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Chapter 6 · Application of Derivatives

Exercise 6.2 · Q6: Strictly increasing or decreasing: five functions

Exercise 6.2Increasing and decreasing functions · Section 6.3
The question

Find the intervals in which the following functions are strictly increasing or decreasing:

(a) x2+2x−5x^2 + 2x - 5     (b) 10−6x−2x210 - 6x - 2x^2

(c) −2x3−9x2−12x+1-2x^3 - 9x^2 - 12x + 1     (d) 6−9x−x26 - 9x - x^2

(e) (x+1)3(x−3)3(x + 1)^3 (x - 3)^3

Exam answer

(a) Let f(x)=x2+2x−5f(x) = x^2 + 2x - 5⁠. Then

f′(x)=2x+2=2(x+1)f'(x) = 2x + 2 = 2(x + 1)

f′(x)=0f'(x) = 0 gives x=−1x = -1⁠.

In (−∞,−1)(-\infty, -1)⁠, f′(x)<0f'(x) < 0⁠. In (−1,∞)(-1, \infty)⁠, f′(x)>0f'(x) > 0⁠.

So ff is strictly decreasing on (−∞,−1)(-\infty, -1) and strictly increasing on (−1,∞)(-1, \infty)⁠.

(b) Let f(x)=10−6x−2x2f(x) = 10 - 6x - 2x^2⁠. Then

f′(x)=−6−4x=−2(2x+3)f'(x) = -6 - 4x = -2(2x + 3)

f′(x)=0f'(x) = 0 gives x=−32x = -\frac32⁠.

In (−∞,−32)\left(-\infty, -\frac32\right)⁠, f′(x)>0f'(x) > 0⁠. In (−32,∞)\left(-\frac32, \infty\right)⁠, f′(x)<0f'(x) < 0⁠.

So ff is strictly increasing on (−∞,−32)\left(-\infty, -\frac32\right) and strictly decreasing on (−32,∞)\left(-\frac32, \infty\right)⁠.

(c) Let f(x)=−2x3−9x2−12x+1f(x) = -2x^3 - 9x^2 - 12x + 1⁠. Then

f′(x)=−6x2−18x−12=−6(x2+3x+2)=−6(x+1)(x+2)\begin{aligned} f'(x) &= -6x^2 - 18x - 12 \\ &= -6(x^2 + 3x + 2) \\ &= -6(x + 1)(x + 2) \end{aligned}

f′(x)=0f'(x) = 0 gives x=−2x = -2 or x=−1x = -1⁠.

In (−∞,−2)(-\infty, -2)⁠, f′(x)<0f'(x) < 0⁠. In (−2,−1)(-2, -1)⁠, f′(x)>0f'(x) > 0⁠. In (−1,∞)(-1, \infty)⁠, f′(x)<0f'(x) < 0⁠.

So ff is strictly increasing on (−2,−1)(-2, -1)⁠, and strictly decreasing on (−∞,−2)(-\infty, -2) and (−1,∞)(-1, \infty)⁠.

(d) Let f(x)=6−9x−x2f(x) = 6 - 9x - x^2⁠. Then

f′(x)=−9−2x=−(2x+9)f'(x) = -9 - 2x = -(2x + 9)

f′(x)=0f'(x) = 0 gives x=−92x = -\frac92⁠.

In (−∞,−92)\left(-\infty, -\frac92\right)⁠, f′(x)>0f'(x) > 0⁠. In (−92,∞)\left(-\frac92, \infty\right)⁠, f′(x)<0f'(x) < 0⁠.

So ff is strictly increasing on (−∞,−92)\left(-\infty, -\frac92\right) and strictly decreasing on (−92,∞)\left(-\frac92, \infty\right)⁠.

(e) Let f(x)=(x+1)3(x−3)3f(x) = (x + 1)^3(x - 3)^3⁠. By the product rule,

f′(x)=3(x+1)2(x−3)3+3(x+1)3(x−3)2=3(x+1)2(x−3)2(2x−2)=6(x+1)2(x−3)2(x−1)\begin{aligned} f'(x) &= 3(x + 1)^2(x - 3)^3 \\ &\quad + 3(x + 1)^3(x - 3)^2 \\ &= 3(x + 1)^2(x - 3)^2(2x - 2) \\ &= 6(x + 1)^2(x - 3)^2(x - 1) \end{aligned}

f′(x)=0f'(x) = 0 gives x=−1x = -1⁠, 11 or 33⁠.

