Loav · VCE Specialist Mathematics · Units 3 & 4
S34-D02 · Recommended sequence: Unit 3
1. The derivative has a rate of change too
The first derivative f′ tells you how the function's value changes as its input increases. The second derivative f″ tells you how that first derivative changes. These answer different questions: a graph can be increasing while its gradient is decreasing. For example, ln x rises for positive x, but becomes progressively flatter. Its first derivative is positive and its second derivative is negative.
Equivalent notation includes y″, f″(x) and d²y/dx². The last expression means differentiating twice with respect to x. It is not the square of dy/dx, and it is not obtained by separately squaring a numerator and denominator. If y is measured in metres and x in seconds, y′ has units m/s and y″ has units m/s².
To calculate f″, first find a correct first derivative, simplify if useful, and differentiate again. Use product, quotient and chain rules afresh. A simpler first-derivative form often reduces errors in the second calculation. Check that both expressions remain defined on the interval being analysed. Domain exclusions from the original function remain in force even when a derivative appears to simplify.
2. Concavity describes changing slopes
On an interval where f″ > 0, f′ is increasing and the graph is concave up. On an interval where f″ < 0, f′ is decreasing and the graph is concave down. Imagine moving along the graph from left to right: increasing slope can mean negative slopes becoming less negative, passing through zero and becoming positive. Therefore concave up does not mean the graph is increasing everywhere.
A useful local picture is the relation to a tangent. For a suitably smooth concave-up graph, nearby points lie above its tangent; for a concave-down graph, they lie below. This picture explains why tangent-based numerical approximations can systematically underestimate or overestimate a curve over a short step. It also helps you distinguish bending from vertical position.
The four sign combinations are all possible. Positive f′ and positive f″ mean rising with increasingly steep slope. Positive f′ and negative f″ mean rising while flattening. Negative f′ and positive f″ mean falling while flattening. Negative f′ and negative f″ mean falling with increasingly negative slope. Make a sketch of each before trying to memorise a verbal rule.
Concavity is an interval property. A single numerical value of f″ supplies local information when the derivative behaves continuously nearby, but does not establish the shape over a whole requested domain. For polynomial and simple rational examples, factor f″ and analyse signs on intervals separated by its zeros and undefined inputs.
Worked example 1: A nonstationary inflection between two extrema
Let f(x) = x³ − 3x. Then f′(x) = 3x² − 3 = 3(x − 1)(x + 1), and f″(x) = 6x. The first derivative is positive on (−∞, −1), negative on (−1, 1) and positive on (1, ∞). Hence (−1, 2) is a local maximum and (1, −2) is a local minimum.
The second derivative is negative for x < 0 and positive for x > 0, so concavity changes at (0, 0). It is an inflection point. Its gradient is f′(0) = −3, so it is nonstationary, with tangent y = −3x. An inflection does not have to be flat.
The graph crosses the horizontal axis at −√3, 0 and √3 and is odd. Combining the intercepts, stationary points, concavity and end behaviour determines a coherent sketch. The first-derivative sign chart and second-derivative sign chart describe different features and should not be substituted for each other.
3. Classify a stationary point with justified evidence
A stationary point has f′(a) = 0 at an input a in the domain. If f′ changes from positive to negative there, the point is a local maximum. If it changes from negative to positive, it is a local minimum. If the derivative retains its sign, it is not a turning point. Use the function value f(a) to provide the full coordinates.
The second-derivative test can be a shorter classification when its conditions apply. At a stationary point, f″(a) > 0 implies a local minimum and f″(a) < 0 implies a local maximum, for the smooth functions considered here. If f″(a) = 0, the test is inconclusive. It does not prove that the point is an inflection, nor that no extremum exists.
Compare three functions at zero. For x⁴, both first and second derivatives vanish, yet zero is a minimum: nearby values are nonnegative and f′ changes from negative to positive. For −x⁴, zero is a maximum. For x³, zero is a stationary inflection: f′ = 3x² stays nonnegative while f″ = 6x changes sign. The same pair of derivative values at the point can lead to three different classifications.
A local extremum is relative to nearby allowed inputs; a global extremum is relative to the entire stated domain. On a closed interval, endpoints can provide the greatest or least value without being stationary. For a maximum/minimum question on a restricted interval, compare endpoint values with all valid interior candidates. Do not accidentally solve a local classification problem when the question asks for an absolute bound.
4. Inflection requires a genuine change of concavity
For a smooth function, zeros of f″ are candidates for inflection. Test signs on both sides. If f″ merely touches zero without changing sign, there is no inflection. The function x⁴ has f″ = 12x², positive on both sides of zero, so the concavity does not change. The equation f″ = 0 identifies where to investigate, not the final answer.
Also consider points where the second derivative is undefined but the function itself exists. Such points require direct local concavity reasoning; a rule that only solves f″ = 0 is not a universal definition. The standard polynomial and rational problems here can usually be decided with sign charts, but the conceptual requirement is always a change in concavity at a point on the graph.
A vertical asymptote is not an inflection point. For f(x) = 1/x, f″ = 2/x³ has opposite signs across zero, but f(0) does not exist. A removable hole is also not a point of inflection in the original function. Always intersect candidate inputs with the original domain before giving coordinates or classifications.
Worked example 2: Second derivatives of an implicit curve
For x² + 4y² = 16, find y″ at (0, 2) and interpret it. First differentiate to obtain 2x + 8yy′ = 0, hence y′ = −x/(4y) where y ≠ 0. Differentiate the unsolved first relation again: 2 + 8[(y′)² + yy″] = 0.
Therefore y″ = −[1 + 4(y′)²]/(4y). At (0, 2), y′ = 0, so y″ = −1/8. The upper branch is concave down there, and the horizontal tangent at that point is a local maximum of the upper branch.
