01 · Select the target
Which equation should you solve to locate stationary points of F(x) = eˣ + x³/3 − 2x?
Hint
Differentiate once.
Worked solution
Solve F′(x) = eˣ + x² − 2 = 0. Its derivative for Newton is F″(x) = eˣ + 2x.
Understand · explore · practise
Finding stationary and inflection points numerically
Use numerical methods on first and second derivatives to find stationary and inflection points. Choose the correct Newton denominator, restore original coordinates, verify accuracy and classify with sign changes.
Before you startDifferentiation; stationary and inflection points; Newton–Raphson
01 / Solve an equation for the feature you want
A root of a derivative describes the original curve.
Stationary input: F′(x) = 0. Inflection candidate: F″(x) = 0, where these derivatives exist.
Numerical root methods can be applied to a derivative just as they can to an original function. Name the target G clearly: G = F′ for stationary inputs, or G = F″ for smooth inflection candidates.
After finding a candidate input α, calculate its ordinate from the original F and classify the point. A zero second derivative alone is not enough to establish an inflection.
Choose the right function to solveExplore
Stationary target: G(x) = F′(x) = exp(x) + x² − 2.
Start at x₀ = −1.5.
Restore the original ordinate with F(xₙ).
Newton uses G/G′, so this stationary-point calculation divides F′ by F″.
The approximate input must be classified before it is called a maximum, minimum or inflection point.
The two graph views share their scales. Values are computed at full stored precision. A displayed residual of zero at a late step reflects machine precision; the written sign checks justify the feature.
02 / Differentiate the target in Newton’s formula
The denominator changes when the equation changes.
For stationary points, set G(x) = F′(x).Worked example
G′(x) = F″(x)
Differentiate the equation actually being solved.
xₙ₊₁ = xₙ − F′(xₙ)/F″(xₙ)
This is Newton’s method for F′ = 0.
For an inflection candidate, instead set G = F″
Then G′ = F‴.
The inflection-candidate update is xₙ₊₁ = xₙ − F″(xₙ)/F‴(xₙ)
Where this quotient is defined; classification still follows separately.
02 · Select the inflection denominator
For the same F, write F‴(x) and the Newton update for an inflection candidate.
Hint
Differentiate F″, not F again from scratch.
Worked solution
F‴(x) = eˣ + 2. Thus xₙ₊₁ = xₙ − [exp(xₙ) + 2xₙ]/[exp(xₙ) + 2].
03 / Find the left stationary input
Bracket a derivative root before refining it.
Use F(x) = eˣ + x³/3 − 2x.Worked example
F′(−1.5) > 0 and F′(−1) < 0
There is a stationary input between −1.5 and −1.
Start x₀ = −1.5 and apply Newton to F′
Use denominator F″ = eˣ + 2x.
x₁ ≈ −1.329617451507
First substitution.
x₂ ≈ −1.316061843802; x₃ ≈ −1.315973781517
The input approaches about −1.31597378.
Calculate F at the refined input
The ordinate is approximately 2.14049849.
Move from a derivative root back to its original curve point
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 · Restore the coordinate
If a root of F′ is α ≈ −1.31597378, is the stationary point (α, 0)?
Hint
The zero belongs to F′, not to F.
Worked solution
No. The stationary point is (α, F(α)), approximately (−1.3160, 2.1405). The point (α, 0) lies on the derivative graph.
04 / Use another bracket for the other stationary input
One converged input does not find every stationary point.
For the same function, inspect F′ on [0, 1].Worked example
F′(0) = −1 and F′(1) = e − 1 > 0
There is another stationary input between 0 and 1.
Start x₀ = 0.5
Use the same stationary-target Newton formula.
x₁ ≈ 0.538236839195
First substitution.
x₂ ≈ 0.537275065501; x₃ ≈ 0.537274449174
The input is approximately 0.53727445.
F(x) is approximately 0.68848450 there
The original curve point is approximately (0.5373, 0.6885).
There are no further stationary inputs: F‴ = eˣ + 2 is positive, so F″ is strictly increasing from a negative to a positive value and has one zero. Therefore F′ first decreases and then increases. Since F′ tends to +∞ at both ends and its minimum is negative, its two bracketed roots are the only ones.