In (−∞,−1)(-\infty, -1) and (−1,1)(-1, 1)⁠, f′(x)<0f'(x) < 0⁠. In (1,3)(1, 3) and (3,∞)(3, \infty)⁠, f′(x)>0f'(x) > 0⁠.

So ff is strictly increasing on (1,3)(1, 3) and (3,∞)(3, \infty)⁠, and strictly decreasing on (−∞,−1)(-\infty, -1) and (−1,1)(-1, 1)⁠.

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

The one idea this problem is built on

Five functions, one method: factorise the derivative, find its zeros, and read its sign on each interval between them. Positive f′f' means strictly increasing, negative means strictly decreasing. Part (e) adds a twist: its derivative is zero at three points, but it changes sign at only one of them, so the function keeps falling (or rising) straight through the other two.

Given, to find, and the plan
Given
Five polynomial functions of a real variable xx⁠.
To find
For each, the intervals on which it is strictly increasing and those on which it is strictly decreasing.
Plan
For each part: (1) differentiate and factorise; (2) solve f′(x)=0f'(x) = 0⁠; (3) sign chart; (4) conclude by Theorem 1 and its proof (strict inequality because f′f' is strictly positive or negative inside each interval).

Basics you need

1Strictly increasing and strictly decreasing (Definition 1 (iv), (v))

On an interval I: strictly increasing means x1<x2⇒f(x1)<f(x2)x_1 < x_2 \Rightarrow f(x_1) < f(x_2)⁠; strictly decreasing means x1<x2⇒f(x1)>f(x2)x_1 < x_2 \Rightarrow f(x_1) > f(x_2)⁠, for all x1,x2x_1, x_2 in I. The question asks for these strict versions. Lesson: increasing and decreasing functions

2Theorem 1 and its proof

Let ff be continuous on [a,b][a, b] and differentiable on (a,b)(a, b)⁠. The Mean Value step in the book's proof, f(x2)−f(x1)=f′(c)(x2−x1)f(x_2) - f(x_1) = f'(c)(x_2 - x_1)⁠, shows: if f′>0f' > 0 on (a,b)(a, b) then ff is strictly increasing on [a,b][a, b]⁠; if f′<0f' < 0⁠, strictly decreasing. For an interval like (−∞,−1)(-\infty, -1)⁠, apply this on [x1,x2][x_1, x_2] for every pair inside.

3Factorising

Take out common factors first (including a minus sign), then factorise the quadratic that is left. Basics: factorising and the discriminant

4Sign of a product

A product is positive when it has an even number of negative factors, negative when odd. A square such as (x+1)2(x + 1)^2 is never negative: it is positive except at its root, where it is 0. Basics: sign of a product

5Product rule and the power rule

(uv)′=u′v+uv′(uv)' = u'v + uv' (Section 5.3), and ddx(x−a)n=n(x−a)n−1\frac{d}{dx}(x - a)^n = n(x - a)^{n-1} (power rule with the chain rule, inner function x−ax - a of derivative 1). Needed in part (e). Basics: the chain rule

Solution

All five functions are polynomials, so each is continuous and differentiable at every real number. Theorem 1 can therefore be used on any closed interval, and only the sign of f′f' needs work.

Part (a): f(x)=x2+2x−5f(x) = x^2 + 2x - 5

  1. Derivative and its zero

    f′(x)=2x+2=2(x+1),f′(x)=0  ⟺  x=−1.f'(x) = 2x + 2 = 2(x + 1), \qquad f'(x) = 0 \iff x = -1.
    Power rule term by term; the constant −5-5 has derivative 0. Taking out the 2 shows that the sign of f′f' is the sign of x+1x + 1⁠.

  2. Sign and conclusion

    For x<−1x < -1⁠: x+1<0x + 1 < 0⁠, so f′(x)<0f'(x) < 0 (test: f′(−2)=−2f'(-2) = -2⁠). For x>−1x > -1⁠: f′(x)>0f'(x) > 0 (test: f′(0)=2f'(0) = 2⁠).

    x −1 f′(x) = 2(x + 1) − + 0 f decreasing increasing
    Part (a), not to scale: f′(x)=2(x+1)f'(x) = 2(x + 1) changes from negative to positive at x=−1x = -1⁠.

