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Chapter 6 · Application of Derivatives

Example 27: Absolute extrema of 2x3−15x2+36x+12x^3-15x^2+36x+1 on [1,5][1,5]

ExamplesAbsolute extrema · Section 6.4.1
The question

Find the absolute maximum and minimum values of a function ff given by

f(x)=2x3−15x2+36x+1 on the interval [1,5].f(x) = 2x^3 - 15x^2 + 36x + 1 \text{ on the interval } [1, 5].
Exam answer f′(x)=6x2−30x+36=6(x−2)(x−3)\begin{aligned} f'(x) &= 6x^2 - 30x + 36 \\ &= 6(x - 2)(x - 3) \end{aligned}

f′(x)=0⇒x=2,3f'(x) = 0 \Rightarrow x = 2, 3⁠, both in [1,5][1, 5]⁠.

Values at the critical points and the end points:

f(1)=2−15+36+1=24f(2)=16−60+72+1=29f(3)=54−135+108+1=28f(5)=250−375+180+1=56\begin{aligned} f(1) &= 2 - 15 + 36 + 1 = 24 \\ f(2) &= 16 - 60 + 72 + 1 = 29 \\ f(3) &= 54 - 135 + 108 + 1 = 28 \\ f(5) &= 250 - 375 + 180 + 1 = 56 \end{aligned}

Hence the absolute maximum value is 56, at x=5x = 5⁠, and the absolute minimum value is 24, at x=1x = 1⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

On a closed interval, the highest and lowest points of a smooth graph can only be at two kinds of places: where the graph is flat (f′=0f' = 0⁠), or at the two ends. So we do not need to search everywhere: list those few candidate points, compute ff at each, and pick the largest and smallest. Local hills and valleys inside the interval may be beaten by the ends, as happens here.

Given, to find, and the plan
Given
f(x)=2x3−15x2+36x+1f(x) = 2x^3 - 15x^2 + 36x + 1 for 1≤x≤51 \le x \le 5⁠.
To find
The absolute maximum (greatest) and absolute minimum (least) values of ff on [1,5][1, 5]⁠, and where they occur.
Plan
Follow the four steps of the Working Rule of Section 6.4.1 (rule step 1: critical points in the interval; rule step 2: the end points; rule step 3: the values of ff there; rule step 4: the largest and smallest). Below, "rule step" means a step of the Working Rule and "Step" a step of this solution.

Basics you need

1Absolute maximum and minimum

The absolute maximum of ff on II is the largest value ff takes anywhere on II⁠; the absolute minimum is the smallest (Section 6.4.1). Lesson: absolute extrema on a closed interval

2Theorems 5 and 6 (Section 6.4.1)

Theorem 5: a continuous function on [a,b][a, b] attains an absolute maximum and an absolute minimum. Theorem 6: if a differentiable ff attains its absolute maximum or minimum at an interior point cc⁠, then f′(c)=0f'(c) = 0⁠.

3The Working Rule (Section 6.4.1)

Critical points in the interval, plus the end points; evaluate ff at all of them; the largest value is the absolute maximum and the smallest is the absolute minimum.

4Factorising a quadratic

x2−5x+6=(x−2)(x−3)x^2 - 5x + 6 = (x - 2)(x - 3)⁠: two numbers with product 6 and sum 5. Basics: factorising and the discriminant

Solution

  1. Why the answer exists and where to look

    ff is a polynomial, so it is continuous on [1,5][1, 5] and differentiable everywhere. By Theorem 5, ff has an absolute maximum and an absolute minimum on [1,5][1, 5]⁠. By Theorem 6, if one of them is at an interior point cc⁠, then f′(c)=0f'(c) = 0⁠. So each must be at an end point or at a point where f′=0f' = 0⁠.

    This is the reason the Working Rule works: it is not a guess but a consequence of Theorems 5 and 6. It reduces an infinite search to a handful of points.
    Theorems 5 and 6, Section 6.4.1 (Basic 2). Lesson: absolute extrema on a closed interval

  2. Critical points (rule step 1)

    f′(x)=6x2−30x+36=6(x2−5x+6)=6(x−2)(x−3).\begin{aligned} f'(x) &= 6x^2 - 30x + 36 \\ &= 6(x^2 - 5x + 6) = 6(x - 2)(x - 3). \end{aligned}

    f′(x)=0f'(x) = 0 gives x=2x = 2 and x=3x = 3⁠. Both lie in [1,5][1, 5]⁠. There are no points where ff fails to be differentiable.

