The first derivative test for increase, decrease and local extrema; concavity as the sign of the second derivative and as the direction the tangents lie; points of inflection where the concavity changes; and the second derivative test with the three functions that show why it can be inconclusive.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 4 — Applications of Derivatives
Derivatives and the Shape of a Graph
Objectives
Five outcomes. Section 4.4 supplied the licence for the first; the rest build a complete reading of a graph on it.
OpenStax Calculus Volume 1, §4.5 Derivatives and the Shape of a Graph §4.5, pp. 339-353 — the section these objectives are drawn from
Warm-up
Section 4.3 found critical points but could not classify them, and Section 4.4 proved that the sign of the derivative controls whether a function rises or falls.
Discussion prompt
The cubic x cubed minus 3x has critical points at plus and minus 1. Using only the sign of the derivative, decide which is a maximum and which a minimum.
Hint: Test the derivative's sign on each of the three intervals.
Answer:
\[ f'(x) = 3x^{2}-3 = 3(x-1)(x+1) \]
For x below negative 1 both factors are negative, so the product is positive and f rises. Between negative 1 and 1 the factors differ in sign, so f falls. Above 1 both are positive and f rises again.
So the function rises, turns down at negative 1, and turns back up at 1 — a maximum at negative 1 and a minimum at 1. That reasoning is the first derivative test, and Section 4.4's corollary is what makes it valid rather than merely plausible.
Concept
The sign of the first derivative says whether the curve rises or falls; the sign of the second says which way it bends. Together they determine the shape completely, and both are read from sign charts.
the shape of a graph — Determined by two independent readings: the first derivative's sign giving the direction of travel, and the second derivative's sign giving the concavity.
\[ f' \text{ sign} \to \text{rising or falling}; \qquad f'' \text{ sign} \to \text{concave up or down} \]
The two questions are genuinely independent. A function can be rising while levelling off, or falling while steepening, and the four combinations give four distinct shapes.
Figure (svg): The four combinations of the two derivative signs, each with the shape it produces
OpenStax Calculus Volume 1, §4.5 Derivatives and the Shape of a Graph §4.5, pp. 339-342
Section
Section 1
Concept
A function increases where its derivative is positive and decreases where it is negative. At a critical point, a change from positive to negative gives a local maximum and from negative to positive a local minimum; no change gives neither.
the first derivative test — At a critical point c, if f prime changes from positive to negative there is a local maximum; from negative to positive, a local minimum; and if the sign does not change, neither.
\[ f' : + \to - \;\Rightarrow\; \text{max}; \qquad f' : - \to + \;\Rightarrow\; \text{min} \]
The test works at every critical point, including corners and cusps where the derivative does not exist — because it only requires the sign on either side, not a value at the point.
Figure (svg): A cubic above a sign chart for its derivative, with the sign changes aligned to the turning points
OpenStax Calculus Volume 1, §4.5 Derivatives and the Shape of a Graph §4.5, pp. 339-344 — the first derivative test
Picture it
The cubic and its derivative's sign chart.
Figure (svg): A cubic above a sign chart for its derivative, with the sign changes aligned to the turning points
Every sign change on the chart sits directly beneath a turning point on the curve. Reading the chart and reading the graph are the same act performed twice.
Worked example
Example 4.26. Chart the sign, then classify.
\[ \text{Find the intervals of increase and decrease and classify the extrema of } f(x)=x^{3}-3x. \]
Differentiate and factor
Why: Take out 3.
\[ f'(x) = 3(x - 1) (x + 1) \]
Find the critical points
Why: Set to zero.
\[ x = -1\text{ and } x = 1 \]
Test the sign on each interval
Why: Pick one point in each.
\[ \text{at } -2: +;\text{ at } 0: -;\text{ at } 2: + \]
Read the increase and decrease
Why: Positive means rising.
Classify each critical point
Why: Plus to minus, then minus to plus.
\[ \max\text{ at } -1, \min\text{ at } 1 \]
Figure (svg): A cubic above a sign chart for its derivative, with the sign changes aligned to the turning points
\[ \text{max } f(-1)=2; \quad \text{min } f(1)=-2 \]
Verify: check the values and the symmetry
Why: The maximum value is 2 and the minimum negative 2, and the function is odd so the two are symmetric about the origin — as they must be. Note the test classified BOTH critical points without evaluating the function at all; the values were computed afterwards only to report them. That separation is useful, because the classification depends only on signs and is often much easier than the arithmetic.
