Critical points and local versus absolute extrema
If a function has a peak or a valley somewhere inside its domain, what must its derivative be doing there, and how does that shrink the search for the largest and smallest values to a handful of points?
Where you are 4 in Maxima, minima and critical points
Why this exists
The The extreme value theorem promises that a function continuous on a closed interval [a, b] reaches a largest and a smallest value. It does not say where. There are infinitely many inputs in [a, b], and we cannot test them one at a time.
The derivative cuts the search down. At the top of a smooth hill the tangent line is horizontal, so the derivative is 0. Turned into a theorem, this says that a peak or valley inside the domain can only sit where f'(c) = 0 or where f'(c) does not exist. Those inputs, the critical points, are usually few, and they can be found by solving an equation. Algebra read peaks and valleys from graphs and tables (Turning points, maximums and minimums); here the formula locates them, with proof that none is missed.
Definition
Let f have domain D. A number c is an interior point of D if some open interval (c - \delta, c + \delta) with \delta > 0 lies inside D. The endpoints of [a, b] are not interior points of [a, b].
Absolute extrema. f has an absolute maximum at c if f(c) \ge f(x) for every x in D, and an absolute minimum at c if f(c) \le f(x) for every x in D. When we restrict f to an interval I, “every x in D” becomes “every x in I”. Local extrema. f has a local maximum at c if c is an interior point of D and f(c) \ge f(x) for every x in some open interval containing c. A local minimum is the same with f(c) \le f(x). The words relative maximum and relative minimum mean the same thing. A local extremum only has to beat its neighbors; an absolute extremum has to beat everyone.
With this definition an endpoint is never a local extremum, although it can be an absolute one. Some books use one-sided intervals and do call an endpoint a local extremum; the theorems below are stated for the definition given here.
Critical points. A critical point of f is an interior point c of the domain at which
Some books say critical number for c and keep “critical point” for the point (c, f(c)) on the graph. Two conditions are built in: c must be in the domain, so an input where f is undefined is never critical, and c must be interior, so an endpoint is never critical.
Fermat’s theorem. If f has a local maximum or a local minimum at c and f'(c) exists, then f'(c) = 0.
Intuition
Figure 1 shows the four kinds of extreme point a function on a closed interval can have. The peak at p beats its neighbors but not the right end. At p and q the curve levels off: just before the peak it rises, just after it falls, and the tangent in between is flat.
A local extremum can also sit at a corner, where there is no tangent at all. Fermat’s theorem says nothing about a corner, because it assumes f'(c) exists. Hence the two parts of the definition of a critical point. Figure 2 shows both, and why the theorem goes one way only: a flat tangent does not force a peak or a valley.
Results
Fermat’s theorem. If f has a local extremum at c and f'(c) exists, then f'(c) = 0.
Local extrema happen at critical points. If f has a local extremum at c, then c is a critical point of f. Either f'(c) exists, and then it is 0 by Fermat’s theorem, or it does not exist.
The converse is false. A critical point need not give a local extremum. For f(x) = x^3, f'(0) = 0 but f has no extremum at 0 (Example 3).
Where absolute extrema live. If f is defined on [a, b] and has an absolute extremum at c, then c is a critical point of f in (a, b) or c is an endpoint, a or b. An endpoint need not be critical to be the absolute maximum or minimum.
With the extreme value theorem, this reduces the search for absolute extrema on [a, b] to comparing a few values, which is Extrema on a closed interval. Deciding which critical points give peaks, which give valleys and which give neither is the subject of Increasing and decreasing from the derivative and The first and second derivative tests.
Why it is true
A fact about limits. Suppose q(h) \le 0 for every small h > 0 and \lim_{h \to 0^+} q(h) = L. Then L \le 0. Indeed, if L > 0, the definition of the limit with tolerance \displaystyle \frac{L}{2} gives \displaystyle q(h) > \frac{L}{2} > 0 for every small enough h > 0, which contradicts q(h) \le 0. The same holds for left-hand limits and for the inequality \ge: a limit cannot jump across a bound that every nearby value respects.
