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LatineSummary
This lesson opens the last unit of the course, which passes from what the derivative says at a point to what it says on an interval. It presents the path through the unit: the Weierstrass theorem guarantees that a function continuous on a closed and bounded interval attains its maximum and its minimum; Fermat’s theorem locates them where the derivative vanishes; the theorems of Rolle, Lagrange and Cauchy turn the sign of the derivative into information about a whole interval, and from them come L’Hôpital’s rule, monotonicity, the tests for extrema, concavity, curvature, optimization and the Newton-Raphson method. It reconstructs the history of these results, from Fermat’s method for maxima to Darboux’s theorem, in an order almost the reverse of that of the unit; it formulates the problem that runs through it, the passage from the local to the global, and shows with an example in the rational numbers that this passage depends on the completeness of the reals. It ends by pointing out where these applications are used in other courses of the site and in science and technology.
Learning objectives
- Describe the path through the unit and explain what each lesson proves and which earlier results it relies on.
- Explain why global conclusions about a function, such as the existence of a maximum, monotonicity or the convergence of a method, require, besides the derivative, the completeness of the real numbers.
- Place the results of the unit in their historical context, from Fermat’s method for maxima to Darboux’s theorem, and distinguish the order in which they were discovered from the order in which they are proved.
- Point out where the applications of derivatives intervene in other courses and in scientific and technical applications.
Entry assessment
Before working through the unit, it is advisable to measure the starting point. This self-assessment is not graded: it serves to know where to pause. The first nine questions cover, in order, the nine content lessons of the unit, one each, and measure the understanding of their central result (what it guarantees, under which hypotheses, what it does not guarantee), not a statement to be remembered; the same skills are measured again at the end, in the synthesis lesson, with other data. The last five check what this unit takes as known from the two preceding ones: Bolzano’s theorem, the sign of the derivative at a point, the second derivative, implicit differentiation and equivalent infinitesimals.
Before starting the unit, measure your starting point. It is a short self-assessment: it is not graded and its result is stored only in your browser.
Question 1In which of these cases does the Weierstrass theorem guarantee that the function attains its maximum value on the indicated set?
The function is continuous and bounded, but is not a bounded interval, and the theorem does not apply. In fact, there is no maximum: the values approach without attaining it.The interval is not closed: it lacks the endpoint , where the maximum would be. The values approach without attaining it.The interval is closed and bounded, but the function is not continuous at : its limit from the left is and its value is . The values approach without attaining it.The function is continuous, being the absolute value of a polynomial, and the interval is closed and bounded. The theorem does not require differentiability: is not differentiable at , where it attains its minimum, , and it attains its maximum, , at .Question 2A function is differentiable on , with and for every . What can be guaranteed about ?
It has been assumed that the derivative equals on the whole interval; the hypothesis only bounds it. The constant function satisfies the conditions and equals at .The value has been bounded instead of the increment: the mean value inequality bounds , not . The function satisfies the hypotheses and equals at .There is no need to know it: the mean value theorem turns a bound on the derivative into a bound on the increment, without knowing at which point the slope of the tangent equals that of the chord.By the mean value theorem, for some in , and : this is the mean value inequality. Both extreme values are attained, with and with , so nothing more can be guaranteed.Question 3Compute .
L’Hôpital’s rule states that, if the quotient of the derivatives has a limit, that of the functions has the same one; if it has none, the rule says nothing, and the limit may exist.That factor oscillates, but it is bounded, and it is multiplied by , which tends to faster than .has been replaced by , as if it were an equivalent infinitesimal. The equivalence of the sine with its argument holds when the argument tends to , and does not tend to .For close to , the quotient is : the first factor tends to , and the second to , because . L’Hôpital’s rule is inconclusive: the quotient of the derivatives, , has no limit.Review: L’Hôpital’s rule
Question 4On , is the function strictly increasing?
What prevents strict monotonicity is not the number of zeros of the derivative, but its vanishing on a whole interval; here the zeros are isolated.A positive derivative suffices for strict increase, but it is not necessary: is strictly increasing on and its derivative vanishes at .The cosine oscillates, but it never goes down faster than goes up: the derivative of the sum, , is never negative.The derivative vanishes only at the isolated points , with an integer. By the proposition on strict monotonicity with a nonnegative derivative, is strictly increasing on .Question 5A function is continuous on and differentiable at each point of except at , and its derivative vanishes only at . If , , , and , what are its absolute maximum and minimum values on ?
The points where the function is not differentiable have been forgotten, and they are also critical points: at the function equals .Only the values at the endpoints of the interval have been compared; the absolute extrema may also lie at the critical points of the interior.The values at the candidates suffice: the proposition guarantees that the absolute extrema are attained at one of them, and it gives all the candidates.By the proposition on absolute extrema on a closed and bounded interval, they exist and are attained at , at or at a critical point: at , where there is no derivative, or at , where it vanishes. It suffices to compare , , and ; the value at , which is not a candidate, plays no role.Review: Maxima and minima of a function
Question 6What are the inflection points of ?