    By Theorem 1 and its proof, ff is strictly decreasing on (−∞,−1)(-\infty, -1) and strictly increasing on (−1,∞)(-1, \infty).

    An upward parabola with vertex at x=−b2a=−1x = -\frac{b}{2a} = -1⁠: falling, then rising, as the chart says.
    A positive constant factor (here 2) never changes the sign. Basics: sign of a product

Part (b): f(x)=10−6x−2x2f(x) = 10 - 6x - 2x^2

  1. Derivative and its zero

    f′(x)=−6−4x=−2(2x+3),f′(x)=0  ⟺  x=−32.f'(x) = -6 - 4x = -2(2x + 3), \qquad f'(x) = 0 \iff x = -\tfrac32.
    Taking out −2-2 (not 22⁠) leaves 2x+32x + 3⁠, whose sign is easy to read. The minus sign in front then flips that sign.

  2. Sign and conclusion

    For x<−32x < -\frac32⁠: 2x+3<02x + 3 < 0⁠, so f′(x)=−2(2x+3)>0f'(x) = -2(2x + 3) > 0 (test: f′(−2)=2f'(-2) = 2⁠). For x>−32x > -\frac32⁠: 2x+3>02x + 3 > 0⁠, so f′(x)<0f'(x) < 0 (test: f′(0)=−6f'(0) = -6⁠).

    x −3/2 2x + 3 − + 0 f′(x) + − 0 f increasing decreasing
    Part (b), not to scale: the factor 2x+32x + 3 goes from negative to positive at −32-\frac32⁠; the factor −2-2 reverses both signs.

    So ff is strictly increasing on (−∞,−32)\left(-\infty, -\frac32\right) and strictly decreasing on (−32,∞)\left(-\frac32, \infty\right).

    A downward parabola (coefficient of x2x^2 is −2<0-2 < 0⁠) rises up to its vertex at x=−b2a=−−62(−2)=−32x = -\frac{b}{2a} = -\frac{-6}{2(-2)} = -\frac32 and falls after it.
    A negative constant factor reverses every sign in the chart. Basics: sign of a product

Part (c): f(x)=−2x3−9x2−12x+1f(x) = -2x^3 - 9x^2 - 12x + 1

  1. Derivative, factorised

    f′(x)=−6x2−18x−12=−6(x2+3x+2)=−6(x+1)(x+2).\begin{aligned} f'(x) &= -6x^2 - 18x - 12\\ &= -6(x^2 + 3x + 2)\\ &= -6(x + 1)(x + 2). \end{aligned}

    Check: (x+1)(x+2)=x2+3x+2(x + 1)(x + 2) = x^2 + 3x + 2⁠. ✓ So f′(x)=0  ⟺  x=−2f'(x) = 0 \iff x = -2 or x=−1x = -1⁠.

    For x2+3x+2x^2 + 3x + 2 we need two numbers with product 2 and sum 3: they are 1 and 2. Taking out −6-6 first keeps the quadratic monic and easy to factorise.
    Product 2, sum 3: the pair 1, 2. Basics: factorising and the discriminant

  2. Sign chart

    • x<−2x < -2⁠: (x+2)(x+1)(x + 2)(x + 1) has two negative factors, so it is positive; times −6-6⁠: f′<0f' < 0⁠. (Test: f′(−3)=−12f'(-3) = -12⁠.)
    • −2<x<−1-2 < x < -1⁠: one negative factor, product negative; times −6-6⁠: f′>0f' > 0⁠. (Test: f′(−1.5)=1.5f'(-1.5) = 1.5⁠.)
    • x>−1x > -1⁠: product positive; times −6-6⁠: f′<0f' < 0⁠. (Test: f′(0)=−12f'(0) = -12⁠.)
    x −2 −1 x + 2 − + + 0 x + 1 − − + 0 f′(x) − + − 0 0 f decreasing increasing decreasing
    Part (c), not to scale: f′(x)=−6(x+2)(x+1)f'(x) = -6(x + 2)(x + 1)⁠. The factor −6-6 is not drawn; it turns the sign of (x+2)(x+1)(x + 2)(x + 1) round.
    Keep track of the −6-6⁠: it is the easiest sign to lose, and losing it swaps every answer.

