01 · Distinguish the features
For this cubic, is its y-intercept a stationary point?
Hint
Evaluate f′ at x = 0.
Worked solution
No. The y-intercept (0, 0) is an inflection with gradient −3. The stationary points occur at x = ±1.
Understand · explore · practise
Differentiation curve problems
Combine differentiation rules to analyse curves, classify turning points, find increasing intervals and inflections, sketch graphs and prove differential identities. Includes original mixed practice with worked solutions.
Before you startDifferentiation rules; second derivatives; inflections; transformations
01 / Turn calculations into a coherent sketch
Each feature should come from a checked statement.
Domain → intercepts → first derivative → second derivative → sketch
For a curve problem, keep the original function, its gradient and its change of gradient separate. Solve the relevant equations, check the domain and use sign information to connect the features.
The interactive graph is a worked example. Hide its features and predict them first; then reveal the results you want to check.
Build a complete curve sketchExplore
x = −0.5; f = 1.375; f′ = −2.25; f″ = −3.
The function is decreasing and bending downwards here.
Choose which calculated features to reveal.
Try finding the stationary points, inflection and intercepts on paper before revealing them. The original curve and your selected tangent stay visible.
02 / Analyse a polynomial completely
Use exact features before drawing the final shape.
f(x) = x³ − 3xWorked example
Domain: all real x. Roots: x(x² − 3) = 0
The intercepts are x = −√3, 0, √3.
f′ = 3(x − 1)(x + 1)
The function increases for x < −1 and x > 1; it decreases for −1 < x < 1.
Stationary points: (−1, 2), a maximum; (1, −2), a minimum
The first derivative changes + to − and − to + respectively.
f″ = 6x; inflection at (0, 0)
Concavity changes from down to up, and the inflection gradient is −3.
As x → −∞, f → −∞; as x → ∞, f → ∞
Combine the end behaviour with the intercepts and turning points.
Build the cubic sketch from its key features
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Analyse a product with an exponential
A positive factor can simplify a sign argument.
f(x) = x e⁻ˣWorked example
f′ = e⁻ˣ(1 − x)
The exponential is always positive.
f′ > 0 for x < 1; f′ < 0 for x > 1
The unique stationary point is a maximum at (1, e⁻¹).
f″ = e⁻ˣ(x − 2)
The inflection is (2, 2e⁻²), with gradient −e⁻².
The curve passes through (0, 0) and approaches y = 0 from above as x → ∞
The horizontal asymptote is an end-behaviour statement, not a stationary point.
02 · A changed decay constant
Find and classify the stationary point of y = x e^(−2x).
Hint
y′ = e^(−2x)(1 − 2x).
Worked solution
x = 1/2 gives a maximum at (1/2, 1/(2e)). The derivative is positive before 1/2 and negative after it.
03 · Find its inflection
Find the inflection of y = x e^(−2x).
Hint
Differentiate again to get y″ = 4e^(−2x)(x − 1).
Worked solution
The second derivative changes sign at x = 1. The inflection is (1, e⁻²), with gradient −e⁻², so it is non-stationary.
04 / Keep denominator exclusions in a sign chart
A cancelled expression can still have an excluded input.
f(x) = x + 4/x, with x ≠ 0Worked example
f′ = 1 − 4/x² = (x² − 4)/x²
The denominator is positive on the domain.
Stationary inputs: x = ±2
The points are (−2, −4) and (2, 4).
f′ is positive for |x| > 2 and negative for 0 < |x| < 2
There is a local maximum at (−2, −4) and a local minimum at (2, 4).
f″ = 8/x³
The two domain intervals have opposite concavity, but there is no inflection at the missing input zero.
04 · Increasing intervals
State the increasing intervals of f(x) = x + 9/x, x ≠ 0.
Hint
f′ = 1 − 9/x².
Worked solution
The function is increasing on (−∞, −3) and (3, ∞). It decreases on (−3, 0) and (0, 3). The domain gap must remain separate.
05 · Classify both turning points
Find and classify the stationary points of x + 9/x.
Hint
The stationary inputs are x = ±3.
Worked solution
(−3, −6) is a local maximum; (3, 6) is a local minimum. Their nature follows from the first-derivative sign changes or f″ = 18/x³.
05 / Analyse a logarithmic quotient
State the positive domain before using logarithms.
f(x) = (ln x)/x, x > 0Worked example
f′ = (1 − ln x)/x²
Use the quotient rule or a product with x⁻¹.
f′ = 0 at x = e
The curve increases before e and decreases after e.
The maximum point is (e, 1/e)
Substitute into f.
f″ = (2 ln x − 3)/x³
The inflection occurs at x = e^(3/2), with y = 3/[2e^(3/2)].
06 · A scaled log quotient
Find the maximum of f(x) = (ln x)/x², for x > 0.
Hint
f′ = (1 − 2 ln x)/x³.
Worked solution
The derivative changes from positive to negative at x = √e. The maximum value is 1/(2e), so the maximum point is (√e, 1/(2e)).
07 · Verify its inflection
Find the inflection of (ln x)/x².
Hint
f″ = (6 ln x − 5)/x⁴.
Worked solution
The inflection is at x = e^(5/6), y = 5/[6e^(5/3)]. The second derivative changes from negative to positive there.
06 / Solve stationary equations over the full stated interval
Include endpoints separately from interior turning points.
f(x) = sin x + cos x on 0 ≤ x ≤ 2πWorked example
f′ = cos x − sin x = 0
This gives x = π/4 and 5π/4 in the interval.
f″ = −sin x − cos x
At π/4 it is negative; at 5π/4 it is positive.
Maximum point (π/4, √2); minimum point (5π/4, −√2)
These are the interior stationary extrema.