Alternatively substitute y′ = −x/(4y) to get y″ = −(4y² + x²)/(16y³) = −1/y³, using the original ellipse equation. This is not obtained by simply differentiating x while holding y constant. Throughout implicit differentiation, y changes with x on the selected local branch.
5. A complete curve investigation follows a sequence
Start with the function's domain, intercepts, symmetry and asymptotes. These establish the graph's framework. Next use f′ for stationary points and increasing/decreasing intervals. Then use f″ for concavity and inflections. Finally combine all the evidence in one sketch. When the graph is restricted to an interval, mark whether each endpoint is included.
Keep three types of input in your sign charts: derivative zeros, derivative singularities and original domain exclusions. They divide the line into intervals, but do not all represent points on the graph. Use one exact test value in each interval when factor signs are not immediate. If a factor has an even power, its sign usually does not change through its zero; an odd power usually changes sign.
Coordinates should be exact when possible. An inflection at x = √3 belongs at height f(√3), not at a rounded height read from a small plotting window. A smooth drawing should respect the gradient direction at each turning point and the correct side of each asymptote. A well-labelled graph is a summary of reasoning, not a replacement for it.
Worked example 3: Three inflections in a simple quotient
Let g(x) = x/(x² + 1). Its domain is all real numbers. The first derivative is (1 − x²)/(x² + 1)² and the second derivative is 2x(x² − 3)/(x² + 1)³. Since the denominator of g″ is always positive, only x(x² − 3) determines its sign.
The sign is negative on (−∞, −√3), positive on (−√3, 0), negative on (0, √3), and positive on (√3, ∞). Therefore all three candidates are inflections: (−√3, −√3/4), (0, 0) and (√3, √3/4). None is stationary, because g′ is −1/8 at the outer two and 1 at zero.
The stationary points instead occur at x = ±1. The graph has minimum (−1, −1/2) and maximum (1, 1/2). This example shows why counting stationary points does not tell you how many inflections to expect, and why a broad plot may hide changes in bending near a horizontal asymptote.
6. Use technology to challenge a claim
Ask a CAS to simplify your derivative and compare it algebraically with a second independently obtained form. Then graph f, f′ and f″ in separate aligned windows. The zero crossings of f″ should correspond to changes in the direction of f′, not necessarily to zeros of f′. A numerical plot can suggest the sign intervals but should not replace exact reasoning when the factors are available.
As a non-routine investigation, consider fa(x) = x⁴ − ax². Its stationary equation is 2x(2x² − a) = 0. When a > 0, there are three stationary points; when a ≤ 0, only zero remains. At a = 0 the second-derivative test fails, yet zero is still a minimum. Varying a on a graph helps locate the structural change; factoring and sign reasoning explain why it happens.
Common errors include equating concavity with increasing behaviour, using f″ = 0 as a complete inflection test, losing an implicit chain-rule factor and calling an excluded input a critical point on the curve. Repair each by naming the property being tested and writing the evidence that actually establishes it.
Practise, then check
- Find both derivatives of f(x) = x²ex.
- Classify the stationary point of y = x⁴ at zero, and decide whether it is an inflection.
- Find and classify the stationary point of y = x³ + 2.
- For y = ln x, state its domain, increasing/decreasing behaviour and concavity.
- Find the inflection point and its tangent for y = x³ + x.
- Find the absolute minimum and maximum of x³ − 3x on [0, 2].
- For fa(x) = x⁴ − ax² with a > 0, classify all stationary points.
Fully explained answers
- Product differentiation gives f′ = ex(x² + 2x). Differentiate that product again to obtain f″ = ex(x² + 4x + 2). The derivative of the polynomial factor must be added to the original factor; the exponential alone is not the entire derivative.
- The first derivative 4x³ changes from negative to positive, so (0, 0) is a minimum. The second derivative 12x² is positive on both sides, so there is no inflection. Its value zero at the point makes the second-derivative test inconclusive, not contradictory.
- The only stationary point is (0, 2), since y′ = 3x². The derivative does not change sign, so it is not an extremum. Since y″ = 6x changes from negative to positive, the point is a stationary inflection.
- The domain is x > 0. Because y′ = 1/x > 0, the function increases throughout its domain. Since y″ = −1/x² < 0, it is concave down throughout. Rising and bending downward are compatible properties.
- Here y″ = 6x changes sign at zero, so the inflection is (0, 0). Its slope is y′(0) = 1, giving tangent y = x. The point is nonstationary because its first derivative is not zero.
- Check the allowed stationary input x = 1 and both endpoints. Values are f(0) = 0, f(1) = −2 and f(2) = 2. Thus the absolute minimum is −2 at x = 1 and maximum is 2 at x = 2. The endpoint maximum is not stationary.
- The stationary inputs are 0 and ±√(a/2). Since f″ = 12x² − 2a, at zero it is −2a < 0, giving a maximum (0, 0). At the other two inputs it is 4a > 0, giving minima (±√(a/2), −a²/4). The parameter condition is essential to the existence of the outer points.
Curriculum and sources
VCE Specialist Mathematics Units 3 and 4, Areas of Study 2 and 4 as applicable, with Outcomes 1–3. Scope follows the VCAA Mathematics Study Design 2023–2027 and the current Specialist Mathematics FAQ. Curriculum checked 15 October 2026. Teaching reference: Cambridge Senior Mathematics VCE Specialist Mathematics 3&4, second edition (2023), sections 8D–8E (pp331–345);8G (pp354–362). All teaching prose, examples, practice questions and diagrams are original. Independent teaching material; not a VCAA publication.