OpenStax Calculus Volume 1, §4.5 Derivatives and the Shape of a Graph §4.5, pp. 341-342
Sorting
Read the derivative's sign on either side.
Sort into buckets
Sort each sign pattern.
The corner case matters: the first derivative test needs only the signs on either side, so it classifies extrema at points where the derivative does not exist. The second derivative test, coming later, cannot do that at all.
Worked example
Checkpoint 4.26. No sign change, no extremum.
\[ \text{Classify the critical point of } f(x)=x^{3} \text{ at the origin.} \]
Differentiate
Why: Power rule.
\[ f'(x) = 3 x ^{2} \]
Find the critical point
Why: Set to zero.
\[ x = 0 \]
Test the sign on the left
Why: A square is positive.
\[ f'(-1) = 3 > 0 \]
Test the sign on the right
Why: Also positive.
\[ f'(1) = 3 > 0 \]
Conclude
Why: No sign change.
Figure (svg): The solution to Worked example a critical point that is neither shown as a ladder of expressions, one row per legal move
\[ f' > 0 \text{ on both sides} \;\Longrightarrow\; \text{no extremum} \]
Verify: confirm against the values
Why: The function takes negative values to the left of the origin and positive values to the right, so 0 is neither the largest nor the smallest value nearby — it is simply passed through. The graph flattens momentarily and continues climbing, which is what a zero derivative without a sign change looks like. This is the standing counterexample from Section 4.3, now classified by a method that handles it correctly.
OpenStax Calculus Volume 1, §4.5 Derivatives and the Shape of a Graph §4.5, pp. 342-343
Trap
\[ f'(0) = 0 \text{ for } f(x)=x^{3} \]
Conclude there is an extremum
Why: The student treats a critical point as an extremum.
\[ \text{max or min at } 0 \quad \text{(wrong)} \]
The derivative is positive on both sides, so the function increases straight through the origin.
\[ f' : + \to + \;\Longrightarrow\; \text{neither} \]
Test the sign on BOTH sides of every critical point
Why: It is the change, not the zero, that produces a turn.
This is the same one-way implication as Fermat's theorem in Section 4.3, now with a method attached. Every critical point must be tested, and the test is two sign evaluations — which is cheap enough that there is no excuse for skipping it.
Fill the middle
The cubic's derivative, evaluated between its two critical points.
Fill in the blanks
f'(0) = 3(0-1)(0+1) = -3
Why: The derivative is negative between the critical points, so the function decreases there. Combined with the positive signs outside, that gives a maximum at negative 1 and a minimum at 1.
Ranking
Classifying critical points by the first derivative.
Put in order
Why: Testing one point per interval is enough because the derivative is continuous between critical points and so cannot change sign without passing through zero — which is the Intermediate Value Theorem of Section 2.4 quietly at work.
Prediction
Commit before reasoning.
Predict first
Why is testing the derivative at a single point in each interval enough?
Correct: Because the derivative cannot change sign between critical points without vanishing there.
\[ f' \text{ continuous and non-vanishing on an interval} \;\Longrightarrow\; \text{constant sign} \]
Why: By the Intermediate Value Theorem a continuous derivative that changed sign would have to pass through zero, which would make that point critical — and by construction there are no critical points strictly inside the interval. So the sign is constant throughout and one test point reveals it. The argument needs the derivative to be continuous, which is why intervals must also be split at points where the derivative fails to exist.
Section
Section 2
Concept
A graph is concave up where it lies above its tangent lines and concave down where it lies below them. Equivalently, it is concave up where the derivative is increasing — that is, where the second derivative is positive.
concavity — A function is concave up on an interval if its derivative is increasing there, equivalently if its graph lies above its tangent lines; concave down if the derivative is decreasing and the graph lies below its tangents.
\[ f'' > 0 \;\Longleftrightarrow\; f' \text{ increasing} \;\Longleftrightarrow\; \text{concave up} \]
The two readings are worth holding together. The tangent-line reading explains Section 4.2's error directions; the increasing-derivative reading is what makes the second derivative the right tool.
Figure (svg): Concave up and concave down, with tangents drawn to show which side the curve lies on
OpenStax Calculus Volume 1, §4.5 Derivatives and the Shape of a Graph §4.5, pp. 344-347 — concavity and points of inflection
Picture it
Concave up and concave down, with tangents.
Figure (svg): Concave up and concave down, with tangents drawn to show which side the curve lies on
On the left the tangents lie beneath the curve everywhere; on the right they lie above. That is the same fact Section 4.2 used to predict whether a linear estimate over- or undershoots.