Proof of Fermat’s theorem. Suppose f has a local maximum at c and f'(c) exists. There is an open interval around c on which f(x) \le f(c). For every h \ne 0 small enough that c + h lies in that interval,
Divide by h. When h > 0, dividing keeps the inequality, so the difference quotient is \le 0, and by the fact above so is its limit:
When h < 0, dividing by a negative number reverses the inequality, so the quotient is \ge 0 and
Because f'(c) exists, both one-sided limits equal f'(c) (see The derivative at a point). So f'(c) \le 0 and f'(c) \ge 0, which leaves f'(c) = 0. Figure 3 shows the two kinds of secant. If f has a local minimum at c, then -f has a local maximum at c, so -f'(c) = 0 and again f'(c) = 0.
The proof needs points on both sides of c, and that is where “interior” is used. At an endpoint only one of the two inequalities is available, and it does not force the derivative to be 0.
Proof of where absolute extrema live. Suppose f has an absolute maximum on [a, b] at c, and c is not an endpoint. Then c is in (a, b), which is an open interval containing c, and f(c) \ge f(x) for every x in it. So f has a local maximum at c, and c is a critical point. The minimum is the same.
Worked examples
Day 1: Ogulnar’s cubic on a closed interval
Ogulnar studies f(x) = x^3 - 27x on [-4, 7]. Show that f has a local maximum at -3 and a local minimum at 3, find the absolute extreme values, and compare each with the value of f' there.
SolutionThe values we need are f(-4) = 44, f(-3) = 54, f(3) = -54 and f(7) = 154.
Local maximum at -3. Compare every value with 54. Factoring,
which can be checked by expanding. For x < 6 the factor x - 6 is negative and (x + 3)^2 \ge 0, so 54 - f(x) \ge 0. On the open interval (-4, 6), which contains -3, we have f(x) \le 54: a local maximum with value 54.
Local and absolute minimum at 3. Likewise
For every x in [-4, 7] the factor x + 6 is positive, so f(x) \ge -54 on the whole interval, with equality only at x = 3. The value -54 is the absolute minimum, and since 3 is interior it is a local minimum too.
Absolute maximum at 7. Here
The quadratic has discriminant 49 - 88 = -39 < 0 and positive leading coefficient, so it is always positive, and 7 - x \ge 0 on the interval. So f(x) \le 154 on [-4, 7]: the absolute maximum value is 154, at the end x = 7. The local maximum value 54 is not the absolute maximum.
The derivative. f'(x) = 3x^2 - 27, so f'(-3) = 0 and f'(3) = 0, as Fermat’s theorem requires at interior extrema. At the end, f'(7) = 120. There is no conflict: 7 is not interior, and the absolute maximum sits there while the graph is still climbing.
Day 1: Merdan’s corner
Merdan looks at g(x) = \lvert x^2 - 4 \rvert on [-1, 3]. Find its critical points, and its local and absolute extrema.
SolutionRemove the absolute value. On [-1, 2], x^2 \le 4, so g(x) = 4 - x^2. On [2, 3], g(x) = x^2 - 4.
Critical points. On (-1, 2), g'(x) = -2x, which is 0 at x = 0. On (2, 3), g'(x) = 2x, never 0 there. At x = 2 the one-sided derivatives are
the slopes of the left and right pieces. They differ, so g'(2) does not exist: the graph has a corner (Differentiability: corners, cusps and piecewise functions). The critical points are 0, where g'(0) = 0, and 2, where g'(2) does not exist.
Extrema. Every value of g is at least 0 and g(2) = 0, so the absolute minimum value is 0, at x = 2; it is interior, so it is also a local minimum. On the open interval (-1, 2), g(x) = 4 - x^2 \le 4 = g(0), so g has a local maximum value 4 at 0. On [2, 3] the values x^2 - 4 run up to g(3) = 5, which beats every value on [-1, 2], so the absolute maximum value is 5, at the end x = 3. The other end gives g(-1) = 3, neither.