04 · Root search completeness
A Newton calculation has found the positive stationary input. What supports searching for a second one on the negative side?
Hint
Use the derivative signs already found.
Worked solution
F′(−1.5) > 0 and F′(−1) < 0 bracket a distinct negative root of F′. A single successful Newton run cannot rule it out.
05 / Classify the stationary points
Use the original function’s derivative information.
Classify the two stationary points of F.Worked example
At the negative input, F″ ≈ −2.36373454 < 0
The stationary point is a local maximum.
At the positive input, F″ ≈ 2.78588506 > 0
The stationary point is a local minimum.
Alternatively, inspect the sign of F′ around each input
Positive to negative means a local maximum; negative to positive means a local minimum.
If F″ is zero, this simple second-derivative classification is inconclusive
Use a sign test or further analysis.
05 · An inconclusive second derivative
For H(x) = x⁴, H′(0) = H″(0) = 0. Is the point unclassifiable?
Hint
Check H′ on either side.
Worked solution
No. H′(x) = 4x³ changes from negative to positive, so (0, 0) is a local minimum. Only the simple second-derivative test was inconclusive.
06 · A stationary point without a turn
For H(x) = x³, explain why solving H′ = 0 gives neither a maximum nor a minimum.
Hint
H′(x) = 3x² is positive on both sides of zero.
Worked solution
The curve increases on both sides, so it does not turn. Since H″(x) = 6x changes sign, (0, 0) is a stationary inflection.
06 / Solve the second derivative equation
Locate a candidate, then check the change of concavity.
For F, seek a root of G(x) = F″(x) = eˣ + 2x.Worked example
G(−0.5) < 0 and G(0) = 1 > 0
There is a candidate between −0.5 and 0.
G′(x) = eˣ + 2 > 0
The target is strictly increasing, so this root is unique.
Start x₀ = −0.5 and use xₙ₊₁ = xₙ − G(xₙ)/G′(xₙ)
The denominator is the third derivative of F.
x₁ ≈ −0.349044806428; x₂ ≈ −0.351732769537
Continue with stored precision.
x₃ ≈ −0.351733711249
This gives the inflection candidate input.
07 · Wrong target
A student uses xₙ₊₁ = xₙ − F(xₙ)/F′(xₙ) to find an inflection point. What equation is that iteration actually solving?
Hint
Look at the numerator.
Worked solution
It is Newton’s method for F(x) = 0, seeking a horizontal-axis intercept. For a smooth inflection candidate it should target F″ = 0 instead.
07 / Verify that the candidate is an inflection
The second derivative must change sign across it.
Let β ≈ −0.351733711249 be the root of F″.Worked example
F″ is strictly increasing and crosses zero at β
It is negative immediately to the left and positive immediately to the right.
The original curve changes from concave down to concave up
This proves an inflection, rather than just a zero second derivative.
F(β) ≈ 1.392429744951
Restore the original ordinate.
F′(β) ≈ −1.17281597, which is not zero
The inflection is nonstationary.
The point is approximately (−0.3517, 1.3924)
State coordinates on the original curve.
08 · A false inflection candidate
For H(x) = x⁴, H″(0) = 0. Why is zero not an inflection input?
Hint
H″(x) = 12x².
Worked solution
The second derivative is positive on both sides of zero, so concavity does not change. The zero alone is insufficient.
08 / Use the original ordinate and original slope
Do not confuse the target graph with the curve being studied.
At the nonstationary inflection of F.Worked example
β ≈ −0.351733711249
This is the root of the second derivative target.
The original ordinate is F(β) ≈ 1.392429744951
It is not F″(β), which is approximately zero.
The tangent gradient is F′(β) ≈ −1.172815973872
It is not the derivative of the root target.
A tangent equation is y − F(β) = F′(β)(x − β)
Use unrounded values until the requested final accuracy.
09 · A stationary tangent
What is the tangent equation at the local minimum of F, using its ordinate rounded to 4 d.p.?