Worked example
Example 4.28. Chart the second derivative's sign.
\[ \text{Find where } f(x)=x^{3}-3x \text{ is concave up and concave down.} \]
Differentiate twice
Why: Two applications of the power rule.
\[ f' = 3 x ^{2} - 3, f'' = 6 x \]
Find where the second derivative vanishes
Why: Set to zero.
\[ x = 0 \]
Test the sign on the left
Why: Negative x gives negative.
\[ f''(-1) = -6 < 0 \]
Test the sign on the right
Why: Positive x gives positive.
\[ f''(1) = 6 > 0 \]
State the intervals
Why: Down then up.
\[ \text{concave down on } x < 0,\text{ up on } x > 0 \]
Figure (svg): The solution to Worked example finding the concavity shown as a ladder of expressions, one row per legal move
\[ f'' = 6x: \; \text{down on } (-\infty,0), \text{ up on } (0,\infty) \]
Verify: check against the graph and against the first derivative
Why: The cubic does bend downward on the left and upward on the right, meeting at the origin. Equivalently, the first derivative 3x squared minus 3 is a parabola that decreases for negative x and increases for positive x — and 'the derivative is increasing' is exactly what concave up means. Both readings agree, which is a useful check whenever the second derivative's sign is in doubt.
OpenStax Calculus Volume 1, §4.5 Derivatives and the Shape of a Graph §4.5, pp. 345-346
Matching
Direction from the first, bending from the second.
Match the pairs
Why: All four occur and all four are distinguishable on a graph. The middle two are the ones that get conflated, and they are opposites: one is a curve flattening as it climbs, the other flattening as it drops.
Worked example
Checkpoint 4.28. The two questions do not interact.
\[ \text{Where is } f(x)=x^{3}-3x \text{ falling and concave up at once?} \]
Recall where the function falls
Why: From the first derivative test.
\[ \text{on } (-1, 1) \]
Recall where it is concave up
Why: From the second derivative.
\[ \text{on } (0, \infty) \]
Intersect the two conditions
Why: Both must hold.
\[ \text{on } (0, 1) \]
Describe the shape there
Why: Falling, but levelling off.
Figure (svg): The four combinations of the two derivative signs, each with the shape it produces
\[ (-1,1) \cap (0,\infty) = (0,1) \]
Verify: confirm the description matches the graph
Why: Between 0 and 1 the cubic is descending toward its minimum at 1, and it flattens as it approaches — falling but with the fall easing. That is exactly what a negative first derivative and a positive second derivative describe together. The two conditions are genuinely independent, which is why all four combinations occur and why the shape needs both to be pinned down.
OpenStax Calculus Volume 1, §4.5 Derivatives and the Shape of a Graph §4.5, pp. 346-347
Error analysis
A student describes a function's bending.
Annotate
On: \( f' > 0 \text{ on an interval} \;\Longrightarrow\; f \text{ is concave up there} \)
The two derivatives answer two separate questions. Confusing them collapses four distinct shapes into two and makes the reading of any graph unreliable.
Fill the middle
The cubic's second derivative, set to zero.
Fill in the blanks
f''(x) = 6x = 0 \;\Longrightarrow\; x = 0
Why: The second derivative vanishes at the origin and changes sign there, so the concavity flips from down to up. That makes the origin a point of inflection, which is the next idea.
Two truths and a lie
All three are about concavity.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and it confuses the two derivatives. The squaring function is concave up everywhere and yet decreases for negative x. Direction and bending are independent questions answered by different derivatives, which is exactly why four combinations exist.
Prediction
Commit before reasoning.
Predict first
Why does a concave up function's linear approximation undershoot?
Correct: Because the graph lies above its tangent lines.
\[ f'' > 0 \;\Longrightarrow\; f(x) > L(x) \text{ near } a \]
Why: The linear estimate is a height on the tangent, and if the curve is above the tangent then the true value exceeds the estimate. That is the tangent-line reading of concavity doing the work, and it explains the direction on both sides of the anchor. Whether the function is increasing is irrelevant, and the size of the second derivative affects how much the estimate is off but not which way.
Section
Section 3
Concept
A point of inflection is where the concavity changes from up to down or from down to up. The second derivative vanishing is necessary but not sufficient — the sign must actually change.
point of inflection — A point on the graph where the function is continuous and the concavity changes. At such a point the second derivative is zero or undefined, and it must change sign there.
\[ f'' \text{ changes sign at } c \;\Longrightarrow\; \text{inflection at } c \]
The parallel with the first derivative test is exact. A vanishing second derivative marks a candidate, and only a sign change confirms it — with the quartic providing the standard counterexample.