The minimum at the corner does not contradict Fermat’s theorem, which only speaks when the derivative exists.
Day 1: critical points that are not extrema
Show that each function has a critical point at 0, and decide whether f(0) is a local extreme value. (a) f(x) = x^3. (b) f(x) = x^{1/3}. (c) A constant function f(x) = k.
Solution(a) f'(x) = 3x^2, so f'(0) = 0 and 0 is critical. But x^3 < 0 = f(0) for x < 0 and x^3 > 0 for x > 0. Every open interval around 0 holds inputs on both sides, so f(0) is neither the largest nor the smallest value on it. There is no local extremum. The tangent at 0 is the horizontal line y = 0, and the graph crosses it (Figure 2, C).
(b) The domain is all real numbers, so 0 is interior. The difference quotient is
which grows without bound as h \to 0, so f'(0) does not exist (the graph has a vertical tangent) and 0 is critical. As in (a), x^{1/3} is negative to the left of 0 and positive to the right, so there is no extremum.
(c) f'(c) = 0 for every c, so every number is a critical point. At each c both f(c) \ge f(x) and f(c) \le f(x) hold for all x, with equality, so every c is a local maximum and a local minimum at once. The definitions use \ge and \le, not strict inequalities.
Day 2: critical points of a polynomial
Find the critical points of p(x) = x^4 - 2x^3 - 2x^2 + 7 and the points they give on the graph.
SolutionA polynomial is defined and differentiable at every real number (The power rule and basic differentiation rules), so every number is interior and p' always exists. The only critical points are the solutions of p'(x) = 0:
A product is 0 only when a factor is, so the critical points are \displaystyle x = -\frac{1}{2}, x = 0 and x = 2. The values are \displaystyle p\left(-\frac{1}{2}\right) = \frac{109}{16} (from \displaystyle \frac{1}{16} + \frac{1}{4} - \frac{1}{2} + 7), p(0) = 7 and p(2) = -1, so the critical points on the graph are \displaystyle \left(-\frac{1}{2}, \frac{109}{16}\right), (0, 7) and (2, -1).
Every peak and valley of p is among these three points; which is which is a later question.
Day 2: rational powers, and an endpoint
Find the critical points of (a) f(x) = x^{2/3}(x - 4) and (b) \displaystyle k(x) = \sqrt{x}\,(x - 3). For (b), show that the largest value of k on [0, 3] is at an input that is not a critical point.
Solution(a) The domain is all real numbers. Distribute to get f(x) = x^{5/3} - 4x^{2/3}, and for x \ne 0 use the power rule for rational exponents:
The second form comes from writing both terms over 3x^{1/3}, since x^{2/3} \cdot x^{1/3} = x. A fraction is 0 when its numerator is 0 and its denominator is not, so f'(x) = 0 at \displaystyle x = \frac{8}{5}.
The power rule does not cover x = 0, where x^{-1/3} is undefined, so we test the definition directly:
As h \to 0^+ the numerator approaches -4 and the denominator is a small positive number, so the quotient decreases without bound; as h \to 0^- it increases without bound. So f'(0) does not exist (the graph has a cusp), and 0 is interior. The critical points are 0 and \displaystyle \frac{8}{5}.
(b) The domain is [0, \infty). For x > 0, k(x) = x^{3/2} - 3x^{1/2} and
This is 0 only at x = 1, with k(1) = -2. The formula is undefined at 0, but 0 is an endpoint of the domain, so it is not a critical point whatever the derivative does. The only critical point is 1.
On [0, 3]: k(0) = 0, and for 0 < x < 3 the factor \displaystyle \sqrt{x} is positive and x - 3 is negative, so k(x) < 0. The largest value of k on [0, 3] is 0, reached at the endpoints 0 and 3, neither of which is a critical point.
