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Chapter 6 · Application of Derivatives

Exercise 6.3 · Q1: Maximum and minimum values of (2x−1)2+3(2x-1)^2+3⁠, 9x2+12x+29x^2+12x+2⁠, −(x−1)2+10-(x-1)^2+10⁠, x3+1x^3+1

Exercise 6.3Maximum and minimum values · Section 6.4
The question

1. Find the maximum and minimum values, if any, of the following functions given by

(i) f(x)=(2x−1)2+3f(x)=(2x-1)^2+3    (ii) f(x)=9x2+12x+2f(x)=9x^2+12x+2

(iii) f(x)=−(x−1)2+10f(x)=-(x-1)^2+10    (iv) g(x)=x3+1g(x)=x^3+1

Exam answer

(i) For all xx⁠,

f(x)=(2x−1)2+3≥3f(x) = (2x - 1)^2 + 3 \ge 3

and f(12)=3f\left(\frac12\right) = 3⁠. Also f(x)→∞f(x) \to \infty as x→∞x \to \infty⁠.

Minimum value 33⁠; no maximum value.

(ii) For all xx⁠,

f(x)=9x2+12x+2=(3x+2)2−2≥−2\begin{aligned} f(x) &= 9x^2 + 12x + 2 \\ &= (3x + 2)^2 - 2 \ge -2 \end{aligned}

and f(−23)=−2f\left(-\frac23\right) = -2⁠. Also f(x)→∞f(x) \to \infty as x→∞x \to \infty⁠.

Minimum value −2-2⁠; no maximum value.

(iii) For all xx⁠,

f(x)=−(x−1)2+10≤10f(x) = -(x - 1)^2 + 10 \le 10

and f(1)=10f(1) = 10⁠. Also f(x)→−∞f(x) \to -\infty as x→∞x \to \infty⁠.

Maximum value 1010⁠; no minimum value.

(iv) g(x)→∞g(x) \to \infty as x→∞x \to \infty⁠, and g(x)→−∞g(x) \to -\infty as x→−∞x \to -\infty⁠.

gg has neither a maximum nor a minimum value.

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

The one idea this problem is built on

A square is never negative, and it is zero at exactly one place. So a function written as "a square plus a constant" can never go below that constant, and reaches it exactly where the square is zero: that constant is the minimum. The square itself grows without limit, so there is no maximum. A minus sign in front of the square turns everything upside down. A cubic such as x3+1x^3+1 keeps climbing for ever in both directions, so it has neither.

Given, to find, and the plan
Given
Four functions defined for every real number xx (no interval is stated, so the domain is all of R\R⁠).
To find
For each one, its maximum value and its minimum value on R\R in the sense of Definition 3 (Section 6.4), or a proof that one or both do not exist.
Plan
(1) Rewrite each quadratic as a constant plus or minus a perfect square. (2) Read off the bound and the point where it is reached. (3) Prove that the other extreme does not exist by showing that, for every candidate point cc⁠, some other point gives a larger (or smaller) value. (4) Cross-check with derivatives: the point found must be a critical point, and the second derivative test must agree. (5) For the cubic, show it rises strictly everywhere.

Basics you need

1Maximum and minimum value (Definition 3)

ff has a maximum value on an interval I if there is a point cc in I with f(c)≥f(x)f(c)\ge f(x) for all xx in I; the number f(c)f(c) is the maximum value. The minimum value is defined the same way with ≤\le⁠. The book prints these with the signs >\gt and <\lt⁠, but x=cx=c is itself in I and f(c)>f(c)f(c)\gt f(c) is impossible, so they must be read as ≥\ge and ≤\le (exactly as Example 14 uses them: "f(x)≥0f(x)\ge0 for all xx" gives minimum value 00⁠). Lesson: maxima and minima

2A square is never negative

For every real tt⁠, t2≥0t^2\ge0⁠, and t2=0t^2=0 only when t=0t=0⁠. Basics: factorising and completing the square

3Completing the square

ax2+bx+c=a(x+b2a)2+c−b24a.ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}.

Expanding the right side gives back the left side, so the two forms are equal for every xx⁠.

4Showing "no maximum"

To show there is no maximum value we must defeat every possible candidate: for each point cc⁠, produce a point xx with f(x)>f(c)f(x)\gt f(c)⁠. Then no f(c)f(c) is at least as big as all the values.