Figure (svg): A cubic with its inflection point, where the concavity changes and the tangent crosses
OpenStax Calculus Volume 1, §4.5 Derivatives and the Shape of a Graph §4.5, pp. 346-349 — points of inflection
Picture it
A cubic at its inflection point.
Figure (svg): A cubic with its inflection point, where the concavity changes and the tangent crosses
The shaded regions show the two concavities meeting, and the tangent at the join passes through the curve rather than staying on one side. That crossing happens only at an inflection.
Worked example
Example 4.29. Zero, then a sign change.
\[ \text{Find the points of inflection of } f(x)=x^{3}-3x. \]
Compute the second derivative
Why: Two differentiations.
\[ f''(x) = 6 x \]
Find the candidates
Why: Where it vanishes or fails.
\[ x = 0 \]
Test the sign on the left
Why: Negative.
\[ f''(-1) = -6 \]
Test the sign on the right
Why: Positive.
\[ f''(1) = 6 \]
Confirm and report the point
Why: The sign changed; get the height.
\[ \text{inflection at } (0, 0) \]
Figure (svg): A cubic with its inflection point, where the concavity changes and the tangent crosses
\[ (0,0): \; f'' \text{ changes from negative to positive} \]
Verify: check that the tangent crosses there
Why: The tangent at the origin has slope negative 3, and the cubic passes from above that line to below it as x increases through 0 — the tangent cuts through the curve. That crossing is the visual signature of an inflection and happens nowhere else on this graph. Note that the point was reported with both coordinates, since an inflection is a point on the curve rather than an input.
OpenStax Calculus Volume 1, §4.5 Derivatives and the Shape of a Graph §4.5, pp. 347-348
Sorting
Check whether the second derivative changes sign.
Sort into buckets
Sort each candidate at the origin.
The odd powers give inflections and the even ones do not, which is the same parity pattern that decided extrema in Section 4.3 — shifted one derivative along. The squaring function is the case where the second derivative never vanishes at all.
Worked example
Checkpoint 4.29. The candidate that fails.
\[ \text{Does } f(x)=x^{4} \text{ have a point of inflection at the origin?} \]
Differentiate twice
Why: Power rule twice.
\[ f' = 4 x ^{3}, f'' = 12 x ^{2} \]
Check the candidate
Why: The second derivative vanishes.
\[ f''(0) = 0 \]
Test the sign on the left
Why: A square is non-negative.
\[ f''(-1) = 12 > 0 \]
Test the sign on the right
Why: Also positive.
\[ f''(1) = 12 > 0 \]
Conclude
Why: No sign change.
Figure (svg): The solution to Worked example a zero second derivative with no inflection shown as a ladder of expressions, one row per legal move
\[ f'' \ge 0 \text{ everywhere} \;\Longrightarrow\; \text{concave up throughout} \]
Verify: confirm the graph is concave up on both sides
Why: The quartic is bowl-shaped throughout, lying above every tangent including the horizontal one at the origin. The second derivative touches zero there and immediately returns to positive, so the bending never reverses. This is exactly the same structure as the cubic's failure in the first derivative test, one derivative higher — a vanishing derivative marks a candidate and only a sign change confirms it.
OpenStax Calculus Volume 1, §4.5 Derivatives and the Shape of a Graph §4.5, pp. 348-349
Trap
\[ f(x)=x^{4}: \; f''(0)=0 \]
Conclude there is a point of inflection
Why: The student treats the zero as sufficient.
\[ f''(x)=12x^{2} \ge 0 \text{ on both sides} \]
The concavity is upward throughout, so nothing reverses. The second derivative merely touches zero.
\[ \text{inflection} \iff f'' \text{ CHANGES SIGN} \]
Test the sign on both sides of every candidate
Why: Exactly as in the first derivative test, one derivative higher.
The two tests have identical structure, and recognising that halves what needs remembering: in both cases a vanishing derivative is a candidate and a sign change is the confirmation. The quartic and the cubic are the two standard counterexamples, one for each test.
Fill the middle
The quartic, whose second derivative is twelve x squared.
Fill in the blanks
f''(-1) = 12 \textdoes not change f''(1) = 12: \quad \text___ \; ___
Why: Both are positive, so the concavity is upward on both sides and the origin is not an inflection point. The second derivative touches zero and returns, exactly as the cubic's first derivative did.
