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Chapter 6 · Application of Derivatives

Example 17: Local extrema of x3−3x+3x^3-3x+3

ExamplesFirst derivative test · Section 6.4
The question

Find all points of local maxima and local minima of the function ff given by

f(x)=x3−3x+3.f(x) = x^3 - 3x + 3.
Exam answer f′(x)=3x2−3=3(x−1)(x+1)\begin{aligned} f'(x) &= 3x^2 - 3 \\ &= 3(x - 1)(x + 1) \end{aligned}

f′(x)=0⇒x=−1,1f'(x) = 0 \Rightarrow x = -1, 1⁠.

f′′(x)=6xf''(x) = 6x

f′′(−1)=−6<0f''(-1) = -6 < 0⁠, so x=−1x = -1 is a point of local maxima.

f′′(1)=6>0f''(1) = 6 > 0⁠, so x=1x = 1 is a point of local minima.

f(−1)=−1+3+3=5f(1)=1−3+3=1\begin{aligned} f(-1) &= -1 + 3 + 3 = 5 \\ f(1) &= 1 - 3 + 3 = 1 \end{aligned}

Hence x=−1x = -1 is a point of local maxima, with local maximum value 5, and x=1x = 1 is a point of local minima, with local minimum value 1.

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

The one idea this problem is built on

A local maximum is the top of a hill: the graph rises up to it and falls after it, so f′f' changes from ++ to −-⁠. A local minimum is the bottom of a valley: f′f' changes from −- to ++⁠. So we find the only places where f′f' can change sign (where f′(x)=0f'(x) = 0⁠; here ff is differentiable everywhere) and look at the sign of f′f' just to the left and just to the right of each. That is the first derivative test.

Given, to find, and the plan
Given
f(x)=x3−3x+3f(x) = x^3 - 3x + 3 for all real xx⁠.
To find
All points of local maxima and of local minima (and the local maximum and minimum values).
Plan
Differentiate and factorise; the critical points are where f′(x)=0f'(x) = 0⁠; make a sign chart; apply Theorem 3 at each critical point; cross-check with the second derivative.

Basics you need

1Local maxima and minima (Definition 4, Section 6.4)

An interior point cc is a point of local maxima if there is an h>0h > 0 with f(c)≥f(x)f(c) \ge f(x) for all xx in (c−h,c+h)(c-h, c+h)⁠; a point of local minima if f(c)≤f(x)f(c) \le f(x) there. "Local" means: compared only with nearby points. Lesson: maxima and minima

2Critical points and Theorem 2

If ff has a local maximum or minimum at cc⁠, then f′(c)=0f'(c) = 0 or ff is not differentiable at cc (Theorem 2). Such points are called critical points (Note before Theorem 3). Only critical points can be local extrema.

3First Derivative Test (Theorem 3, Section 6.4)

Let ff be continuous at a critical point cc⁠. (i) If f′f' changes sign from positive to negative as xx increases through cc⁠, then cc is a point of local maxima. (ii) If it changes from negative to positive, cc is a point of local minima. (iii) If it does not change sign, cc is neither (a point of inflection). Lesson: first and second derivative tests

4Sign of a product

x−ax - a is negative left of aa and positive right of aa⁠; multiply the signs of the factors. Basics: sign of a product

Solution

  1. Differentiate and factorise

    By the power, constant-multiple and sum rules,

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

    using the difference of squares x2−1=(x−1)(x+1)x^2 - 1 = (x-1)(x+1)⁠.

    A factorised derivative shows at once where it is zero and what its sign is on each side, which is all the first derivative test needs.

  2. Find the critical points

    ff is a polynomial, so it is differentiable at every real number: there are no critical points of the "not differentiable" kind. The others are where f′(x)=0f'(x) = 0⁠:

    3(x−1)(x+1)=0  ⟺  x=1  or  x=−1.3(x-1)(x+1) = 0 \iff x = 1 \ \text{ or } \ x = -1.

    So x=±1x = \pm1 are the only critical points, and by Theorem 2 the only points that could be local maxima or minima.

    Theorem 2 narrows an infinite search to two points. It does not say these points are extrema (the Remark after Theorem 2 gives x3x^3 at 0 as a counterexample), so each must still be tested.

  3. Sign chart of f′(x)=3(x+1)(x−1)f'(x) = 3(x+1)(x-1)

    −∞ +∞ −1 1 x + 1 − + + x − 1 − − + f′(x) + − + f(x) increasing decreasing increasing at the point local max local min
    Sign chart (not to scale). f′f' is ++ on (−∞,−1)(-\infty,-1)⁠, −- on (−1,1)(-1,1) and ++ on (1,∞)(1,\infty)⁠.

