |
| (1,∞) | + | − | increasing, concave down |
At zero, f′ changes from negative to positive, giving a local minimum 0. Concavity changes across the excluded inputs, so there are no inflection points. The middle branch approaches +∞ at both boundaries; the outer sides approach −∞ near their asymptotes.
Recover a function from geometric conditions
Curve information can determine unknown coefficients. Translate each condition into an equation: a point uses f, a tangent slope uses f′, and an inflection candidate uses f′′. Then verify an actual concavity change.
A point, a slope and an inflection
Find a and b so that \(f(x)=2x^3+ax^2+bx\) has an inflection at (−1,1) and slope −3 there.
Show worked solution
f′′=12x+2a, so f′′(−1)=0 gives a=6. The point equation −2+a−b=1 gives b=3. Then f′(−1)=6−12+3=−3, as required. Finally f′′=12(x+1) changes sign at −1, confirming the inflection.
Transformations provide useful checks on a sketch. Adding a constant changes height but not either derivative. Adding a linear term changes slopes but not concavity. Multiplication by a negative constant reverses maxima and minima, and upward and downward concavity.
A tilted cubic
Compare \(f(x)=x^3\) and \(g(x)=x^3+2x\).
Show worked solution
Both second derivatives are 6x, so both change concavity at zero. But f′(0)=0 while g′(0)=2. An inflection is not necessarily a stationary point.
Curvature of a differentiable inverse
When an inverse g exists locally and the needed derivatives exist with f′(c)≠0, its second derivative is g′′(f(c))=−f′′(c)/(f′(c))³.
Differentiate g′(f(x))f′(x)=1, then substitute g′(f(x))=1/f′(x). The cube in the denominator keeps the orientation sign; do not assume inverse concavity always reverses without checking f′.
Interpret shape within the model domain
Advertising and predicted sales
Monthly sales are modeled by \(S(x)=3x^3-x^4/4+200\) for 0≤x≤10, where x is advertising spending in thousands of dollars. Determine where sales increase and decrease.
Show worked solution
Sales increase on (0,9) and decrease on (9,10). The derivative is zero at 0 as well as 9, but 0 is a domain endpoint. The model’s behavior outside [0,10] is not justified by this problem.
Keep in mind
A graph is a synthesis of evidence, not a collection of disconnected calculations. Label the domain exclusions, extrema, inflection points and asymptotes, then connect branches consistently.
Point of diminishing returns
For an increasing sales model, a transition from concave upward to concave downward marks a point of diminishing returns: marginal sales change from increasing to decreasing. It is not necessarily the input that maximizes sales or profit.
Sales still rise after marginal gains peak
Suppose \(N(x)=100+12x^2-x^3\) for 0≤x≤8, with x in thousands of pesos of advertising. Find where marginal sales peak.
Show worked solution
N″ changes from positive to negative at x=4, so marginal sales reach their maximum N′(4)=48 units per thousand pesos there. The point on the sales curve is (4,228). Sales still increase on (4,8), but at a decreasing rate. Maximum sales occur at 8 within this domain, not at 4.
References
Raymond A. Barnett, Michael R. Ziegler, Karl E. Byleen and Christopher J. Stocker, Calculus for Business, Economics, Life Sciences, and Social Sciences, 14th Global Edition, chapters on limits, differentiation and applications. Mark D. Tomenes, Applied Calculus for Business and Economics I, teaching notes and assessment materials (2025–2026).
Further reading: Gilbert Strang and Edwin Herman, Calculus, Volume 1, OpenStax, chapters 2–4.