Two truths and a lie
All three are about inflection points.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The quartic has a vanishing second derivative at the origin and is concave up on both sides, so nothing reverses. The structure is identical to the first derivative test: a zero is a candidate, a sign change is the confirmation.
Prediction
Commit before reasoning.
Predict first
Why does locating inflection points work just like locating extrema?
Correct: Because both ask where a derivative changes sign, one level apart.
\[ \text{extremum}: f' \text{ flips}; \qquad \text{inflection}: f'' \text{ flips} \]
Why: An extremum is where f prime changes sign, and an inflection is where f double prime changes sign. The reasoning, the counterexamples and the sign-chart method are identical, shifted one derivative along — and recognising that halves what must be learned. The cubic fails the first test and the quartic the second, for exactly the same structural reason.
Section
Section 4
Concept
At a point where the derivative vanishes, a positive second derivative means the graph is bowl-shaped there and the point is a minimum; a negative one means dome-shaped and a maximum. A zero second derivative gives no information.
the second derivative test — If f prime of c is zero and f double prime of c is positive, c gives a local minimum; if f double prime of c is negative, a local maximum. If f double prime of c is zero the test is inconclusive.
\[ f'(c)=0, f''(c) > 0 \;\Rightarrow\; \text{min}; \qquad f''(c) < 0 \;\Rightarrow\; \text{max} \]
Its advantage is that it needs only one evaluation rather than two sign tests, which matters when the derivative is awkward to sign. Its cost is the third case and the requirement that the second derivative exist.
Figure (svg): The second derivative test, with the two conclusive cases and the inconclusive one
OpenStax Calculus Volume 1, §4.5 Derivatives and the Shape of a Graph §4.5, pp. 349-352 — the second derivative test
Picture it
The test's outcomes.
Figure (svg): The second derivative test, with the two conclusive cases and the inconclusive one
The first two are quick and decisive. The third is a genuine gap rather than an oversight, and the next worked example shows exactly why it cannot be closed.
Worked example
Example 4.31. One evaluation per critical point.
\[ \text{Classify the critical points of } f(x)=x^{3}-3x \text{ with the second derivative test.} \]
Find the critical points
Why: From the first derivative.
\[ x = -1\text{ and } x = 1 \]
Compute the second derivative
Why: Differentiate again.
\[ f''(x) = 6 x \]
Evaluate at the first critical point
Why: Negative.
\[ f''(-1) = -6 < 0 \]
Conclude for it
Why: Concave down there.
Evaluate at the second
Why: Positive.
\[ f''(1) = 6 > 0, a\text{ local minimum} \]
Figure (svg): The solution to Worked example applying the test shown as a ladder of expressions, one row per legal move
\[ f''(-1) < 0: \text{ max}; \quad f''(1) > 0: \text{ min} \]
Verify: compare with the first derivative test's answer
Why: The first derivative test gave the same classification by testing three signs; this needed only two evaluations. Both are correct and the second is quicker here, because the second derivative is trivial. When the second derivative is messy or the critical point is a corner, the first derivative test is the better tool — the choice is tactical rather than a matter of one being superior.
OpenStax Calculus Volume 1, §4.5 Derivatives and the Shape of a Graph §4.5, pp. 350-351
Sorting
At a critical point, read the second derivative's sign.
Sort into buckets
Sort each case.
The fourth case is worth noticing: a tiny positive value is still positive, so the test is conclusive. Only exactly zero fails, which is why the test is more useful in practice than the existence of a third case suggests.
Worked example
Checkpoint 4.31. Three functions, identical conditions.
\[ \text{Compare } x^{4}, -x^{4} \text{ and } x^{3} \text{ at the origin.} \]
Compute both derivatives at the origin for each
Why: All vanish.
\[ f'(0) = f''(0) = 0\text{ in all three} \]
Examine the quartic
Why: It is bowl-shaped.
Examine its negative
Why: Dome-shaped.
Examine the cubic
Why: It passes straight through.
Conclude
Why: Same conditions, three outcomes.
Figure (svg): Three functions with identical first and second derivatives at the origin and three different behaviours
\[ f'(0)=f''(0)=0 \text{ in all three, with three different answers} \]
Verify: confirm the first derivative test does distinguish them
Why: For the quartic, f prime is 4x cubed, negative then positive — a minimum. For its negative, positive then negative — a maximum. For the cubic, positive on both sides — neither. So the first derivative test separates all three cleanly. That is what 'inconclusive' means here: not that the answer is unknowable, but that these two numbers do not carry it and a different test must be used.
