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Chapter 6 · Application of Derivatives

Exercise 6.3 · Q3: Local maxima and minima of eight functions, from x2x^2 to x1−xx\sqrt{1-x}

Exercise 6.3Local maxima and minima · First and second derivative tests · Section 6.4
The question

3. Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:

(i) f(x)=x2f(x)=x^2    (ii) g(x)=x3−3xg(x)=x^3-3x

(iii) h(x)=sin⁡x+cos⁡x, 0<x<π2h(x)=\sin x+\cos x,\ 0\lt x\lt\frac{\pi}{2}

(iv) f(x)=sin⁡x−cos⁡x, 0<x<2πf(x)=\sin x-\cos x,\ 0\lt x\lt 2\pi

(v) f(x)=x3−6x2+9x+15f(x)=x^3-6x^2+9x+15    (vi) g(x)=x2+2x, x>0g(x)=\dfrac{x}{2}+\dfrac{2}{x},\ x\gt0

(vii) g(x)=1x2+2g(x)=\dfrac{1}{x^2+2}    (viii) f(x)=x1−x, 0<x<1f(x)=x\sqrt{1-x},\ 0\lt x\lt1

Exam answer

(i) f′(x)=2x=0f'(x) = 2x = 0 gives x=0x = 0⁠.

f′′(x)=2f''(x) = 2⁠, so f′′(0)=2>0f''(0) = 2 > 0⁠: x=0x = 0 is a point of local minima.

Local minimum value f(0)=0f(0) = 0⁠; no local maximum.

(ii) Here

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

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

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

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

Local minimum value g(1)=−2g(1) = -2⁠; local maximum value g(−1)=2g(-1) = 2⁠.

(iii) h′(x)=cos⁡x−sin⁡x=0h'(x) = \cos x - \sin x = 0 gives tan⁡x=1\tan x = 1⁠, so x=π4x = \frac{\pi}{4} in (0,π2)\left(0, \frac{\pi}{2}\right)⁠.

h′′(x)=−sin⁡x−cos⁡xh''(x) = -\sin x - \cos x⁠, so h′′(π4)=−2<0h''\left(\frac{\pi}{4}\right) = -\sqrt2 < 0⁠: x=π4x = \frac{\pi}{4} is a point of local maxima.

Local maximum value h(π4)=2h\left(\frac{\pi}{4}\right) = \sqrt2⁠; no local minimum.

(iv) f′(x)=cos⁡x+sin⁡x=0f'(x) = \cos x + \sin x = 0 gives tan⁡x=−1\tan x = -1⁠, so x=3π4x = \frac{3\pi}{4} or 7π4\frac{7\pi}{4} in (0,2π)(0, 2\pi)⁠.

f′′(x)=−sin⁡x+cos⁡xf''(x) = -\sin x + \cos x⁠.

f′′(3π4)=−2<0f''\left(\frac{3\pi}{4}\right) = -\sqrt2 < 0⁠, so x=3π4x = \frac{3\pi}{4} is a point of local maxima.

f′′(7π4)=2>0f''\left(\frac{7\pi}{4}\right) = \sqrt2 > 0⁠, so x=7π4x = \frac{7\pi}{4} is a point of local minima.

Local maximum value f(3π4)=2f\left(\frac{3\pi}{4}\right) = \sqrt2⁠; local minimum value f(7π4)=−2f\left(\frac{7\pi}{4}\right) = -\sqrt2⁠.

(v) Here

f′(x)=3x2−12x+9=3(x−1)(x−3)\begin{aligned} f'(x) &= 3x^2 - 12x + 9 \\ &= 3(x - 1)(x - 3) \end{aligned}

f′(x)=0f'(x) = 0 gives x=1x = 1 or x=3x = 3⁠.

f′′(x)=6x−12f''(x) = 6x - 12⁠. f′′(1)=−6<0f''(1) = -6 < 0⁠, so x=1x = 1 is a point of local maxima.

f′′(3)=6>0f''(3) = 6 > 0⁠, so x=3x = 3 is a point of local minima.

Local maximum value f(1)=19f(1) = 19⁠; local minimum value f(3)=15f(3) = 15⁠.

(vi) g′(x)=12−2x2=0g'(x) = \frac12 - \frac{2}{x^2} = 0 gives x2=4x^2 = 4⁠, so x=2x = 2 (as x>0x > 0⁠).

g′′(x)=4x3g''(x) = \frac{4}{x^3}⁠, so g′′(2)=12>0g''(2) = \frac12 > 0⁠: x=2x = 2 is a point of local minima.