  3. Conclusion

    ff is strictly increasing on (−2,−1)(-2, -1) and strictly decreasing on (−∞,−2)(-\infty, -2) and on (−1,∞)(-1, \infty).

    "On (−∞,−2)(-\infty, -2) and on (−1,∞)(-1, \infty)": separately. On their union ff is not decreasing: −2.1-2.1 lies in the first interval and −0.9-0.9 in the second, yet f(−2.1)=5.032<5.968=f(−0.9)f(-2.1) = 5.032 < 5.968 = f(-0.9)⁠. The values went up in between, on (−2,−1)(-2, -1)⁠, where ff climbs from f(−2)=5f(-2) = 5 to f(−1)=6f(-1) = 6⁠.

Part (d): f(x)=6−9x−x2f(x) = 6 - 9x - x^2

  1. Derivative and its zero

    f′(x)=−9−2x=−(2x+9),f′(x)=0  ⟺  x=−92.f'(x) = -9 - 2x = -(2x + 9), \qquad f'(x) = 0 \iff x = -\tfrac92.

  2. Sign and conclusion

    For x<−92x < -\frac92⁠: 2x+9<02x + 9 < 0⁠, so f′(x)>0f'(x) > 0 (test: f′(−5)=1f'(-5) = 1⁠). For x>−92x > -\frac92⁠: f′(x)<0f'(x) < 0 (test: f′(0)=−9f'(0) = -9⁠).

    x −9/2 2x + 9 − + 0 f′(x) + − 0 f increasing decreasing
    Part (d), not to scale: f′(x)=−(2x+9)f'(x) = -(2x + 9) changes from positive to negative at −92-\frac92⁠.

    So ff is strictly increasing on (−∞,−92)\left(-\infty, -\frac92\right) and strictly decreasing on (−92,∞)\left(-\frac92, \infty\right).

    Downward parabola, vertex at x=−b2a=−−92(−1)=−92x = -\frac{b}{2a} = -\frac{-9}{2(-1)} = -\frac92⁠: rising before, falling after.

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

  1. Product rule

    Take u=(x+1)3u = (x + 1)^3 and v=(x−3)3v = (x - 3)^3⁠. Then u′=3(x+1)2u' = 3(x + 1)^2 and v′=3(x−3)2v' = 3(x - 3)^2⁠, so

    f′(x)=u′v+uv′=3(x+1)2(x−3)3+3(x+1)3(x−3)2.\begin{aligned} f'(x) &= u'v + uv'\\ &= 3(x + 1)^2(x - 3)^3 + 3(x + 1)^3(x - 3)^2. \end{aligned}
    Expanding (x+1)3(x−3)3(x + 1)^3(x - 3)^3 first would give a degree-6 polynomial that is hard to factorise back. The product rule keeps the factors visible.
    Chain rule with inner function x+1x + 1⁠, whose derivative is 1: ddx(x+1)3=3(x+1)2⋅1\frac{d}{dx}(x + 1)^3 = 3(x + 1)^2 \cdot 1⁠. Basics: the chain rule

  2. Take out the common factor

    Both terms contain 3(x+1)2(x−3)23(x + 1)^2(x - 3)^2⁠:

    f′(x)=3(x+1)2(x−3)2[(x−3)+(x+1)]=3(x+1)2(x−3)2(2x−2)=6(x+1)2(x−3)2(x−1).\begin{aligned} f'(x) &= 3(x + 1)^2(x - 3)^2\big[(x - 3) + (x + 1)\big]\\ &= 3(x + 1)^2(x - 3)^2(2x - 2)\\ &= 6(x + 1)^2(x - 3)^2(x - 1). \end{aligned}

    So f′(x)=0  ⟺  x=−1,1f'(x) = 0 \iff x = -1, 1 or 33⁠.

    The first term has one extra factor (x−3)(x - 3) and the second one extra (x+1)(x + 1)⁠; those are what is left in the bracket. Then 2x−2=2(x−1)2x - 2 = 2(x - 1)⁠.