5Second derivative test (Theorem 4)

If f′(c)=0f'(c)=0 and f′′(c)>0f''(c)\gt0⁠, then cc is a point of local minima; if f′(c)=0f'(c)=0 and f′′(c)<0f''(c)\lt0⁠, a point of local maxima. We use it only as a cross-check here. Lesson: first and second derivative tests

Part (i): f(x)=(2x−1)2+3f(x)=(2x-1)^2+3

  1. A lower bound that is actually reached

    For every real xx⁠, (2x−1)2≥0(2x-1)^2\ge0⁠. Adding 33 to both sides,

    f(x)=(2x−1)2+3 ≥ 3for all x∈R.f(x)=(2x-1)^2+3\ \ge\ 3\quad\text{for all } x\in\R.

    Equality needs (2x−1)2=0(2x-1)^2=0⁠, that is 2x−1=02x-1=0⁠, so x=12x=\tfrac12⁠. Indeed f ⁣(12)=02+3=3f\!\left(\tfrac12\right)=0^2+3=3⁠.

    So the point c=12c=\tfrac12 satisfies f(c)≤f(x)f(c)\le f(x) for all xx⁠: the minimum value is 33⁠, attained at x=12x=\tfrac12.

    Definition 3 needs two things: a number that no value of ff goes below, and a point where ff equals that number. The inequality supplies the first; x=12x=\tfrac12 supplies the second. A bound that is never reached would not be a minimum value.
    t2≥0t^2\ge0⁠, with equality only at t=0t=0⁠; here t=2x−1t=2x-1⁠. Basics: completing the square

  2. There is no maximum value

    Write the function around its lowest point: since 2x−1=2(x−12)2x-1=2\left(x-\tfrac12\right)⁠,

    f(x)=4(x−12)2+3.f(x)=4\left(x-\tfrac12\right)^2+3.

    So ff grows as the distance d=∣x−12∣d=\left|x-\tfrac12\right| from 12\tfrac12 grows: f=4d2+3f=4d^2+3⁠. Take any point cc and let d=∣c−12∣d=\left|c-\tfrac12\right|⁠. Choose x=c+1x=c+1 if c≥12c\ge\tfrac12 and x=c−1x=c-1 if c<12c\lt\tfrac12⁠. Either way xx is one unit further from 12\tfrac12 than cc⁠, so its distance is d+1d+1⁠, and

    f(x)−f(c)=4(d+1)2−4d2=4(2d+1) > 0.\begin{aligned} f(x)-f(c)&=4(d+1)^2-4d^2\\ &=4(2d+1)\ \gt\ 0. \end{aligned}

    So no point cc has f(c)≥f(x)f(c)\ge f(x) for all xx⁠: ff has no maximum value.

    "No maximum" is a statement about every candidate point, so we beat an arbitrary cc⁠, not just a few examples. Moving one unit away from the bottom of the parabola, on the same side as cc⁠, always gives a larger value. Because 2d+1≥12d+1\ge1⁠, the difference is at least 44⁠.
    To disprove "f(c)f(c) is the largest value", one larger value is enough; to do it for every cc⁠, give a recipe for that larger value.

  3. Cross-check with derivatives

    By the chain rule (outer function u2u^2⁠, inner function u=2x−1u=2x-1⁠, u′=2u'=2⁠),

    f′(x)=2(2x−1)⋅2=4(2x−1),f′′(x)=8.\begin{aligned} f'(x)&=2(2x-1)\cdot2=4(2x-1),\\ f''(x)&=8. \end{aligned}

    f′(x)=0f'(x)=0 only at x=12x=\tfrac12⁠, and f′′ ⁣(12)=8>0f''\!\left(\tfrac12\right)=8\gt0⁠, so by the second derivative test (Theorem 4) x=12x=\tfrac12 is a point of local minima with value 33⁠. This agrees with Part (i), Step 1.

    The algebra in Part (i), Step 1 already proves the answer. The derivatives confirm that x=12x=\tfrac12 is the only critical point, so there is no other "turning" point that could hide a different extreme value.
    Theorem 4 (Second Derivative Test), Section 6.4. Lesson: first and second derivative tests
x y O −1 1/2 1 2 3 6 9 12 (1/2, 3) y = 3
f(x)=(2x−1)2+3f(x)=(2x-1)^2+3⁠: the whole graph lies on or above the dashed line y=3y=3 and touches it only at x=12x=\tfrac12⁠. Both arms rise without end, so there is no highest point.

