Extrema, curvature and curve analysis

The same height can occur at a turn and at an endpoint

For f(x)=x³−3x, the exact readings at x=−2,−1,0,1,2 are −2,2,0,−2,2. The curve turns near −1 and 1. A table suggests those turns, but five readings alone do not prove what happens between them or outside the window.

Its tangent slope will be f′(x)=3x²−3=3(x−1)(x+1). This is positive outside [−1,1] and negative inside (−1,1). We will prove why those signs force rising, then falling, then rising. The exact turns are (−1,2) and (1,−2).

On the whole real line these are local extrema, not global ones: x³−3x grows without bound in both signed directions. On the restricted domain [−2,2], the largest value is 2 at both −1 and 2, and the smallest is −2 at both −2 and 1. The domain changes the question; the formula does not.

Start with the domain and a slope

The published preparation links practise function domains, quadratic factorisation and inequality signs. All derivative and bending notation needed here is defined locally. First follow the concrete curves and sign tables; return to the completeness proof after seeing why the sign tests need it.

Write f:D→ℝ for a function with allowed real inputs D and finite real outputs. Continuity at c∈D means that for every ε>0 there is δ>0 such that x∈D and |x−c|<δ imply |f(x)−f(c)|<ε. This promise includes the actual point value and every allowed nearby input.

At an interior point c of an interval domain, form Q(h)=[f(c+h)−f(c)]/h for small nonzero h of either sign. The finite derivative f′(c)=m means: for every ε>0, all sufficiently small such h satisfy |Q(h)−m|<ε. A tangent is the line y=f(c)+m(x−c), through the actual curve point.

A derivative at an included domain endpoint can be defined using only the inward side, and must be named one-sided. The interior two-sided derivative tests below do not silently apply there. At an isolated domain point there is no approaching quotient under this convention. The second derivative f″(c) is the derivative of the slope function f′ at c, when that function is defined nearby.

Compute the slopes directly

For x² at c, [(c+h)²−c²]/h=2c+h, which approaches 2c. For x³ the quotient is 3c²+3ch+h²; for x⁴ it is 4c³+6c²h+4ch²+h³. Their derivatives are 2x,3x²,4x³. The remaining terms tend to zero by bounding |h|<1 and then choosing a sufficiently small radius.

The quotient for a constant is zero, and the quotient for x is one. Difference quotients add and scale: if two errors are each less than ε/2, their sum error is less than ε; a nonzero fixed multiplier k uses error ε/|k|. Thus f=x³−3x has f′=3x²−3 and f″=6x. These calculations also give the second derivatives 2,6x,12x² of x²,x³,x⁴.

For 1/x at c≠0, its quotient equals −1/[c(c+h)]. With |h|<|c|/2, denominators stay nonzero and the error from −1/c² is at most 2|h|/|c|³. Hence f′(x)=−1/x². The quotient for −1/x² at c is (2c+h)/[c²(c+h)²], whose error from 2/c³ is at most 16|h|/|c|⁴ when |h|<|c|/2. This follows by subtracting the fractions: the numerator becomes −3ch−2h² over c³(c+h)². A further bound |h|<ε|c|⁴/16 proves f″(x)=2/x³.

Differentiability implies continuity: f(c+h)−f(c)=hQ(h), and a convergent finite quotient is bounded near zero. Given ε, make |Q(h)|<|m|+1 and |h|<ε/(|m|+1); the product error is then less than ε. This argument uses the actual value f(c), not an omitted point on a nearby formula.

An extremum compares values on a stated set

A local minimum at c∈D means that some radius r>0 gives f(c)≤f(x) for every x∈D with |x−c|<r. A local maximum reverses the inequality. A strict local minimum requires f(c)<f(x) for those inputs x≠c; similarly for a strict maximum. These are comparisons relative to the stated domain.

A global minimum compares f(c) with every f(x) on D; a global maximum does the reverse. Every global extremum is local under these definitions, but a local extremum need not be global. A constant function has non-strict local and global minima and maxima everywhere, with no strict extremum on a nondegenerate interval.

An interior critical input is an included interior input where f′ is zero or where the finite two-sided derivative does not exist. A stationary input is one with f′(c)=0. Endpoints are separate candidates. Excluded inputs are not candidates because there is no f(c) to compare.