  3. Sign chart: only one factor can change sign

    6>06 > 0⁠, and the squares (x+1)2(x + 1)^2 and (x−3)2(x - 3)^2 are positive except at x=−1x = -1 and x=3x = 3 (where they are 0). So, away from −1,1,3-1, 1, 3⁠, the sign of f′(x)f'(x) is the sign of x−1x - 1⁠:

    • f′(x)<0f'(x) < 0 on (−∞,−1)(-\infty, -1) and on (−1,1)(-1, 1) (tests: f′(−2)=−450f'(-2) = -450⁠, f′(0)=−54f'(0) = -54⁠);
    • f′(x)>0f'(x) > 0 on (1,3)(1, 3) and on (3,∞)(3, \infty) (tests: f′(2)=54f'(2) = 54⁠, f′(4)=450f'(4) = 450⁠).
    x −1 1 3 (x + 1)² + + + + 0 (x − 3)² + + + + 0 x − 1 − − + + 0 f′(x) − − + + 0 0 0 f decreasing decreasing increasing increasing
    Part (e), not to scale: f′(x)=6(x+1)2(x−3)2(x−1)f'(x) = 6(x + 1)^2(x - 3)^2(x - 1)⁠. The squared factors touch 0 but never go negative, so f′f' changes sign only at x=1x = 1⁠.
    At x=−1x = -1 and x=3x = 3 the derivative is 0 but keeps the same sign on both sides. The graph is momentarily flat there without turning.
    A square is never negative. Basics: sign of a product

  4. Conclusion, and a stronger statement

    By Theorem 1 and its proof, ff is strictly decreasing on (−∞,−1)(-\infty, -1) and on (−1,1)(-1, 1), and strictly increasing on (1,3)(1, 3) and on (3,∞)(3, \infty). This is the answer key's form.

    In fact more is true. ff is continuous at −1-1⁠, so Theorem 1 on [x1,x2][x_1, x_2] gives strict decrease on (−∞,−1](-\infty, -1] and on [−1,1][-1, 1]⁠. Pairs inside one of these two pieces are covered already, and for a pair on opposite sides, x1<−1<x2≤1x_1 < -1 < x_2 \le 1⁠, we get f(x1)>f(−1)>f(x2)f(x_1) > f(-1) > f(x_2)⁠. So ff is strictly decreasing on all of (−∞,1](-\infty, 1], and in the same way strictly increasing on all of [1,∞)[1, \infty).

    x y −1 1 2 3 −60 −30 20 (−1, 0) (1, −64) (3, 0)
    y=(x+1)3(x−3)3y = (x + 1)^3(x - 3)^3⁠: falling (rose) all the way down to (1,−64)(1, -64)⁠, then rising (green). At (−1,0)(-1, 0) and (3,0)(3, 0) the curve flattens but does not turn. Vertical scale compressed.
    A neat cross-check: (x+1)(x−3)=x2−2x−3=(x−1)2−4(x + 1)(x - 3) = x^2 - 2x - 3 = (x - 1)^2 - 4⁠, so f(x)=[(x−1)2−4]3f(x) = \left[(x - 1)^2 - 4\right]^3⁠. For x<1x < 1⁠, as xx grows towards 1 the quantity (x−1)2(x - 1)^2 shrinks, so the bracket shrinks, and cubing keeps the order; so ff falls. For x>1x > 1 the bracket grows, so ff rises. Exactly the conclusion above.
Result

(a) Strictly decreasing on (−∞,−1)(-\infty, -1)⁠, strictly increasing on (−1,∞)(-1, \infty)⁠.
(b) Strictly increasing on (−∞,−32)\left(-\infty, -\frac32\right)⁠, strictly decreasing on (−32,∞)\left(-\frac32, \infty\right)⁠.
(c) Strictly increasing on (−2,−1)(-2, -1)⁠; strictly decreasing on (−∞,−2)(-\infty, -2) and on (−1,∞)(-1, \infty)⁠.
(d) Strictly increasing on (−∞,−92)\left(-\infty, -\frac92\right)⁠, strictly decreasing on (−92,∞)\left(-\frac92, \infty\right)⁠.
(e) Strictly increasing on (1,3)(1, 3) and (3,∞)(3, \infty)⁠, strictly decreasing on (−∞,−1)(-\infty, -1) and (−1,1)(-1, 1)⁠; in fact strictly decreasing on (−∞,1](-\infty, 1] and strictly increasing on [1,∞)[1, \infty)⁠.
(Textbook answer: (a) decreasing for x<−1x < -1⁠, increasing for x>−1x > -1⁠; (b) decreasing for x>−32x > -\frac32⁠, increasing for x<−32x < -\frac32⁠; (c) increasing for −2<x<−1-2 < x < -1⁠, decreasing for x<−2x < -2 and x>−1x > -1⁠; (d) increasing for x<−92x < -\frac92⁠, decreasing for x>−92x > -\frac92⁠; (e) increasing in (1,3)(1, 3) and (3,∞)(3, \infty)⁠, decreasing in (−∞,−1)(-\infty, -1) and (−1,1)(-1, 1)⁠.)