Part (ii): f(x)=9x2+12x+2f(x)=9x^2+12x+2

  1. Complete the square

    The first two terms are the start of (3x+2)2=9x2+12x+4(3x+2)^2=9x^2+12x+4⁠. So

    f(x)=(9x2+12x+4)−4+2=(3x+2)2−2.\begin{aligned} f(x)&=(9x^2+12x+4)-4+2\\ &=(3x+2)^2-2. \end{aligned}

    Check by expanding: (3x+2)2−2(3x+2)^2-2 =9x2+12x+4−2=9x^2+12x+4-2 =9x2+12x+2=9x^2+12x+2⁠. ✓

    In this form the smallest possible value is visible at a glance: the square part is never negative, so the whole thing is never below −2-2⁠.
    Completing the square: 9x2+12x=(3x)2+2(3x)(2)9x^2+12x=(3x)^2+2(3x)(2)⁠, so add and subtract 22=42^2=4⁠. Basics: completing the square

  2. Minimum value −2-2⁠, reached at x=−23x=-\tfrac23

    Since (3x+2)2≥0(3x+2)^2\ge0 for all xx⁠, f(x)≥−2f(x)\ge-2 for all xx⁠. Equality needs 3x+2=03x+2=0⁠, i.e. x=−23x=-\tfrac23⁠, and then

    f ⁣(−23)=9⋅49−12⋅23+2=4−8+2=−2.f\!\left(-\tfrac23\right)=9\cdot\tfrac49-12\cdot\tfrac23+2=4-8+2=-2.

    So the minimum value is −2-2⁠, at x=−23x=-\tfrac23.

    We also evaluated the original formula at x=−23x=-\tfrac23⁠, not only the rewritten one. That guards against an error in the completing-the-square step.
    Definition 3(b), Section 6.4.

  3. No maximum value

    Since 3x+2=3(x+23)3x+2=3\left(x+\tfrac23\right)⁠, f(x)=9(x+23)2−2f(x)=9\left(x+\tfrac23\right)^2-2⁠, which is 9d2−29d^2-2⁠, where dd is the distance from xx to −23-\tfrac23⁠. For any point cc at distance dd⁠, move one more unit away from −23-\tfrac23 on the same side (x=c+1x=c+1 if c≥−23c\ge-\tfrac23⁠, x=c−1x=c-1 otherwise). Then

    f(x)−f(c)=9(d+1)2−9d2=9(2d+1)>0.f(x)-f(c)=9(d+1)^2-9d^2=9(2d+1)\gt0.

    Every candidate is beaten, so there is no maximum value.

    Same argument as Part (i), Step 2: the value depends only on the distance from the lowest point, and grows with it.

  4. Cross-check with derivatives

    f′(x)=18x+12=6(3x+2)f'(x)=18x+12=6(3x+2)⁠, zero only at x=−23x=-\tfrac23⁠; f′′(x)=18>0f''(x)=18\gt0⁠. By Theorem 4, x=−23x=-\tfrac23 is a point of local minima, value −2-2⁠. ✓

    Power rule term by term: the derivative of 9x29x^2 is 18x18x⁠, of 12x12x is 1212⁠, of the constant 22 is 00⁠.
    Theorem 4 (Second Derivative Test). Lesson: first and second derivative tests
x y O −1 1/2 −2 2 6 10 (−2/3, −2)
f(x)=9x2+12x+2=(3x+2)2−2f(x)=9x^2+12x+2=(3x+2)^2-2⁠: lowest point (−23,−2)\left(-\tfrac23,-2\right) on the dashed line y=−2y=-2⁠; the arms rise without end.

Part (iii): f(x)=−(x−1)2+10f(x)=-(x-1)^2+10

  1. An upper bound that is reached

    Since (x−1)2≥0(x-1)^2\ge0⁠, multiplying by −1-1 reverses the inequality: −(x−1)2≤0-(x-1)^2\le0⁠. Adding 1010⁠,

    f(x)=10−(x−1)2 ≤ 10for all x∈R,f(x)=10-(x-1)^2\ \le\ 10\quad\text{for all } x\in\R,

    with equality only when x−1=0x-1=0⁠, i.e. x=1x=1⁠: f(1)=−0+10=10f(1)=-0+10=10⁠. So the maximum value is 1010⁠, at x=1x=1.