Local minimum value g(2)=2g(2) = 2⁠; no local maximum.

(vii) g′(x)=−2x(x2+2)2=0g'(x) = -\frac{2x}{(x^2 + 2)^2} = 0 gives x=0x = 0⁠.

In (−∞,0)(-\infty, 0)⁠, g′(x)>0g'(x) > 0⁠. In (0,∞)(0, \infty)⁠, g′(x)<0g'(x) < 0⁠. So x=0x = 0 is a point of local maxima.

Local maximum value g(0)=12g(0) = \frac12⁠; no local minimum.

(viii) By the product rule, for 0<x<10 < x < 1⁠,

f′(x)=1−x−x21−x=2−3x21−x\begin{aligned} f'(x) &= \sqrt{1 - x} - \frac{x}{2\sqrt{1 - x}} \\ &= \frac{2 - 3x}{2\sqrt{1 - x}} \end{aligned}

f′(x)=0f'(x) = 0 gives x=23x = \frac23⁠.

In (0,23)\left(0, \frac23\right)⁠, f′(x)>0f'(x) > 0⁠. In (23,1)\left(\frac23, 1\right)⁠, f′(x)<0f'(x) < 0⁠. So x=23x = \frac23 is a point of local maxima.

f(23)=2313=239f\left(\tfrac23\right) = \tfrac23\sqrt{\tfrac13} = \frac{2\sqrt3}{9}

Local maximum value 239\dfrac{2\sqrt3}{9}⁠, at x=23x = \dfrac23⁠; no local minimum.

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

The one idea this problem is built on

Every function here is differentiable at every point of its (open) domain, so by Theorem 2 a local maximum or minimum can only occur where the derivative is zero. So we solve f′(x)=0f'(x)=0 to get the few candidate points, and then decide each candidate with a test: the second derivative test (sign of f′′f'' at the point) when f′′f'' is easy, or the first derivative test (does f′f' change sign, and which way?) when it is not.

Given, to find, and the plan
Given
Eight functions with the stated domains (all of R\R where none is given).
To find
The points of local maxima and minima and the local maximum and minimum values (Definition 4, Section 6.4).
Plan
For each part: (1) differentiate, naming each rule; (2) solve f′(x)=0f'(x)=0 inside the domain, keeping only solutions in it; (3) apply Theorem 4 (second derivative test) or Theorem 3 (first derivative test) to each critical point; (4) evaluate ff there.

Basics you need

1Local maxima and minima (Definition 4)

An interior point cc of the domain is a point of local maxima if f(c)≥f(x)f(c)\ge f(x) for all xx in some interval (c−h,c+h)(c-h,c+h)⁠; f(c)f(c) is then the local maximum value. Local minima: the same with ≤\le⁠. Lesson: maxima and minima

2Critical points (Theorem 2 and the Note after it)

If ff has a local maximum or minimum at cc (in an open interval), then f′(c)=0f'(c)=0 or ff is not differentiable at cc⁠. Such points are called critical points. Here every function is differentiable on its domain, so only f′(c)=0f'(c)=0 matters.

3First derivative test (Theorem 3)

At a critical point cc where ff is continuous: f′f' changes from ++ to −- means local maxima; from −- to ++ means local minima; no change of sign means neither (a point of inflection). Lesson: first and second derivative tests

4Second derivative test (Theorem 4)

If f′(c)=0f'(c)=0⁠: f′′(c)<0f''(c)\lt0 means local maxima; f′′(c)>0f''(c)\gt0 means local minima; f′′(c)=0f''(c)=0 means the test fails and we go back to the first derivative test.

5Signs of products and quotients

A product or quotient is positive when it has an even number of negative factors. A sign chart records the sign of each factor on each interval. Basics: sign of a product

6Exact trigonometric values

sin⁡π4=cos⁡π4=22\sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{\sqrt2}{2}⁠; in the second quadrant sine is positive and cosine negative, in the fourth the reverse. Basics: signs of sin and cos

Part (i): f(x)=x2f(x)=x^2

  1. Critical point

    f′(x)=2xf'(x)=2x (power rule), which is 00 only at x=0x=0⁠.

    ff is a polynomial, differentiable everywhere, so by Theorem 2 the only possible local extreme point is where f′=0f'=0⁠.
    Theorem 2, Section 6.4. Lesson: tests

  2. Test it

    f′′(x)=2f''(x)=2⁠, so f′′(0)=2>0f''(0)=2\gt0⁠: by the second derivative test, x=0x=0 is a point of local minima, and the local minimum value is f(0)=0f(0)=0⁠. The first derivative test agrees: f′(x)=2xf'(x)=2x is negative for x<0x\lt0 and positive for x>0x\gt0⁠.