Fermat’s necessary condition: at an interior local extremum where the finite derivative exists, f′(c)=0. At a local maximum, right quotients are ≤0 and left quotients are ≥0. A common finite limit must satisfy both inequalities and is zero. Reverse them for a minimum. This proof requires both sides; it says nothing about endpoint derivatives or nonexistent derivatives.

The converse fails. At zero, x³ has derivative zero but negative values to the left and positive values to the right, so it has no extremum there. The corner |x| has a strict minimum at zero, but its left quotient is −1 and its right quotient is 1, so its two-sided derivative does not exist.

Why derivative signs control a whole interval

The closed-interval extreme-value fact used here is: a continuous real function on a nonempty closed bounded interval attains both a largest and a smallest value. A proof from completeness is given in the final proof section. Closedness, boundedness and continuity matter; x on (0,1) has neither attained extreme value.

Rolle’s theorem: if a<b, f is continuous on [a,b], differentiable on (a,b), and f(a)=f(b), then some c∈(a,b) has f′(c)=0. If f is constant this is immediate. Otherwise an attained maximum or minimum differs from the equal endpoint value, lies in the interior, and Fermat’s condition applies.

Let s=[f(b)−f(a)]/(b−a) and subtract the secant line: g(x)=f(x)−f(a)−s(x−a). Under the same continuity and differentiability conditions, g(a)=g(b)=0. Rolle gives g′(c)=f′(c)−s=0. Thus f(b)−f(a)=f′(c)(b−a) for some interior c. This is the mean value theorem, not a claim that every tangent equals the secant slope.

Now let f be differentiable on an open interval I. For any x<y in I, apply that theorem on [x,y]. If f′>0 everywhere on I, then f(y)>f(x), so f is strictly increasing; f′<0 gives strictly decreasing. The weak conditions f′≥0 and f′≤0 give nondecreasing and nonincreasing functions. If f′=0 everywhere on I, the function is constant there.

The interval condition is essential. For 1/x, f′<0 on each of (−∞,0) and (0,∞), so each branch decreases. But −1<1 and f(−1)=−1<f(1)=1; it is not decreasing on the whole disconnected domain. Nor does strict increase require a positive derivative at every input: x³ is strictly increasing while f′(0)=0.

Check signs throughout both nearby intervals

Suppose c is inside an interval, f is continuous at c and differentiable on each side in a neighbourhood of c. If f′>0 throughout the left side and f′<0 throughout the right side, then c is a strict local maximum. If the signs are negative then positive, c is a strict local minimum. The derivative at c itself need not exist.

For the minimum case, take any nearby x<c and use the mean value theorem on [x,c]. Continuity holds at c and differentiability is needed only in (x,c); the negative derivative gives f(c)−f(x)<0. For x>c apply it on [c,x], obtaining f(x)−f(c)>0. This proves strictness for every nearby input. The maximum proof reverses the inequalities.

If f′ has the same strict sign on both sides, one side has values below f(c) and the other above it, so c is neither a local minimum nor a local maximum. These are sufficient tests with full side conditions, not an unrestricted equivalence for all extrema. A constant plateau has non-strict extrema without either strict sign switch.

For x³−3x, the factors (x−1)(x+1) determine the sign on every interval separated by −1 and 1. At −1 the switch is + to −; at 1 it is − to +. A chosen sample helps display that established sign. One sampled derivative value by itself cannot guarantee a sign throughout an interval.

Continuity at c cannot be dropped. Set f(x)=x² for x≠0 but f(0)=1. The slopes are negative on the left and positive on the right, yet zero is a strict local maximum, because nearby nonzero squares are less than one. The punctured rule’s minimum is not the actual assigned point.

A nonzero second derivative can settle a stationary point

Suppose f′ is defined on an open neighbourhood of c, f′(c)=0, and the finite derivative f″(c)=A exists. If A>0, c is a strict local minimum; if A<0, c is a strict local maximum. If A=0, this test is inconclusive. No continuity of f″ throughout a larger interval is required by this version.

The proof is local: [f′(c+h)−f′(c)]/h=f′(c+h)/h approaches A. For A>0, request error A/2, so this quotient is positive for every sufficiently small nonzero h. Therefore f′ is negative when h<0 and positive when h>0. Differentiability at c gives continuity there, and the first sign test proves a strict minimum. For A<0, request |A|/2 and reverse the signs.