    The minus sign in front of the square turns "never below" into "never above". Multiplying an inequality by a negative number flips its direction.
    Definition 3(a), Section 6.4. Basics: signs of products

  2. No minimum value

    With d=∣x−1∣d=|x-1|⁠, f(x)=10−d2f(x)=10-d^2⁠: the further from 11⁠, the smaller. For any point cc at distance dd from 11⁠, take x=c+1x=c+1 if c≥1c\ge1 and x=c−1x=c-1 if c<1c\lt1⁠, at distance d+1d+1⁠. Then

    f(x)−f(c)=−(d+1)2+d2=−(2d+1)<0.f(x)-f(c)=-(d+1)^2+d^2=-(2d+1)\lt0.

    So every candidate for the minimum is undercut: there is no minimum value.

    The graph is a parabola opening downwards; its arms fall without end, so there is no lowest point.

  3. Cross-check with derivatives

    By the chain rule, f′(x)=−2(x−1)f'(x)=-2(x-1)⁠, zero only at x=1x=1⁠; f′′(x)=−2<0f''(x)=-2\lt0⁠. By Theorem 4, x=1x=1 is a point of local maxima, value 1010⁠. ✓

    The inner function x−1x-1 has derivative 11⁠, so the chain rule adds no extra factor here.
    Theorem 4(i). Lesson: first and second derivative tests
x y O −2 1 2 4 4 10 highest point (1, 10)
f(x)=10−(x−1)2f(x)=10-(x-1)^2⁠: highest point (1,10)(1,10) on the dashed line y=10y=10⁠; the arms fall without end.

Part (iv): g(x)=x3+1g(x)=x^3+1

  1. gg rises strictly everywhere

    Take any real cc⁠. Compare gg at c+1c+1 and at cc⁠:

    g(c+1)−g(c)=(c+1)3−c3=3c2+3c+1=3(c+12)2+14.\begin{aligned} g(c+1)-g(c)&=(c+1)^3-c^3\\ &=3c^2+3c+1\\ &=3\left(c+\tfrac12\right)^2+\tfrac14. \end{aligned}

    The last line is a square times 33 plus 14\tfrac14⁠, so it is at least 14>0\tfrac14\gt0⁠. Hence g(c+1)>g(c)g(c+1)\gt g(c) for every cc⁠. Applying the same identity at the point c−1c-1 gives g(c)>g(c−1)g(c)\gt g(c-1)⁠.

    Expanding: (c+1)3=c3+3c2+3c+1(c+1)^3=c^3+3c^2+3c+1⁠, so the c3c^3 and the two +1+1s cancel. Completing the square shows the quadratic 3c2+3c+13c^2+3c+1 is never zero or negative (its discriminant 9−12=−39-12=-3 is negative).
    Completing the square and the discriminant. Basics: factorising and the discriminant

  2. Neither a maximum nor a minimum value

    Any candidate point cc for a maximum is beaten by c+1c+1⁠, since g(c+1)>g(c)g(c+1)\gt g(c)⁠. Any candidate for a minimum is undercut by c−1c-1⁠, since g(c−1)<g(c)g(c-1)\lt g(c)⁠. So gg has neither a maximum value nor a minimum value.

    This is the pattern of Example 16 (f(x)=xf(x)=x on (0,1)(0,1)⁠): a function that rises strictly across an open stretch, with no end points included, can never stop at a highest or lowest value.
    Definition 3, Section 6.4. Lesson: maxima and minima

  3. The trap: g′(0)=0g'(0)=0 but 00 is not an extreme point

    g′(x)=3x2g'(x)=3x^2⁠, which is 00 at x=0x=0⁠. Yet g(−0.1)=0.999g(-0.1)=0.999⁠, g(0)=1g(0)=1 and g(0.1)=1.001g(0.1)=1.001⁠: just left of 00 the values are smaller, just right they are larger. Since g′(x)=3x2>0g'(x)=3x^2\gt0 on both sides of 00⁠, g′g' does not change sign, and by the first derivative test (Theorem 3(iii)) 00 is a point of inflection, not of local maxima or minima.