OpenStax Calculus Volume 1, §4.5 Derivatives and the Shape of a Graph §4.5, pp. 351-352
Error analysis
A student applies the second derivative test to a quartic.
Annotate
On: \( f(x)=x^{4}: \; f'(0)=0 \text{ and } f''(0)=0 \;\Longrightarrow\; \text{no extremum at } 0 \)
An inconclusive result is an instruction to use a different method, not a conclusion. The first derivative test always works where this one fails, and it takes only two extra sign evaluations.
Fill the middle
The cubic's second derivative, evaluated at its left critical point.
Fill in the blanks
f''(-1) = 6(-1) = -6 < 0 \;\Longrightarrow\; \textmaximum \; ___
Why: A negative second derivative means the graph is dome-shaped there, so the critical point is at the top — a local maximum. One evaluation settles it.
Matching
Two tests, different situations.
Match the pairs
Why: The first row is decisive: the second derivative test needs the second derivative to exist at the point, so it says nothing at a corner. The first derivative test works there, which makes it the more general of the two.
Prediction
Commit before reasoning.
Predict first
The second derivative test is inconclusive at a critical point. What follows?
Correct: Nothing yet; use the first derivative test.
\[ x^{4}, -x^{4}, x^{3}: \text{ same } f'(0), f''(0); \text{ three answers} \]
Why: The quartic, its negative and the cubing function all satisfy exactly the same conditions and have a minimum, a maximum and neither respectively. So the two numbers genuinely carry no information — but the sign of the first derivative on either side separates all three cleanly. Inconclusive means this particular test cannot decide, never that the answer does not exist.
Section
Section 5
Concept
Charting the first derivative's sign gives the intervals of increase and decrease and locates the extrema. Charting the second gives the concavity and the inflection points. Together they determine the shape completely.
a complete shape analysis — The combination of both sign charts, partitioning the domain into intervals on which the direction and the bending are each constant, with the critical points and inflection points marking the boundaries.
\[ f' \text{ chart} + f'' \text{ chart} \;\Longrightarrow\; \text{the shape} \]
The two charts generally have different boundaries, so the domain is cut into more regions than either chart alone provides. Each region has one of the four shape combinations.
Figure (svg): A complete sign analysis: intervals, critical points and the resulting shape
OpenStax Calculus Volume 1, §4.5 Derivatives and the Shape of a Graph §4.5, pp. 349-353 — drawing the graph of a function
Picture it
The cubic's first and second derivative sign charts, stacked.
Figure (svg): A complete sign analysis: intervals, critical points and the resulting shape
The upper chart's boundaries are at plus and minus 1 and the lower one's at 0, so together they split the line into four intervals — each with its own combination of direction and bending.
Worked example
Example 4.32. Both charts, then the description.
\[ \text{Analyse the shape of } f(x)=x^{4}-4x^{3}. \]
Differentiate twice
Why: Two applications.
\[ f' = 4 x ^{3} - 12 x ^{2}, f'' = 12 x ^{2} - 24 x \]
Factor and find the critical points
Why: Take out the common factors.
\[ 4 x ^{2}(x - 3) = 0: x = 0, 3 \]
Chart the first derivative's sign
Why: Test three intervals.
Classify
Why: Only one sign change.
\[ \min\text{ at } 3; x = 0\text{ is neither} \]
Chart the second derivative
Why: Factor: 12x(x-2).
\[ \text{inflections at } 0\text{ and } 2 \]
Figure (svg): The solution to Worked example a full analysis shown as a ladder of expressions, one row per legal move
\[ \text{min at } 3; \quad \text{inflections at } 0 \text{ and } 2 \]
Verify: check the unusual point at the origin
Why: At x equal to 0 the first derivative vanishes but does not change sign — it is negative on both sides, since 4x squared is non-negative and x minus 3 is negative there. So the function is decreasing through the origin with a momentary flat spot, which is a critical point that is not an extremum. It is also an inflection point, since the second derivative changes sign there. A single input can be both, and this analysis catches it only because both charts were made.
OpenStax Calculus Volume 1, §4.5 Derivatives and the Shape of a Graph §4.5, pp. 351-353
Ranking
Describing a graph's shape completely.
Put in order
Why: The two charts are made independently and combined only at the end, since their boundaries generally differ. Step e is where the analysis becomes a description rather than a list of points.
Worked example
Checkpoint 4.32. Four regions, four descriptions.
\[ \text{Describe the shape of } f(x)=x^{3}-3x \text{ on each of its four regions.} \]
List the boundaries
Why: From both charts.
\[ x = -1, 0, 1 \]
Describe the leftmost region
Why: Rising, concave down.