    Factorising shows the zeros at once. Both must be checked to lie in the interval: a critical point outside [1,5][1, 5] would be ignored.
    Power rule; factorising (Basic 4). Basics: factorising and the discriminant

  3. End points (rule step 2)

    The end points of [1,5][1, 5] are x=1x = 1 and x=5x = 5⁠. The candidates are therefore

    x=1,2,3,5.x = 1,\quad 2,\quad 3,\quad 5.
    At an end point the graph can be highest or lowest without being flat, because the function simply stops there. That is why ends are always on the list.

  4. Values at the candidates (rule step 3)

    f(1)=2−15+36+1=24,f(2)=16−60+72+1=29,f(3)=54−135+108+1=28,f(5)=250−375+180+1=56.\begin{aligned} f(1) &= 2 - 15 + 36 + 1 = 24, \\ f(2) &= 16 - 60 + 72 + 1 = 29, \\ f(3) &= 54 - 135 + 108 + 1 = 28, \\ f(5) &= 250 - 375 + 180 + 1 = 56. \end{aligned}
    Each line is 2x3−15x2+36x+12x^3 - 15x^2 + 36x + 1 with the number put in: for x=5x = 5⁠, 2⋅125=2502 \cdot 125 = 250⁠, 15⋅25=37515 \cdot 25 = 375⁠, 36⋅5=18036 \cdot 5 = 180⁠.

  5. Pick the largest and smallest (rule step 4)

    Of 24, 29, 28 and 56, the largest is 56 (at x=5x = 5⁠) and the smallest is 24 (at x=1x = 1⁠). So the absolute maximum value is 56, at x=5x = 5⁠, and the absolute minimum value is 24, at x=1x = 1.

    1 2 3 4 5 24 28 40 56 x y (1, 24) least (2, 29) (3, 28) (5, 56) greatest
    ff on [1,5][1, 5]⁠; the vertical axis starts at 20. The local maximum (2,29)(2, 29) and local minimum (3,28)(3, 28) are not the absolute ones: the ends win.
    Note what the critical points turned out to be: f′f' is ++ on (1,2)(1, 2)⁠, −- on (2,3)(2, 3) and ++ on (3,5)(3, 5)⁠, so x=2x = 2 is a local maximum and x=3x = 3 a local minimum (first derivative test). Neither is absolute: the graph climbs to 56 at the right end and starts lower, at 24, at the left end.
    Working Rule step 4; Theorem 3 for the local nature. Lesson: absolute extrema on a closed interval
Result

On [1,5][1, 5]⁠: absolute maximum value 56, at x=5x = 5⁠; absolute minimum value 24, at x=1x = 1⁠. This agrees with the book.

Check it with numbers

A few other points of the interval: f(1.5)=28f(1.5) = 28⁠, f(2.5)=28.5f(2.5) = 28.5⁠, f(4)=33f(4) = 33⁠, f(4.5)=41.5f(4.5) = 41.5⁠. All lie between 24 and 56, as they must.

Common mistakes
  1. Forgetting the end points. Here both answers are at end points; without them you would wrongly report 29 and 28.
  2. Confusing local with absolute. f(2)=29f(2) = 29 is a local maximum value, not the greatest value on [1,5][1, 5]⁠.
  3. Arithmetic slips. Evaluate each term separately, as in Step 4.
What to take away
  1. On a closed interval: critical points inside, plus both ends, then compare values. No second derivative test is needed.
  2. Theorems 5 and 6 are why this works: the extremes exist, and at an interior one f′=0f' = 0⁠.
  3. A local maximum can be smaller than a value at an end point.

Every fact used, and where it comes from

FactUsed inSource
Absolute extrema exist; f′=0f' = 0 at interior onesStep 1Theorems 5 and 6, Section 6.4.1
f′(x)=6(x−2)(x−3)f'(x) = 6(x-2)(x-3)Step 2Power rule; factorising
Working Rule, rule steps 1 to 4Steps 2 to 5Section 6.4.1 (Lesson: absolute extrema)
Local maximum at 2, local minimum at 3Step 5First Derivative Test, Theorem 3, Section 6.4
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