    This is exactly the Remark after Theorem 2: a zero derivative does not guarantee a maximum or minimum, and the book's own example there is x3x^3⁠. Adding the constant 11 only lifts the graph. (The second derivative test cannot decide: g′′(0)=6⋅0=0g''(0)=6\cdot0=0⁠.)
    Theorem 2 and the Remark after it; Theorem 3 (First Derivative Test). Lesson: first and second derivative tests
x y O −2 −1 1 2 −7 9 (0, 1) (2, 9) (−2, −7)
g(x)=x3+1g(x)=x^3+1 climbs all the way from left to right. At (0,1)(0,1) it is momentarily flat (g′(0)=0g'(0)=0⁠) but keeps rising, so that point is neither a peak nor a valley.
Result

(i) Minimum value 33 (at x=12x=\tfrac12⁠); no maximum value.

(ii) Minimum value −2-2 (at x=−23x=-\tfrac23⁠); no maximum value.

(iii) Maximum value 1010 (at x=1x=1⁠); no minimum value.

(iv) Neither a maximum nor a minimum value.

(Textbook answer: (i) Minimum Value = 3 (ii) Minimum Value = −2 (iii) Maximum Value = 10 (iv) Neither minimum nor maximum value. The key does not mention the missing extreme in (i)–(iii); the steps above show it does not exist.)

Check it with numbers

(i) f(0.4)=(−0.2)2+3=3.04f(0.4)=(-0.2)^2+3=3.04⁠, f(0.5)=3f(0.5)=3⁠, f(0.6)=3.04f(0.6)=3.04⁠: lowest in the middle. ✓ And f(10)=192+3=364f(10)=19^2+3=364⁠, already far above 33⁠.

(ii) f(−0.6)=−1.96f(-0.6)=-1.96⁠, f(−0.7)=−1.99f(-0.7)=-1.99⁠, f(−23)=−2f(-\tfrac23)=-2⁠: both neighbours are above −2-2⁠. ✓

(iii) f(0.9)=f(1.1)=9.99<10=f(1)f(0.9)=f(1.1)=9.99\lt10=f(1)⁠. ✓

(iv) g(−0.1)=0.999g(-0.1)=0.999⁠, g(0)=1g(0)=1⁠, g(0.1)=1.001g(0.1)=1.001⁠: rising straight through 00⁠. ✓

Common mistakes
  1. Answering only the extreme that exists. "If any" asks about both; say clearly "no maximum value" and give the reason.
  2. Saying x3+1x^3+1 has a minimum or maximum at 00 because g′(0)=0g'(0)=0⁠. A zero derivative only flags a candidate; gg rises on both sides of 00⁠.
  3. Completing the square wrongly in (ii). 9x2+12x+29x^2+12x+2 is (3x+2)2−2(3x+2)^2-2⁠, not (3x+2)2+2(3x+2)^2+2⁠: you add 44 to complete the square, so you must subtract 44 again.
  4. Forgetting to flip the inequality in (iii). (x−1)2≥0(x-1)^2\ge0 gives −(x−1)2≤0-(x-1)^2\le0⁠, an upper bound, not a lower one.
What to take away
  1. For a quadratic, completing the square gives the extreme value and where it occurs, with no calculus at all.
  2. A minimum (or maximum) value needs both a bound and a point where the bound is reached.
  3. To show an extreme value does not exist, beat every candidate point with an explicit recipe.
  4. f′(c)=0f'(c)=0 is a candidate test, not a verdict: x3x^3 at 00 is the standard counterexample.

Every fact used, and where it comes from

FactUsed inSource
Maximum value and minimum value of ff on I (printed with >\gt⁠, <\lt⁠; read as ≥\ge⁠, ≤\le⁠)Every partDefinition 3, Section 6.4 (Lesson: maxima and minima)
t2≥0t^2\ge0⁠, equality only at t=0t=0⁠; completing the squareParts (i)–(iii), Steps 1–2; Part (iv), Step 1Algebra; compare Example 14 (x2x^2 has minimum 00⁠, no maximum)
Expansion (c+1)3=c3+3c2+3c+1(c+1)^3=c^3+3c^2+3c+1Part (iv), Step 1Binomial expansion
A strictly rising function on an open stretch has no maximum or minimum valuePart (iv), Step 2Example 16, Section 6.4
Chain rule, power ruleParts (i)–(iii), last step; Part (iv), Step 3Chapter 5 (Continuity and Differentiability)
Second derivative testParts (i)–(iii), last stepTheorem 4, Section 6.4 (Lesson: tests)
f′(c)=0f'(c)=0 need not give an extremum (example x3x^3⁠); first derivative test, point of inflectionPart (iv), Step 3Theorem 2 and the Remark after it; Theorem 3(iii), Section 6.4
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