That vanishes is necessary, but not sufficient: the concavity must change, and shows at that it may not change.They are the critical points, where the tangent is horizontal. An inflection point is a point where the concavity changes, and that is decided by the second derivative, not by the first.The third derivative test gives a sufficient condition, and when it does not apply it decides nothing. At it does apply: , which is not zero.The second derivative, , changes sign at ; at it does not, because it is negative on both sides. Near the function is concave on both sides, and is not an inflection point, although vanishes there.Question 7What is the osculating circle to the graph of at the point ?
It has the same tangent and the correct radius, but it curves toward the other side: its center lies below the point, and since , the center of curvature lies above it.The number is the curvature, not the radius: the radius of curvature is its reciprocal.has been used as the radius, with the center on the vertical through the point, which is valid only if the tangent is horizontal. Here the slope is , and the factor is not .With and , the radius is , and the center lies on the normal, which is the line , at that distance from the point and on the side toward which the graph curves: above the point, because .Question 8Which points of the parabola are closest to the point ?
It is the lowest point of the parabola, but not the closest to : its distance is , and that of the points with ordinate is . The critical point of is a local maximum.The negative root has been discarded as if were a length. It is an abscissa, which may be negative, and the point symmetric with respect to the vertical axis is at the same distance.The proposition on the distance from a point to a graph guarantees the minimum for a function continuous on an interval that contains its endpoints, such as : the distant points are far away, and the minimum is sought on a closed and bounded interval.It suffices to minimize the square of the distance, , which attains its minimum on by the proposition on the distance from a point to a graph. Its derivative, , vanishes at and at , and the value at these two, , is less than .Review: Optimization problems
Question 9The function is increasing and convex on , and its root is , close to . What do the Newton-Raphson iterates do starting from ?
To the left of the root, the tangent of a convex increasing function crosses the axis to the right of the root, because it lies below the graph. Monotone approximation requires starting from a point where is positive.The theorem does not apply from , but it does from : the corollary shows that the first iteration leads to the right side.Only the first iteration crosses the root; from on, all the iterates stay to its right.At the function is negative: this is the wrong side. By the corollary on a starting point on the wrong side, the first iterate lands where is positive, and from it the monotone convergence theorem gives iterates that decrease toward the root; the second is , close to .Review: The Newton-Raphson method
Question 10From earlier unitsA function is continuous on , with and . What can be guaranteed?
The theorem guarantees existence, not uniqueness: satisfies the hypotheses and vanishes at three points. Uniqueness requires something more, for instance that the function be strictly monotonic.Two values do not decide monotonicity: between them the function may go up and down, like .The midpoint is the first one that bisection examines, but the root may be elsewhere: satisfies the hypotheses and vanishes at .This is Bolzano’s theorem: a function continuous on a closed and bounded interval, with values of opposite signs at the endpoints, vanishes at some point of the interior.Review: The intermediate value theorem
Question 11From earlier unitsA function is defined on an open interval containing , and . What can be guaranteed?
The derivative at a point does not control monotonicity around it: the function equal to if and to at has derivative at , and it is not increasing on any open interval containing .The proposition is local: it holds only near . The sine has derivative at , and it equals at .The hypothesis only speaks of the derivative at : may not be differentiable at the nearby points, and even if it is, its derivative may fail to be continuous at .This is the proposition on the sign of the derivative at a point: the difference quotient keeps near the sign of its limit. It compares each nearby value with , not the nearby values with one another.Question 12From earlier unitsCompute for .
It is , the first derivative: one more differentiation is missing.The derivative of the term of , which is , is missing.The second derivatives of the factors have been multiplied, . The second derivative of a product is not the product of the second derivatives.By the product rule, , and differentiating again, ; at the logarithm vanishes.Question 13From earlier unitsThe equation defines near the point a differentiable function , with . What is the value of ?
has been differentiated as if were constant: the term of the product rule is missing.has been differentiated as , without multiplying by : depends on , and the chain rule gives .The sign was lost when solving: from one gets .Differentiating with respect to , with the product rule in and the chain rule in , one obtains ; at the point this becomes .Review: Derivative of the inverse function and implicit differentiation
Question 14From earlier unitsCompute .
has been taken as the equivalent of ; the equivalent is .It is the limit of the reciprocal quotient: the numerator and the denominator have been swapped.That form only indicates that the theorem on the limit of a quotient does not apply; the limit may exist, and here it does.As , and ; replacing the numerator and the denominator by their equivalents, the quotient is .Review: Equivalent infinitesimals and infinitely large functions
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