Day 2: Aylar’s quotient and product
Aylar finds the critical points of (a) r(x) = \dfrac{x^2 - 3}{x - 2} and (b) q(x) = (x^2 - 3)e^x.
Solution(a) The domain is every x \ne 2. By the quotient rule,
The numerator is 0 at x = 1 and x = 3, where the denominator is not 0. The derivative fails to exist only at x = 2, which is not in the domain, so 2 is not a critical point (the graph has a vertical asymptote there). The critical points are 1 and 3, with r(1) = 2 and r(3) = 6.
(b) By the product rule and \dfrac{d}{dx}\big[e^x\big] = e^x (The number e and the natural exponential function),
The factor e^x is positive for every x, so it is never 0 and can be set aside: q'(x) = 0 exactly when x = -3 or x = 1. The derivative exists everywhere, so the critical points are -3 and 1, with q(-3) = 6e^{-3} and q(1) = -2e.
Day 2: a family of cubics
For each real number k, let f(x) = 2x^3 + 6kx. Find the critical points in terms of k, and show that when k > 0 the function has no local extremum and no absolute extremum.
Solutionf is a polynomial, so the critical points are the solutions of f'(x) = 0, where
So f'(x) = 0 exactly when x^2 = -k.
- If k < 0, then -k > 0 and there are two critical points, \displaystyle x = \sqrt{-k} and \displaystyle x = -\sqrt{-k}. For k = -4 they are \pm 2.
- If k = 0, the only critical point is 0, and f(x) = 2x^3 has no extremum there, by the argument of Example 3.
- If k > 0, then f'(x) = 6x^2 + 6k \ge 6k > 0 for every x. There are no critical points.
For k > 0: every local extremum is at a critical point, and there are none, so f has no local extremum anywhere. An absolute extremum on the whole real line would be at an interior point, hence a local extremum, so there is none of those either. For k < 0, what happens at the two candidates is a question for The first and second derivative tests.
Boundaries
- Fermat’s theorem needs f'(c) to exist. At a corner or cusp a function can have an extremum with no derivative at all (\lvert x \rvert at 0). This is why critical points include the inputs where f' does not exist.
- It needs c to be interior. At an endpoint the proof has points on one side only. An absolute extremum at an endpoint can have a nonzero one-sided derivative (f'(7) = 120 in Example 1).
- It is one-way. f'(c) = 0 does not give an extremum (x^3 at 0), and neither does a missing derivative (x^{1/3} at 0). Critical points are candidates only.
- A critical point must be in the domain. An input where f is undefined, such as 2 for \dfrac{x^2 - 3}{x - 2}, is never critical, even though the formula for f' breaks there.
- Fermat’s theorem does not promise an extremum exists. f(x) = x on (0, 1) has no critical points and no extrema. Existence comes from the extreme value theorem, which needs a continuous function on a closed interval.
Where this goes next
Critical points split the number line into pieces, and the sign of f' on each piece says where f rises and falls: Increasing and decreasing from the derivative. From there The first and second derivative tests decides which critical points are peaks, which are valleys and which are neither. With the extreme value theorem, the result on where absolute extrema live becomes Extrema on a closed interval: compare f at the critical points and the two ends. A maximum inside an interval forcing a flat tangent is also the heart of The mean value theorem. The wider family of maxima and minima is in Maxima, minima and critical points.
Exercises
Day 1: local and absolute extrema
For Exercises 1 to 4, find every local and absolute extreme value of the function on the interval, and where each occurs. Justify each with algebra, and check f' at every interior extremum.
- f(x) = x^2 - 6x + 5 on [0, 7]
- f(x) = 10 - x^2 on [-2, 4]
- h(x) = \lvert x - 1 \rvert + 2 on [-2, 2]
- f(x) = x^3 on [-1, 2]
- Let f(x) = x^3 - 75x. Show that f(x) + 250 = (x - 5)^2(x + 10), and use it to show that f has a local minimum at 5. Check that f'(5) = 0.