Check it with numbers

(a) f(−3)=−2f(-3) = -2⁠, f(−1)=−6f(-1) = -6⁠, f(1)=−2f(1) = -2⁠: down then up. (b) f(−3)=10f(-3) = 10⁠, f(−1.5)=14.5f(-1.5) = 14.5⁠, f(0)=10f(0) = 10⁠: up then down. (c) f(−3)=10f(-3) = 10⁠, f(−2)=5f(-2) = 5⁠, f(−1.5)=5.5f(-1.5) = 5.5⁠, f(−1)=6f(-1) = 6⁠, f(0)=1f(0) = 1⁠: down, up, down. (d) f(−6)=24f(-6) = 24⁠, f(−4.5)=26.25f(-4.5) = 26.25⁠, f(−3)=24f(-3) = 24⁠: up then down. (e) Using f(x)=[(x+1)(x−3)]3f(x) = \left[(x + 1)(x - 3)\right]^3⁠: f(−2)=53=125f(-2) = 5^3 = 125⁠, f(−1)=0f(-1) = 0⁠, f(0)=(−3)3=−27f(0) = (-3)^3 = -27⁠, f(1)=(−4)3=−64f(1) = (-4)^3 = -64⁠, f(2)=−27f(2) = -27⁠, f(3)=0f(3) = 0⁠, f(4)=125f(4) = 125⁠. Falling all the way to x=1x = 1 (even through x=−1x = -1⁠), rising after it (even through x=3x = 3⁠).

Common mistakes
  1. Losing a negative constant (the −2-2 in (b), the −6-6 in (c), the −1-1 in (d)). It reverses every sign in the chart.
  2. In (e), treating every zero of f′f' as a change of direction. At x=−1x = -1 and x=3x = 3 the squared factors are 0 but do not change sign.
  3. Expanding (x+1)3(x−3)3(x + 1)^3(x - 3)^3 before differentiating. Legal but long, and the factorised derivative is hard to recover. Use the product rule.
  4. Joining intervals with "∪\cup". In (c), ff is decreasing on each of (−∞,−2)(-\infty, -2) and (−1,∞)(-1, \infty)⁠, not on their union.
What to take away
  1. Factorise f′f' completely and read the sign factor by factor; constants and squares are part of the chart too.
  2. A zero of f′f' from an even power does not change the sign of f′f'⁠: the function keeps its direction through it.
  3. For a quadratic ax2+bx+cax^2 + bx + c⁠, the turning point is x=−b2ax = -\frac{b}{2a}⁠; a>0a > 0 means down-then-up, a<0a < 0 means up-then-down.

Every fact used, and where it comes from

FactUsed inSource
Strictly increasing and strictly decreasingEvery partDefinition 1 (iv), (v), Section 6.3 (Lesson: increasing and decreasing functions)
Derivatives of polynomials; product ruleEvery part; Part (e), Step 1Algebra of derivatives (rules (1), (2)), Section 5.3; Table 5.3
ddx(x+a)3=3(x+a)2\frac{d}{dx}(x + a)^3 = 3(x + a)^2Part (e), Step 1Chain rule, Theorem 4, Section 5.3.1
Sign of f′f' gives strict increase or decreaseEvery partTheorem 1 and its proof, Section 6.3
Joining intervals at a point of continuityPart (e), Step 4Theorem 1 (closed interval) and Definition 1
Interval method with a sign tableEvery partExamples 10 and 11, Section 6.3
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