Describe the next
Why: Falling, concave down.
Describe the third
Why: Falling, concave up.
Describe the last
Why: Rising, concave up.
Figure (svg): The solution to Worked example describing each region shown as a ladder of expressions, one row per legal move
\[ \text{all four of } (\pm f', \pm f'') \text{ appear} \]
Verify: check the description against the graph
Why: The cubic climbs from the lower left, flattening to its maximum at negative 1; drops away steepening to the inflection at 0; continues down but easing to its minimum at 1; then climbs away steepening. All four combinations appear, in that order, which is typical of a cubic. Describing a graph this way is more informative than listing extrema, and it is what Section 4.6 will combine with end behaviour to sketch a curve completely.
OpenStax Calculus Volume 1, §4.5 Derivatives and the Shape of a Graph §4.5, pp. 353-353
Trap
\[ f(x)=x^{4}-4x^{3}: \; f'(x) = 4x^{2}(x-3) = 0 \text{ at } x = 0, 3 \]
Report two extrema
Why: The student classifies both critical points as turning points.
\[ \text{but } f' < 0 \text{ on BOTH sides of } 0 \]
The factor 4x squared is non-negative, so the sign is carried entirely by x minus 3 — which does not change sign at the origin.
\[ x = 3: \text{ a minimum}; \quad x = 0: \text{ critical but not an extremum} \]
Test the sign at every critical point separately
Why: A repeated factor often produces a critical point with no sign change.
An even-power factor in the derivative is the signal to watch for: it contributes a zero without a sign change, exactly as the cubing function's derivative did. Reading the factored form for the parity of each factor predicts the answer before any testing.
Fill the middle
A derivative whose factored form contains an even power.
Fill in the blanks
f'(x) = 4x^does not change(x-3): \; \text___ x=0 \text___ \; ___
Why: The squared factor is non-negative, so the sign is carried entirely by x minus 3 and does not flip at the origin. An even-power factor always produces a zero without a sign change.
Sorting
For the quartic minus four times the cubic, at each marked input.
Sort into buckets
Sort each input.
The origin is the interesting case: a critical point that is not an extremum, and simultaneously a point of inflection. Only making both charts reveals it, which is why a full analysis does both rather than stopping at the extrema.
Prediction
Commit before reasoning.
Predict first
Why is charting f' alone not enough to describe a graph's shape?
Correct: Because it gives direction but not bending, and four shapes need both.
\[ f' > 0 \text{ alone: two possible shapes; with } f'' \text{: one} \]
Why: Knowing a function rises on an interval leaves open whether it steepens or levels off, and those are visibly different curves. The second derivative supplies the missing half, and the two charts together partition the domain into regions where both are constant. Section 4.6 will add end behaviour and asymptotes to complete the picture, but these two charts do most of the work.
Comparison
Fill the blanks. Neither is superior; each has its situations.
Comparison matrix
| First derivative test | Second derivative test | |
|---|---|---|
| What it needs | the sign on both sides | one value of f'' at the point |
| Works at a corner | yes | no: f'' must exist there |
| Can be inconclusive | no, it always decides | yes, when f''(c) = 0 |
| Best when | the derivative's sign is easy to read | the second derivative is easy to evaluate |
The first derivative test always works and the second is often quicker. Using the second and falling back when it fails is the usual practical compromise.
Pattern
Given a function and asked to describe its shape.
Factoring in step one pays off twice: an even-power factor signals a zero with no sign change, so a critical point that is not an extremum can often be predicted before any testing.
Stewart, Calculus: Early Transcendentals 8e, §4.3 How Derivatives Affect the Shape of a Graph §4.3, pp. 293-303
Check
The first derivative test. Sign change, not zero.
Check your understanding
For f(x) = x^3 - 3x, what happens at x = -1?
Answer: A
Why: The derivative 3(x-1)(x+1) is positive to the left and negative to the right of -1.
Check
Inflection points. Sign change again.
Check your understanding
Does f(x) = x^4 have a point of inflection at the origin?
Answer: A
Why: The second derivative is non-negative on both sides, so the concavity never reverses.
Check
The second derivative test. Read the third case correctly.
Check your understanding
At a critical point, f''(c) = 0. What follows?
Answer: A
Why: x^4, -x^4 and x^3 all satisfy these conditions with three different behaviours.
Real world
An epidemiologist tracks the cumulative number of cases in an outbreak. Reports say the total is still rising, but the daily new-case count peaked last week and has been falling since.