    Points close to each critical point, as in the book's table:

    NearSideValue of f′f'Sign
    x=1x = 1right, say 1.11.13(1.21)−3=0.633(1.21) - 3 = 0.63>0> 0
    left, say 0.90.93(0.81)−3=−0.573(0.81) - 3 = -0.57<0< 0
    x=−1x = -1right, say −0.9-0.9−0.57-0.57<0< 0
    left, say −1.1-1.10.630.63>0> 0
    The sign does not hold only at 1.1 or 0.9: by the factor signs, f′<0f' < 0 at every point of (−1,1)(-1,1) and f′>0f' > 0 at every point of (1,∞)(1,\infty) and of (−∞,−1)(-\infty,-1)⁠. That is what "sufficiently close to and to the right/left of cc" in Theorem 3 requires.
    Basics: sign of a product

  4. At x=1x = 1⁠: a point of local minima

    As xx increases through 1, f′(x)f'(x) changes from negative (on (−1,1)(-1,1)⁠) to positive (on (1,∞)(1,\infty)⁠). ff is continuous at 1. By Theorem 3(ii), x=1x = 1 is a point of local minima, and the local minimum value is

    f(1)=1−3+3=1.f(1) = 1 - 3 + 3 = 1.
    Falling until x=1x = 1⁠, rising after: the graph has a valley floor at x=1x = 1⁠. Indeed f(x)−1=(x−1)2(x+2)≥0f(x) - 1 = (x-1)^2(x+2) \ge 0 for all x>−2x > -2⁠, so f(x)≥f(1)f(x) \ge f(1) on the whole neighbourhood (−2,4)(-2, 4)⁠, exactly as Definition 4(b) asks.

  5. At x=−1x = -1⁠: a point of local maxima

    As xx increases through −1-1⁠, f′(x)f'(x) changes from positive to negative. By Theorem 3(i), x=−1x = -1 is a point of local maxima, and the local maximum value is

    f(−1)=−1+3+3=5.f(-1) = -1 + 3 + 3 = 5.
    Rising until x=−1x = -1⁠, falling after: a hilltop. Directly: f(x)−5=(x+1)2(x−2)≤0f(x) - 5 = (x+1)^2(x-2) \le 0 for all x<2x < 2⁠, so f(x)≤f(−1)f(x) \le f(-1) on the neighbourhood (−4,2)(-4, 2)⁠.
    x y O −2 2 −2 2 4 6 8 −1 1 local max (−1, 5) local min (1, 1)
    The graph of f(x)=x3−3x+3f(x) = x^3 - 3x + 3⁠: green where rising (f′>0f' > 0⁠), rose where falling (f′<0f' < 0⁠). The hilltop (−1,5)(-1, 5) and the valley floor (1,1)(1, 1) are the local maximum and minimum. Vertical scale smaller than horizontal.

  6. Cross-check with the second derivative, and "local" versus "largest"

    f′′(x)=6xf''(x) = 6x⁠, so f′′(1)=6>0f''(1) = 6 > 0 and f′′(−1)=−6<0f''(-1) = -6 < 0⁠. By the Second Derivative Test (Theorem 4) this again gives a local minimum at 1 and a local maximum at −1-1⁠.

    These are only local: f(3)=27−9+3=21>5f(3) = 27 - 9 + 3 = 21 > 5 and f(−3)=−27+9+3=−15<1f(-3) = -27 + 9 + 3 = -15 < 1⁠. In fact ff has no maximum or minimum value on all of R\R⁠, since a cubic takes arbitrarily large positive and negative values.

    Definition 4 compares f(c)f(c) only with values on a small interval around cc⁠. A hilltop need not be the highest point of the whole graph.
    Lesson: first and second derivative tests
Result

f′(x)=3(x−1)(x+1)f'(x) = 3(x-1)(x+1)⁠, critical points x=±1x = \pm1⁠. x=−1x = -1 is a point of local maxima, local maximum value f(−1)=5f(-1) = 5⁠; x=1x = 1 is a point of local minima, local minimum value f(1)=1f(1) = 1⁠. There are no other local extrema. This agrees with the book's solution.

Check it with numbers
  • Around x=1x = 1⁠: f(0.9)=1.029f(0.9) = 1.029⁠, f(1)=1f(1) = 1⁠, f(1.1)=1.031f(1.1) = 1.031⁠. The middle value is the smallest: a valley.
  • Around x=−1x = -1⁠: f(−1.1)=4.969f(-1.1) = 4.969⁠, f(−1)=5f(-1) = 5⁠, f(−0.9)=4.971f(-0.9) = 4.971⁠. The middle value is the largest: a hill.
Common mistakes
  1. Stopping at f′(x)=0f'(x) = 0⁠. Solving gives candidates only; the sign change (or f′′f''⁠) decides which is which, and whether each is an extremum at all.
  2. Mixing up the two cases. ++ then −- is a maximum (up, then down); −- then ++ is a minimum.
  3. Giving xx when the value is asked, or the reverse. The point of local maxima is x=−1x = -1⁠; the local maximum value is f(−1)=5f(-1) = 5⁠.
  4. Calling 5 the maximum value of ff⁠. It is only a local maximum value; f(3)=21f(3) = 21 is bigger.
What to take away
  1. First derivative test: critical points, then the sign of f′f' just left and just right of each.
  2. A factorised f′f' with a sign chart answers both "where is ff increasing" and "where are the local extrema" at once.
  3. Local extrema compare a point with its neighbours only.

Every fact used, and where it comes from

FactUsed inSource
Derivative of a polynomial; x2−1=(x−1)(x+1)x^2 - 1 = (x-1)(x+1)Step 1Derivative rules; difference of squares
Local extremum implies f′=0f' = 0 or not differentiable; critical pointStep 2Theorem 2 and the Note before Theorem 3, Section 6.4
Sign of a product of linear factorsStep 3Basics: sign of a product
Point of local maxima / minimaSteps 4–6Definition 4, Section 6.4 (Lesson)
First Derivative TestSteps 4, 5Theorem 3, Section 6.4 (Lesson)
Second Derivative TestStep 6Theorem 4, Section 6.4
Values in the tables and check boxSteps 3–6; CheckRecomputed in the site's verification script
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