Largest and smallest over a closed interval: candidates inside plus the two ends

What to assume they know

• Local maxima and minima, and both derivative tests, from the previous topics
• Critical points, and both ways a point can be one
• Differentiating a fractional power, from Chapter 5
• Evaluating a function at several inputs and comparing the results
• Closed and open intervals, and what including an endpoint means
• Squares and square roots of small fractions

What they should be able to do

• Distinguish the largest value over a whole interval from a value that is only largest nearby, using the chapter's own four names
• Explain why a function on an open interval may attain neither, and why closing the interval fixes it
• State the existence theorem and the single hypothesis it needs
• State the theorem about interior extremes and say what candidate list it produces
• Say what that theorem omits, and repair the candidate list accordingly
• Carry out the four-step working rule on a polynomial over a closed interval
• Include a critical point at which the derivative does not exist among the candidates
• Handle a problem with no endpoints at all, and supply the extra argument that becomes necessary
• Report both the extreme value and the input at which it occurs
• Recognise a problem that is easier by substitution than by differentiation

Where it usually goes wrong

• "The largest value is at a turning point." Often it is at an end. Example 27's largest and smallest are both at ends, and Fig 6.19 is drawn to make that the expected case rather than the surprising one.
• "Critical points means where the derivative is zero." Step 1 of the rule says zero or undefined, and Example 28's smallest value sits at a point of the second kind. A candidate list built from Theorem 6 alone misses it.
• "If the interval is open I use the same method." The existence guarantee is gone, so there may be nothing to find. Example 16 and the §6.4.1 opening both make this point, on two different functions.
• "Local and absolute are the same when the function has one turning point." Fig 6.19 has two turning points and neither is an absolute extreme. Keep the two questions separate and answer the one that was asked.
• "I have to classify each critical point before comparing." You do not. The rule evaluates and compares; no derivative test is needed at all. That is its main economy, and students waste time running the second test on every candidate.
• "With no endpoints I can still just take the single critical point." You can find it, but not classify it. Example 29 adds a comparison at a second input for exactly this reason.
• "The answer is the value." The answer is the value and the input at which it occurs. Every worked example in §6.4.1 reports both, and exercise items ask for both.
• "Every one of these needs calculus." Miscellaneous Exercise Q11 falls to a substitution and a quadratic. Look at the function before differentiating.

Questions to check understanding

• Find the largest and smallest values of a stated function on a stated closed interval, reporting both values and both inputs — the form of Exercise 6.3 Q5 and Q7
• Build the candidate list for a function whose derivative fails to exist somewhere inside the interval
• Find the parameter value that puts a stated largest value at a stated input — the form of Exercise 6.3 Q11
• Report the inputs at which a periodic function reaches its largest value on a stated interval — the form of Exercise 6.3 Q8
• Explain why a function on an open interval need not attain either extreme
• State the two theorems that license the working rule, and the hypothesis each needs
• Solve one of these by substitution instead of differentiation, and say why it worked

Examples worth working on the board

Values marked verified are worked out here from the chapter's own printed data; neither answers file was opened, and this chapter prints no answers to its own exercises.