Hint
A stationary tangent is horizontal.
Worked solution
y ≈ 0.6885. The exact equation is y = F(α), where α is the positive root of F′.
10 · Which graph has ordinate zero?
At β, which of F, F′ and F″ has value zero?
Hint
β was found from the second-derivative equation.
Worked solution
F″(β) = 0. Here F(β) ≈ 1.39243 and F′(β) ≈ −1.17282 are both nonzero.
09 / Bracket the derivative target for input accuracy
The sign check must use the equation that defined the input.
Verify the left stationary input to 3 d.p.Worked example
The proposed input is −1.316
Its halfway boundaries are −1.3165 and −1.3155.
F′(−1.3165) > 0 and F′(−1.3155) < 0
Use the first derivative target, not F.
F′ is continuous and strictly decreasing throughout that bracket
There is one root between these bounds.
The stationary input therefore rounds to −1.316
Calculate the ordinate using a sufficiently refined input, rather than immediately discarding its extra digits.
11 · Inflection accuracy
Which target and halfway boundaries would verify β = −0.352 to 3 d.p.?
Hint
Use F″ and half a thousandth.
Worked solution
Test F″ at −0.3525 and −0.3515. Its values are negative and positive respectively, and the continuous strictly increasing target has a unique root inside.
12 · Rounding the coordinate
Does proving the x-coordinate to 3 d.p. automatically prove that substituting its rounded value gives the y-coordinate to 3 d.p.?
Hint
The original function may amplify an input error.
Worked solution
No. Retain more working precision and verify the requested ordinate accuracy separately if needed. A steep original function can magnify the effect of input rounding.
10 / Include endpoints when finding an absolute extreme
Stationary points are only part of an interval comparison.
Find the absolute extrema of F on [−2, 1].Worked example
Both stationary inputs lie in the interval
Use the two candidates already found.
F(−2) = e⁻² + 4/3 ≈ 1.468668617
Evaluate the left endpoint.
F(1) = e − 5/3 ≈ 1.051615162
Evaluate the right endpoint.
The stationary values are about 2.140498485 and 0.688484502
Compare all four values.
The absolute maximum is at the negative stationary point, and the absolute minimum at the positive one
This conclusion depends on the specified interval.
13 · A restricted interval
On [0, 0.4], where is the maximum of F?
Hint
The positive stationary input is beyond 0.4; F′ is negative throughout this shorter interval.
Worked solution
At x = 0, with F(0) = 1. The curve decreases over [0, 0.4], so the left endpoint is the maximum.
14 · Do not omit endpoints
Must an absolute maximum on a closed interval be stationary?
Hint
Consider a strictly increasing function on [a, b].
Worked solution
No. It can occur at an endpoint. Include allowed endpoints and any other relevant nonsmooth candidates as well as stationary points.
11 / Keep the original curve and target equation separate
Solve, restore coordinates, classify and interpret.
Different rearrangements, inverse-trig branches and starting values may find only some derivative roots. Use the original domain, disjoint brackets and suitable monotonicity arguments to support the completeness of a search. A smooth inflection also need not be stationary, and a stationary derivative root need not give a sign change in F′.
- Choose G = F′ for stationary inputs, or G = F″ for smooth inflection candidates.
- Newton divides G by G′.
- Restore y from F, not G.
- Classify stationary points and verify concavity changes.
- Bracket the correct target for input accuracy.
- Check branches, the requested interval and endpoints.
15 · An even derivative root
Can sign-change bracketing of H′ always locate the stationary point of H(x) = x³?
Hint
H′ = 3x² has the same sign on both sides of zero.
Worked solution
No. The stationary derivative root at zero has even multiplicity and does not change the sign of H′. Equal endpoint signs do not rule out a derivative root.
16 · The complete method
Summarise what is still required after a numerical method returns a root of F″.
Hint
An input alone is not a classified point.
Worked solution
Check the candidate lies in the domain, verify that F″ changes sign, calculate the original ordinate F(x), determine whether F′ is zero if stationary status matters, and justify any requested accuracy.
Section 1 of 11 · Solve an equation for the feature you want