At zero, x³, x⁴ and −x⁴ all have f′=f″=0. Their outcomes differ: x³ has no extremum; x⁴ has a strict minimum; −x⁴ has a strict maximum. Their first derivatives are respectively 3x²,4x³,−4x³, so their full side signs resolve the cases. Zero is a result to investigate, not a failed computation to replace by a guess.

Curvature here means bending of the graph

Rising and bending ask different questions. The first derivative is the tangent slope; the second derivative describes how that slope changes with input. For −x² on (−∞,0), f′=−2x>0 but f″=−2<0: the graph rises while its tangent slope becomes less positive. For x² on (−∞,0), it falls while its slope becomes less negative.

We call f convex, or concave up, on an interval when its graph lies at or below every chord: for a<b and 0≤t≤1, f((1−t)a+tb)≤(1−t)f(a)+tf(b). Concave, or concave down, reverses the inequality. Strict bending uses a strict inequality for a<b and 0<t<1. A straight line has both weak properties but neither strict one.

For a differentiable function, a nondecreasing slope f′ guarantees convexity. If a<x<b, the mean value theorem gives the left and right secant slopes as f′(u) and f′(v), with a<u<x<v<b. Thus [f(x)−f(a)]/(x−a)≤[f(b)−f(x)]/(b−x). Multiplying by positive denominators and rearranging is exactly the chord inequality. Strictly increasing f′ gives strict convexity. Reverse the order for concavity.

The same slope order puts every tangent below a convex graph: for x>c the secant slope is f′(u)≥f′(c); for x<c it is f′(u)≤f′(c), and multiplication by x−c reverses that inequality. Both give f(x)≥f(c)+f′(c)(x−c). For a concave graph every tangent is above it.

If f″≥0 everywhere on an open interval, apply the mean value theorem to f′ to get a nondecreasing slope, hence convexity. If f″>0 everywhere, the bending is strict. Negative signs give the corresponding concavity conclusions. These are whole-interval conditions, not conclusions from one second-derivative reading.

The chapter uses curvature in this qualitative bending sense. It does not compute geometric curvature or identify its magnitude with f″. The graph’s vertical and horizontal scales are stated separately; changing those scales changes the picture’s apparent steepness without changing the sign conclusions.

A zero second derivative is only a candidate

Here an inflection point is an included point (c,f(c)) where f is continuous and the graph changes from convex bending to concave bending, or the reverse, across c. Both nearby sides must exist, and each must genuinely bend rather than be straight throughout a neighbourhood. Our strict side-bending tests avoid counting a straight line as a change of bending.

A sufficient test is continuity at an included c and strictly opposite signs of f″ throughout the left and right nearby intervals. The second derivative at c need not exist. For x³, f″=6x is negative on the left and positive on the right, so (0,0) is an inflection point, although no extremum occurs there.

For x⁴, f″=12x² is positive on both sides of zero. The second derivative vanishes at zero, but bending does not change: there is no inflection point there. Solving f″=0 lists candidates, not answers. Under the side-bending hypotheses, if f″(c) exists at an inflection it must be zero: the ordered side slopes force its difference-quotient limit to be both ≥0 and ≤0.

For f(x)=x|x|, the left branch is −x² and the right branch is x². Its quotient at zero is |h|, so f′(0)=0, and f′(x)=2|x| everywhere. The second derivative is −2 on the left and 2 on the right, but does not exist at zero. The function is continuous, both sides bend oppositely, and (0,0) is an inflection point without a second derivative there.

The reciprocal 1/x has f″=2/x³, negative on its left component and positive on its right component. Zero is excluded and the function has no continuous point there. Its separated branches do not produce an inflection point at zero. An asymptote is not a curve point.

Compare every possible attained global candidate

For a continuous function on [a,b], a<b, global extrema exist by the closed-interval fact. An attained interior extremum is critical by Fermat when its derivative exists; if it does not, it is already a critical input. Thus compare all included endpoints and all interior critical inputs. If there are infinitely many critical inputs, a finite list is not automatically complete.

For x³−3x on [−2,2], the derivative vanishes only at −1 and 1 and exists everywhere. The complete candidate list is therefore −2,−1,1,2, with values −2,2,−2,2. Comparing them gives all attained global extrema, including the endpoint ties. A local turn alone would have missed those ties.