• The §6.4.1 opening (Part I p. 171). The chapter takes the function that adds two to its input, on the open unit interval, and observes it has no largest and no smallest value there and no local ones either. It then closes the interval and both appear, at the two ends. Sections 1 and 2 are this paragraph. Pair it with Example 16 (Part I p. 162), which ran the same argument on the identity function without naming the notions; the two together make the point twice with one idea.
• The four names (Part I p. 171). The chapter names the largest value three times over in a single sentence — the absolute one, the global one, and the greatest — and does the same for the smallest. Section 3 should list all six words, say they name two things, and recommend the pair a student should write in an answer. This chapter uses the first pair in every subsequent example and in the Summary.
• Fig 6.19 (Part I p. 172). A continuous curve over a closed interval with four inputs marked on the axis: the two ends and two interior points, each with a dashed vertical and a labelled height. Read on the printed page. The curve starts high at the left end, falls to the first interior mark, rises to the second, and falls to the right end. Both whole-interval extremes are at the ends, and both interior marks are local extremes that are not global ones. The paragraph beneath says exactly this. Section 4 is this figure, and it is the picture that makes the two-part candidate list feel necessary rather than arbitrary.
• Theorem 5 (Part I p. 172). Stated without proof: an unbroken function on a closed interval attains a largest value and a smallest value, each at least once. The chapter has already declared this result beyond its scope once, in the Remark on Part I p. 162, in almost the same words. Section 5 should say what the hypothesis buys and what happens without it — Example 16 is the counter-case, and it fails only because its interval is open.
• Theorem 6 (Part I p. 172). Stated without proof, for a function that is differentiable across a closed interval, at a point strictly inside it: if the largest value is attained there, the derivative vanishes there, and likewise for the smallest. Section 6 draws the consequence — an extreme is either at an end or at a point where the derivative vanishes — and section 7 says what is missing: the theorem assumes the function differentiable throughout, so it says nothing about a function that has a corner. Example 28, two pages later, is exactly such a function, and its answer is found at a point where the derivative does not exist. The Working Rule repairs this silently in its own first step.
• The Working Rule (Part I p. 172). Four numbered steps: find every critical point inside the interval, meaning every input where the derivative vanishes or fails to exist; take the two ends; evaluate the function at all of them; and pick the largest and the smallest of the values obtained. Section 8 is these four steps, read as printed. Note that Step 1's second clause is the repair Theorem 6 needed, made without comment.
• Example 27 (Part I p. 173). A cubic on a closed interval whose endpoints are stated. Its derivative factors into six times two linear factors, vanishing at two and at three, both inside the interval. Verified: the four candidate values are twenty-four at the left end, twenty-nine at two, twenty-eight at three, and fifty-six at the right end. So the largest value is fifty-six at the right end and the smallest is twenty-four at the left end. Section 9. Two things are worth naming: both extremes turned out to be at the ends, exactly as in Fig 6.19, and the two interior candidates were local extremes that lost.
• Example 28 (Part I pp. 173–174). A difference of two fractional powers of the input, on the closed interval from minus one to one. The chapter differentiates, finds the derivative vanishes at one eighth, and then observes separately that the derivative is not defined at zero, adding zero to the candidate list. Verified: the derivative is twice the quantity eight times the input less one, over the input to the two-thirds power; the numerator vanishes at one eighth and the denominator vanishes at zero. The four candidate values are eighteen at minus one, zero at zero, minus nine quarters at one eighth, and six at one. So the largest value is eighteen at the left end and the smallest is minus nine quarters at one eighth. Section 10, and it is the example that justifies section 7.
• Example 29 (Part I p. 174). A point tracked along a parabola, minimising its distance from a stated fixed point. The chapter minimises the square of the distance, finds the derivative factors with a single real root at one, and then says plainly that there are no endpoints to add, so the candidate list has one member. It then evaluates at a second input to establish which kind of extreme the single candidate is. Verified: the squared distance at one is five and at zero is nine, so the value at one is the smaller and the least distance is the square root of five. Section 11 is this example, and its real content is the extra step: with no ends, the rule alone cannot say whether the one candidate is the largest or the smallest, so something else must.
• Exercise 6.3 Q5 (Part I p. 175). Four functions on four stated closed intervals. Verified: (i) the cube on a symmetric interval has extremes at the two ends, eight and minus eight. (ii) sine plus cosine on a half turn has a critical point at an eighth of a turn where the value is the square root of two, and end values one and minus one, so the extremes are the square root of two and minus one. (iii) four times the input less half its square, on the stated interval, has a critical point at four with value eight; the end values are minus ten and sixty-three eighths, so the extremes are eight and minus ten. (iv) a shifted square plus three on the stated interval has its critical point at the right end; the values are nineteen at the left end and three at the right, so the extremes are nineteen and three.
• Exercise 6.3 Q7, Q10, Q11 and Q12 (Part I pp. 175–176). Verified for Q7: the quartic's derivative factors as twelve times two less than the input, times two plus the square of the input, so the only critical point is two; the three candidate values are twenty-five, minus thirty-nine and sixteen, giving extremes twenty-five and minus thirty-nine. Verified for Q10: on the first stated interval the only critical point is two, and the three values are eighty-five, seventy-five and eighty-nine, so the largest is eighty-nine at the right end; on the second stated interval the critical point is minus two and the three values are one hundred twenty-five, one hundred thirty-nine and one hundred twenty-nine, so the largest is one hundred thirty-nine. Verified for Q11: the derivative at one must vanish for the stated largest value to occur there, which forces the parameter to be one hundred twenty. Verified for Q12: the derivative is one plus twice the cosine of the doubled input, vanishing four times inside a full turn; comparing all six candidate values gives the largest at the right end, a full turn, and the smallest at the left end, zero.
• Exercise 6.3 Q8 (Part I p. 175). The sine of the doubled input on a full turn, with the points asked for rather than the value. Verified: the value reaches one where the doubled input is a right angle or five right angles, so at a quarter of a half turn and at five quarters of a half turn. Worth showing because it inverts the usual question.
• Miscellaneous Exercise Q11 (Part I p. 184). The square of the cosine plus the sine, on a half turn. Verified by substitution rather than by the rule: writing everything in terms of the sine gives one plus the sine less its square, a downward quadratic in a quantity confined between zero and one; it peaks when the sine is a half, giving five quarters, and takes the value one at both ends of that range. So the largest value is five quarters and the smallest is one. Section 12 should show this alongside the four-step route: the substitution is shorter and gets the same answer, and recognising that is a real skill.
• The Summary's Working Rule bullet (Part I p. 186). The four steps restated. Verified against Part I p. 172 word by word: faithful, including Step 1's clause about the derivative failing to exist. This is one Summary bullet in this chapter that a student can safely revise from.

Figures to have open

• The same interval drawn twice for section 1, once with hollow endpoints and once with filled ones. Not in the book.
• A redraw of Fig 6.19 (Part I p. 172): a continuous curve over a closed interval with four marked inputs — the two ends and two interior turns — each with a dashed vertical and a labelled height, the left end highest and the right end lowest. The chapter's own, and the topic's key picture.
• A curve with a corner on it for section 7, with the corner marked and the region Theorem 6 covers shaded. Not in the book; the chapter draws corners only in Fig 6.14, in a different context.
• A four-row candidate table for section 9. Content is Example 27's own values from Part I p. 173; layout is added here. Build it with the repo's DataTable component.
• A graph of the fractional-power function of Example 28 near the origin, showing the cusp. Not in the book; the chapter prints no figure with Example 28, which is a gap worth filling because the shape is unfamiliar.

Where this sits in the book

• NCERT Class 12 Mathematics, Part I, Chapter 6 "Application of Derivatives", §6.4.1 Maximum and Minimum Values of a Function in a Closed Interval, opening and vocabulary, Part I p. 171
• Fig 6.19 and the paragraph beneath it, Theorem 5, Theorem 6 and the Working Rule, Part I p. 172
• Examples 27, 28 and 29, Part I pp. 173–174
• Example 16 and the Remark declaring two results beyond scope, Part I p. 162
• Exercise 6.3, questions 5, 7, 8 and 10 to 12, Part I pp. 175–176
• Miscellaneous Exercise on Chapter 6, question 11, Part I p. 184
• Summary, the Working Rule bullet, Part I p. 186