The endpoint values are f(0) = f(2π) = 1
Compare them too if asked for absolute extrema on the closed interval.
08 · Complete a trig interval
Find and classify all interior stationary points of y = sin(2x) on 0 ≤ x ≤ π.
Hint
y′ = 2 cos(2x).
Worked solution
x = π/4 gives y = 1, a maximum. x = 3π/4 gives y = −1, a minimum. The endpoint values are both zero.
07 / Map turning points under a transformation
Vertical reflection swaps maxima and minima.
f has a maximum at (2, 5). Let g(x) = −3f(x − 1) + 4.Worked example
x − 1 = 2 gives x = 3
The input shift moves the point right by 1.
g(3) = −3·5 + 4 = −11
Transform the output.
The maximum becomes a minimum at (3, −11)
The negative vertical scale reflects the graph.
g′(x) = −3f′(x − 1)
This derivative relation confirms the stationary input.
09 · A horizontal reflection
If f has a local minimum at (−2, 3), identify the corresponding point of h(x) = 2f(−x) − 1 and its nature.
Hint
Solve −x = −2, then transform the output.
Worked solution
The corresponding point is (2, 5), still a local minimum. A horizontal reflection preserves its nature, and the vertical scale is positive.
10 · Transform an inflection
If f has an inflection at (1, 4), where is the corresponding inflection of g(x) = f(2x + 3) − 5?
Hint
Solve 2x + 3 = 1.
Worked solution
It is at (−1, −1). The affine transformations preserve a genuine change of concavity here.
08 / Prove a differential identity by substitution
Differentiate first, then simplify the requested combination.
Show that y = eˣ sin x satisfies y″ − 2y′ + 2y = 0Worked example
y′ = eˣ(sin x + cos x)
Use the product rule.
y″ = 2eˣ cos x
Differentiate and cancel the sine terms.
y″ − 2y′ + 2y = 2eˣcos x − 2eˣ(sin x + cos x) + 2eˣsin x
Substitute all three expressions.
The expression simplifies to zero for every real x
The identity is established.
11 · A polynomial identity
Show that y = x³ + 2x² − 4 satisfies xy″ − 2y′ = −4x.
Hint
y′ = 3x² + 4x and y″ = 6x + 4.
Worked solution
xy″ − 2y′ = x(6x + 4) − 2(3x² + 4x) = −4x.
12 · A trig identity
Show that y = 3 sin(2x) − cos(2x) satisfies y″ + 4y = 0.
Hint
Differentiate twice using the inner-angle factor 2.
Worked solution
y″ = −12 sin(2x) + 4 cos(2x) = −4y, so y″ + 4y = 0.
09 / Prove that a stationary point exists in an interval
A sign change needs continuity; uniqueness needs more.
f(x) = x³/3 − x − e⁻ˣWorked example
f′ = x² − 1 + e⁻ˣ
The derivative is continuous.
f′(1/2) = −3/4 + e^(−1/2) < 0; f′(1) = e⁻¹ > 0
For example e^(1/2) > 1 + 1/2 gives e^(−1/2) < 2/3 < 3/4.
There is at least one root of f′ in (1/2, 1)
Continuity gives the sign-bracket conclusion.
f″ = 2x − e⁻ˣ > 0 on [1/2, 1]
Here 2x ≥ 1 and e⁻ˣ < 1, so f′ is strictly increasing.
There is exactly one stationary point in this interval, and it is a minimum
The derivative crosses from negative to positive.
13 · What a bracket proves
If a continuous derivative is negative at a and positive at b, what can you conclude without any other information?
Hint
Continuity guarantees a zero, but not its uniqueness.
Worked solution
There is at least one stationary input in (a, b). The endpoint signs alone do not identify its exact position, prove uniqueness or classify every stationary point inside.
10 / Use a given stationary condition to find a parameter
Substitute into the derivative, then verify the point.
y = x²e^(−kx), k > 0, has a stationary point at x = 4Worked example
y′ = e^(−kx)(2x − kx²)
Apply the product rule.
At x = 4: 8 − 16k = 0, so k = 1/2
The exponential factor is nonzero.
The point is (4, 16e⁻²)
Substitute into the original function.
Near x = 4, the derivative changes positive to negative
This stationary point is a local maximum; the function also has a stationary point at x = 0.
14 · Retain the other stationary point
For y = x²e^(−x/2), classify the stationary point at x = 0.
Hint
y′ = e^(−x/2)x(2 − x/2).
Worked solution
The derivative is negative immediately before zero and positive immediately after. Thus (0, 0) is a local minimum. Solving only 2 − x/2 = 0 would miss it.
11 / Match each conclusion to the right evidence
A clear sign argument is part of the answer.
- Keep domain restrictions beside the derivative.
- Use f′ = 0 for stationary inputs and signs or f″ to classify them.
- Use a change of concavity for inflections.
- Include endpoints when finding absolute extrema on a closed interval.
- Transform both coordinates and reconsider the point’s nature.
- For identities, substitute complete derivative expressions.
- A sign bracket gives existence only when the relevant function is continuous.
15 · Closed-interval extrema
Find the absolute maximum and minimum of f(x) = x³ − 3x on −2 ≤ x ≤ 3.
Hint
Compare the stationary values with both endpoint values.
Worked solution
f(−2) = −2, f(−1) = 2, f(1) = −2 and f(3) = 18. The absolute maximum is 18 at x = 3. The absolute minimum is −2, attained at x = −2 and x = 1.
16 · An inflection is not a minimum
For y = x e⁻ˣ, explain why its inflection at x = 2 is not stationary.
Hint
Evaluate y′, not only y″.
Worked solution
y′(2) = −e⁻² ≠ 0. The point is a non-stationary inflection on the decreasing part of the curve.
Section 1 of 11 · Turn calculations into a coherent sketch