For x² on [0,2], the global minimum is 0 at the left endpoint, and the maximum is 4 at the right endpoint. Relative to this domain they are also local extrema. The inward derivatives are 0 and 4; Fermat does not require the latter to vanish. For x on (0,1), 0 and 1 are bounds but neither is an attained value.

For x² on ℝ\{0}, all values are positive and arbitrarily close to zero, so zero is a greatest lower bound but there is no minimum. The omitted point is not a critical input of this function. Adding it would define a different domain and create a minimum.

Finish the curve beyond the candidate list

Curve analysis starts with the original domain and its interval components. Mark included endpoints, excluded inputs and any nonsmooth included points before applying derivative tests. Use established factor signs or another proof to cover each full interval; use exact tables to check arithmetic rather than to replace that proof.

For 1/x, the vertical asymptote is x=0: for every M>0, 0<x<1/M gives f(x)>M and −1/M<x<0 gives f(x)<−M. The horizontal asymptote is y=0: for every ε>0, |x|>1/ε gives |f(x)|<ε. Neither limit adds zero to the domain, and neither joins the two branches.

The displayed reciprocal window clips values beyond y=±4. Continuation arrows indicate that the curve extends beyond the frame. For polynomial examples a finite frame also shows only part of the full real-domain graph. A frame edge is not automatically a domain endpoint or a global extreme value.

Find possible turns using f′=0 and missing finite derivatives at included interior inputs; find possible bend changes using f″=0 or its absence. Check the actual sign changes and continuity, compare genuine endpoints for global questions, and report bounds or non-attainment when the domain is open or disconnected.

Predict before revealing the derivative record

Choose a formula, a domain and a target input. Predict the local extremum or inflection conclusion. The original function, input scale, actual point or hole, and exact function readings are visible first. Reveal the derivative formulas, exact sample slopes, full interval sign record and conclusion.

The solid curve is f. When a finite derivative exists at the target, a dashed line shows its tangent through the actual point. Bending comparisons add a dotted chord with square endpoints; an open circle records an omitted value. These line styles and marker shapes keep the relationships visible without relying only on colour.

A derivative sample of zero does not say which side signs hold, and a second derivative sample of zero does not decide bending. Relate each revealed case to its factorisation or side formula. Distinguish the displayed coordinate window from the complete domain and keep local, global and domain-relative claims separate.

Optional proof: completeness supplies an attained extreme value

This proof supplies the closed-interval fact used by Rolle and the global comparison. Real completeness says that every nonempty set of real numbers bounded above has a least upper bound, written sup. Also, for every positive number r there is an integer n with 1/n<r: otherwise the positive integers would have a supremum S, and an integer larger than S−1 would make its successor larger than S.

Nonempty nested closed intervals [aₙ,bₙ] whose lengths tend to zero share exactly one point. The increasing left endpoints have a supremum c. Every aₙ≤c≤bₙ, since each bₙ bounds all left endpoints. Closedness includes c. Any two shared points would be separated by at most every interval length, hence have zero separation.

Every sequence of inputs in a closed bounded interval has a convergent subsequence within it. Bisect the interval and retain a closed half containing infinitely many sequence indices; repeat. The nested lengths tend to zero. Choose strictly increasing indices from these halves. Their distances from the shared point are bounded by the corresponding lengths, so they converge to that point. A singleton interval is immediate.

Continuity preserves these sequence limits: given an output tolerance, its continuity radius eventually contains every subsequence input. If a continuous f on [a,b] were unbounded in absolute value, choose inputs xₙ with |f(xₙ)|>n. A convergent input subsequence would give a finite output limit f(c), making its outputs eventually bounded, a contradiction.

The nonempty output set therefore has a finite supremum S. For each positive integer n choose an input xₙ with S−1/n<f(xₙ)≤S, using the least-upper-bound property. A convergent input subsequence gives outputs tending to f(c) by continuity, while their squeeze between S−1/n and S gives limit S. Thus f(c)=S is an attained maximum. Apply the same argument to −f for an attained minimum.

This proof explains the hypotheses rather than assuming a plotted peak must exist. On (0,1), the input subsequence can approach an excluded endpoint; for a discontinuous function, input convergence need not preserve output values. A first reading can return to the concrete curve analysis now that the supporting fact has been established.

Worked examples

Try it yourself

1. On [0,2], x² has a maximum at 2 and inward derivative 4. Does this contradict Fermat’s condition?

   Show the answer and reasoning

   No. The condition concerns an interior extremum with a finite two-sided derivative.