- Find every local and absolute extreme value of f(x) = x^2 on (-2, 1], or explain why it does not exist.
Day 1: reasoning
- f(x) = \lvert x \rvert has a minimum at 0, but f'(0) is not 0. Explain why this does not contradict Fermat’s theorem.
- Show that g(x) = x^5 has a critical point at 0 but no local extremum there.
- Prove Fermat’s theorem for a local minimum directly from difference quotients, without passing to -f.
- True or false: if f'(c) = 0, then f has a local maximum or a local minimum at c. Explain.
- Give a function on [0, 1] whose absolute maximum is at an endpoint where the one-sided derivative is not 0. Explain why the proof of Fermat’s theorem breaks down there.
Day 2: finding critical points
Find every critical point. Say whether each comes from f'(c) = 0 or from f'(c) not existing.
- f(x) = x^3 - 9x^2 + 15x + 4
- f(x) = 2x^3 + 9x^2 - 60x
- f(x) = x^4 - 18x^2 + 1
- f(x) = x^3 + 6x^2 + 12x - 1
- f(x) = x^{4/5}(x - 9)
- f(x) = x^{1/3}(x + 4)
- f(x) = \dfrac{x^2 + 5}{x + 2}
- f(x) = \dfrac{x - 1}{x^2 + 3}
- f(x) = (x - 4)e^x
- f(x) = \dfrac{e^x}{x + 3}
- f(x) = \dfrac{x^2 + 1}{e^x}
- \displaystyle f(x) = x - 4\sqrt{x}
Day 2: reasoning
- Show that f(x) = x^5 + 2x^3 + x - 4 has no critical points, and explain why it has no local extremum.
- Find the number a for which f(x) = x^3 + ax^2 + 12x has a critical point at x = 2. For that a, show that 2 is the only critical point and that f has no local extremum there.
Answers
- Local and absolute minimum -4 at x = 3, since f(x) = (x - 3)^2 - 4 \ge -4; f'(3) = 0. Absolute maximum 12 at the end x = 7, where (x - 3)^2 is largest; f'(7) = 8. No local maximum: the only interior critical point is 3. f(0) = 5 is neither.
- Local and absolute maximum 10 at x = 0, since x^2 \ge 0; f'(0) = 0. Absolute minimum -6 at the end x = 4, where x^2 is largest. No local minimum. f(-2) = 6 is neither.
- Local and absolute minimum 2 at x = 1, a corner: the one-sided derivatives are -1 and 1, so h'(1) does not exist. Absolute maximum 5 at the end x = -2, since \lvert x - 1 \rvert \le 3 on [-2, 2] with 3 only at -2. No local maximum. h(2) = 3 is neither.
- Absolute minimum -1 at x = -1 and absolute maximum 8 at x = 2, since x^3 is increasing. No local extrema: the only critical point is 0, and x^3 changes sign there.
- Expand: (x - 5)^2(x + 10) is (x^2 - 10x + 25)(x + 10), which is x^3 - 75x + 250. For x > -10 this is \ge 0, so f(x) \ge -250 = f(5) on the open interval (-10, \infty), a local minimum. f'(x) = 3x^2 - 75, so f'(5) = 0.
- Local and absolute minimum 0 at x = 0. No absolute maximum: the values approach 4 as x \to -2^+, and -2 is not in the interval. No local maximum: by Fermat’s theorem one would need f'(c) = 2c = 0, so c = 0, which is a minimum.
- Fermat’s theorem assumes f'(c) exists. Here the one-sided difference quotients at 0 are -1 and 1, so f'(0) does not exist, and the theorem makes no claim. 0 is a critical point of the second kind.