Discussion prompt
Translate both statements into derivative conditions, say what the shape of the cumulative curve is now, and explain what the peak in daily cases corresponds to.
Hint: Cumulative cases is the function; daily new cases is its derivative.
Answer:
Let C be the cumulative case count. Daily new cases is C prime, the rate at which the total grows.
\[ C' > 0 \text{ (total still rising)}, \qquad C'' < 0 \text{ (daily cases falling)} \]
So the cumulative curve is rising but levelling off — one of the four shapes, and the one that means the outbreak is still growing while the growth eases.
The peak in daily cases is where C prime was largest, which is where C double prime changed from positive to negative — a point of inflection on the cumulative curve. It is not a maximum of anything epidemiologically interesting: the total keeps climbing.
That distinction was the source of enormous public confusion during real outbreaks. 'Cases have peaked' refers to the DERIVATIVE peaking, while the cumulative total continues to rise and will never fall. The inflection point is genuinely good news — it means the epidemic has turned — but reading it as a maximum of total cases is one rung wrong on the ladder, exactly the error Section 3.2's transfer example described for inflation.
\[ \text{inflection of } C \;=\; \text{maximum of } C' \;=\; \text{the epidemic's turning point} \]
Commit first
Answer, then rate your confidence honestly.
Predict first
A function has f'(c) = 0 and f''(c) = 0 at an interior point c. What can you conclude?
Correct: Nothing from those two facts; use the first derivative test.
\[ x^{4} \to \text{min}, \quad -x^{4} \to \text{max}, \quad x^{3} \to \text{neither} \]
Why: The quartic, its negative and the cubing function all have exactly these two conditions at the origin, and they have a minimum, a maximum and neither respectively. So the pair of values genuinely carries no information. The first derivative test separates all three by looking at the sign on either side, which is why an inconclusive second derivative test is an instruction rather than a conclusion.
Explain it
They report an inflection point wherever the second derivative vanishes.
Discussion prompt
In four sentences or fewer, show them the counterexample and the fix.
Hint: Use the quartic.
Answer:
Ask them about the fourth power at the origin: its second derivative is twelve x squared, which is zero there. But twelve x squared is positive on both sides, so the curve is bowl-shaped throughout and nothing reverses — there is no inflection.
The fix is the same one they already use for extrema: a vanishing derivative is a candidate and only a SIGN CHANGE confirms it. The two tests have identical structure, one derivative apart, so remembering one gives them the other.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the two questions, remember the first derivative gives direction and the second gives bending, and all four combinations occur. For inflections, test the sign on both sides exactly as for extrema. For the failing test, remember the three functions with identical conditions. For combining, list all boundaries from both charts and describe each region by two signs. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, draw the cubic x cubed minus 3x with its two turning points and its inflection marked, and beneath it draw two sign charts on a shared axis — one for the first derivative with boundaries at plus and minus 1, and one for the second with a boundary at 0. Label each region of the resulting partition with its two signs and a phrase describing the shape. Beside that, draw four small curves illustrating the four sign combinations. In the middle of the page, write both tests as rules, with the second derivative test's three cases and the first's two. Below, draw the three functions x to the fourth, its negative, and x cubed near the origin, and write beneath each what happens there and why the second derivative test cannot tell them apart. At the bottom, take x to the fourth minus 4x cubed, factor both derivatives, and mark on a number line which inputs are extrema, which are inflections, and which are both. In a margin, write the one sentence distinguishing a candidate from a confirmation.
If your two sign charts have the same boundaries, check them — for a cubic the first derivative's zeros and the second's are always different, and it is that mismatch that produces four regions rather than two.
Recap
Five things, and together they read a graph's shape completely from two derivatives.
| If you see | Then |
|---|---|
| f' changing + to - | A local maximum |
| f' vanishing without changing sign | A critical point, not an extremum |
| f'' > 0 | Concave up: above its tangents |
| f'' changing sign | A point of inflection |
| f'' = 0 without a sign change | No inflection |
| f'(c)=0 and f''(c) > 0 | A local minimum |
| f'(c)=0 and f''(c)=0 | Inconclusive: use the first derivative test |
Section 4.6 adds the last piece of a curve sketch: what happens as the input grows without bound. Limits at infinity give horizontal and oblique asymptotes, and with them a graph can be drawn from the algebra alone.
OpenStax Calculus Volume 1, §4.5 Derivatives and the Shape of a Graph §4.5, pp. 339-353 — everything on these slides traces back here
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