   Two is an endpoint of this domain; the comparison is only inward. Endpoint candidates must be checked separately.

2. For x³ at zero, decide whether the zero first derivative gives an extremum.

   Show the answer and reasoning

   No extremum: negative inputs have negative outputs and positive inputs have positive outputs.

   The derivative 3x² is positive on both punctured sides; a horizontal tangent alone does not force a turn.

3. Classify zero for −x⁴ using side signs, and decide whether it is an inflection.

   Show the answer and reasoning

   A strict local and global maximum, with no inflection.

   The slope −4x³ changes + to −. The second derivative −12x² is negative on both sides, so bending stays concave.

4. For f=5 on ℝ, which extrema occur and does a strict first-derivative sign switch occur?

   Show the answer and reasoning

   Every input is a non-strict local and global minimum and maximum; no strict extrema or sign switches occur.

   All output comparisons are equal, and f′=0 everywhere. This prevents treating the strict sign test as a universal equivalence.

5. On (−∞,0), is −x² rising or falling, and convex or concave?

   Show the answer and reasoning

   Strictly rising and strictly concave.

   Its first derivative −2x is positive, while its second derivative −2 is negative. Height and change of slope are different questions.

6. Does f″(0)=0 for x⁴ imply an inflection?

   Show the answer and reasoning

   No: x⁴ is strictly convex on both sides of zero.

   The side second derivatives 12x² are positive. An inflection requires an actual continuous bend change.

7. For x|x|, compute f′(0) and decide whether an inflection can exist without f″(0).

   Show the answer and reasoning

   f′(0)=0; an inflection exists at (0,0), while f″(0) does not exist.

   The quotient is |h|→0. The slope 2|x| has second-derivative side values −2 and 2, so the continuous curve changes strict bending.

8. Does x² on ℝ\{0} have a minimum at zero?

   Show the answer and reasoning

   No minimum is attained; zero is excluded and is only the greatest lower bound of the outputs.

   Every output is positive, and values 1/n² approach zero. Adding (0,0) would change the domain.

9. Find the attained global minimum and maximum of x on (0,1).

   Show the answer and reasoning

   Neither is attained.

   For every allowed x, x/2 is smaller and (x+1)/2 is larger, and both remain inside the domain.

10. A derivative equals 1 at one sampled point in an interval. May you conclude that the function increases throughout that interval?

    Show the answer and reasoning

    No. The sign must be established throughout the interval, with the differentiability conditions used by the mean value theorem.

    One reading does not exclude zeros, negative slopes or missing derivatives elsewhere. Factorisation or another complete argument must control the whole interval.

11. Let f(x)=x² for x≠0 but f(0)=1. The punctured slopes switch − to +. Classify zero.

    Show the answer and reasoning

    A strict local maximum, not the minimum suggested by the punctured rule.

    For 0<|x|<1, f(x)=x²<1=f(0). Continuity at zero fails, so the first sign test’s hypotheses are absent.

12. For x³−3x on [−2,2], list every input where its global maximum is attained.

    Show the answer and reasoning

    At x=−1 and x=2, with value 2.

    The complete candidate values at −2,−1,1,2 are −2,2,−2,2. Endpoints can tie interior turns.

13. For x² at c=1, f″(1)=2>0. Does the second derivative test give a local minimum at one?

    Show the answer and reasoning

    No. Its stationary-point hypothesis fails because f′(1)=2≠0.

    The positive second derivative establishes convex bending, not an extremum at every point.

14. For 1/x, opposite second-derivative signs occur across zero. Is zero an inflection?

    Show the answer and reasoning

    No. Zero is outside the domain and there is no continuous curve point there.

    The two component signs describe separated branches. The vertical asymptote does not supply a value f(0).

What to carry forward

• A local extremum compares nearby allowed values; a global extremum compares every value on the stated domain.
• Fermat’s zero-slope condition is necessary only at interior extrema with an existing finite derivative.
• First-derivative side signs prove turns only with continuity at the target and full nearby interval conditions.
• A nonzero second derivative classifies a stationary point; a zero second derivative leaves the test inconclusive.
• The first derivative controls height change, while the second controls change of slope and bending.
• An inflection needs an included continuous point and a genuine bend change; zero or missing f″ alone is insufficient.
• Check endpoints, domain components, exclusions and non-attainment; finite graph windows and samples do not prove global claims.