- g'(x) = 5x^4, so g'(0) = 0 and 0 is critical. But x^5 < 0 for x < 0 and x^5 > 0 for x > 0, so every open interval around 0 has values above and below g(0) = 0.
- Near c, f(c + h) - f(c) \ge 0. For h > 0 the quotient \dfrac{f(c + h) - f(c)}{h} \ge 0, so f'(c) \ge 0; for h < 0 it is \le 0, so f'(c) \le 0. Hence f'(c) = 0.
- False. f(x) = x^3 has f'(0) = 0 and no extremum at 0.
- f(x) = x: the maximum 1 is at x = 1, where the one-sided derivative is 1. At 1 only h < 0 is allowed, which gives f'(1) \ge 0; the inequality from h > 0 is missing.
- f'(x) = 3x^2 - 18x + 15 = 3(x - 1)(x - 5): critical points 1 and 5, both f' = 0.
- f'(x) = 6x^2 + 18x - 60 = 6(x + 5)(x - 2): critical points -5 and 2, both f' = 0.
- f'(x) = 4x^3 - 36x = 4x(x - 3)(x + 3): critical points -3, 0, 3, all f' = 0.
- f'(x) = 3x^2 + 12x + 12 = 3(x + 2)^2: critical point -2 only, f' = 0.
- f'(x) = \dfrac{9}{5}x^{4/5} - \dfrac{36}{5}x^{-1/5}, which is \dfrac{9(x - 4)}{5x^{1/5}} for x \ne 0. f'(4) = 0; at 0 the quotient \dfrac{f(h) - f(0)}{h} = \dfrac{h - 9}{h^{1/5}} is unbounded, so f'(0) does not exist. Critical points 0 and 4.
- f'(x) = \dfrac{4}{3}x^{1/3} + \dfrac{4}{3}x^{-2/3}, which is \dfrac{4(x + 1)}{3x^{2/3}} for x \ne 0. f'(-1) = 0; f'(0) does not exist (the quotient \dfrac{h + 4}{h^{2/3}} grows without bound). Critical points -1 and 0.
- f'(x) = \dfrac{x^2 + 4x - 5}{(x + 2)^2}, which is \dfrac{(x + 5)(x - 1)}{(x + 2)^2}: critical points -5 and 1, both f' = 0. -2 is not in the domain.
- f'(x) = \dfrac{(x^2 + 3) - 2x(x - 1)}{(x^2 + 3)^2}, which is \dfrac{-(x - 3)(x + 1)}{(x^2 + 3)^2}: critical points -1 and 3, both f' = 0.
- f'(x) = e^x + (x - 4)e^x = (x - 3)e^x: critical point 3, f' = 0.
- f'(x) = \dfrac{e^x(x + 3) - e^x}{(x + 3)^2}, which is \dfrac{e^x(x + 2)}{(x + 3)^2}: critical point -2, f' = 0. -3 is not in the domain.
- f'(x) = \dfrac{2x\,e^x - (x^2 + 1)e^x}{e^{2x}}, which is -\dfrac{(x - 1)^2}{e^x}: critical point 1, f' = 0.
- Domain [0, \infty). For x > 0, \displaystyle f'(x) = 1 - \dfrac{2}{\sqrt{x}} = \dfrac{\sqrt{x} - 2}{\sqrt{x}}, which is 0 at x = 4. 0 is an endpoint, so it is not critical. Critical point 4 only.
- f'(x) = 5x^4 + 6x^2 + 1 \ge 1 for every x, so f' exists everywhere and is never 0: no critical points. Every local extremum is at a critical point, so there is none.
- f'(x) = 3x^2 + 2ax + 12, and f'(2) = 24 + 4a = 0 gives a = -6. Then f'(x) = 3x^2 - 12x + 12 = 3(x - 2)^2, zero only at 2. Also x^3 - 6x^2 + 12x equals (x - 2)^3 + 8, which is below 8 for x < 2 and above 8 for x > 2: no extremum at 2.