    There is no other critical point, so there is no local maximum.

    Theorem 2 rules out a local maximum anywhere else, because nowhere else is f′f' zero. (This matches Example 14, where x2x^2 has minimum value 00 at 00⁠.)
    Theorem 4(ii); Example 14.

Part (ii): g(x)=x3−3xg(x)=x^3-3x

  1. Critical points

    g′(x)=3x2−3=3(x−1)(x+1).g'(x)=3x^2-3=3(x-1)(x+1).

    So g′(x)=0g'(x)=0 at x=1x=1 and x=−1x=-1⁠.

    Take out the common factor 33⁠, then use x2−1=(x−1)(x+1)x^2-1=(x-1)(x+1) (difference of squares).
    Factorising. Basics: factorising

  2. Second derivative test

    g′′(x)=6xg''(x)=6x⁠.

    • g′′(1)=6>0g''(1)=6\gt0⁠: x=1x=1 is a point of local minima; value g(1)=1−3=−2g(1)=1-3=-2⁠.
    • g′′(−1)=−6<0g''(-1)=-6\lt0⁠: x=−1x=-1 is a point of local maxima; value g(−1)=−1+3=2g(-1)=-1+3=2⁠.
    The second derivative tells us which way the graph bends: positive g′′g'' means bending upwards (a valley), negative means bending downwards (a hilltop).
    Theorem 4, Section 6.4. Lesson: tests
x y O −2 −1 1 2 −2 2 local max (−1, 2) local min (1, −2)
g(x)=x3−3xg(x)=x^3-3x⁠: a local hilltop at (−1,2)(-1,2) and a local valley at (1,−2)(1,-2)⁠. Neither is a maximum or minimum of the whole graph, which rises and falls without end; they are only highest or lowest nearby.

Part (iii): h(x)=sin⁡x+cos⁡x, 0<x<π2h(x)=\sin x+\cos x,\ 0\lt x\lt\frac{\pi}{2}

  1. Critical points in (0,π2)\left(0,\frac{\pi}{2}\right)

    h′(x)=cos⁡x−sin⁡xh'(x)=\cos x-\sin x⁠. Setting it to zero: sin⁡x=cos⁡x\sin x=\cos x⁠. On (0,π2)\left(0,\frac{\pi}{2}\right)⁠, cos⁡x>0\cos x\gt0⁠, so we may divide by it: tan⁡x=1\tan x=1⁠. The only angle in (0,π2)\left(0,\frac{\pi}{2}\right) with tan⁡x=1\tan x=1 is x=π4x=\frac{\pi}{4}⁠.

    Dividing by cos⁡x\cos x is safe only because it is not zero on this interval. tan⁡\tan rises strictly from 00 towards +∞+\infty on (0,π2)\left(0,\frac{\pi}{2}\right)⁠, so it takes the value 11 exactly once.
    tan⁡π4=1\tan\frac{\pi}{4}=1⁠. Basics: sin and cos on the unit circle

  2. Second derivative test

    h′′(x)=−sin⁡x−cos⁡xh''(x)=-\sin x-\cos x⁠, so

    h′′ ⁣(π4)=−22−22=−2<0.h''\!\left(\tfrac{\pi}{4}\right)=-\tfrac{\sqrt2}{2}-\tfrac{\sqrt2}{2}=-\sqrt2\lt0.

    So x=π4x=\frac{\pi}{4} is a point of local maxima, with local maximum value h ⁣(π4)=22+22=2h\!\left(\frac{\pi}{4}\right)=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}=\sqrt2⁠. It is the only critical point, so there is no local minimum.

    22+22=222=2\frac{\sqrt2}{2}+\frac{\sqrt2}{2}=\frac{2\sqrt2}{2}=\sqrt2⁠. The end points 00 and π2\frac{\pi}{2} are excluded, and Definition 4 only concerns interior points anyway.
    Theorem 4(i). Lesson: tests

Part (iv): f(x)=sin⁡x−cos⁡x, 0<x<2πf(x)=\sin x-\cos x,\ 0\lt x\lt2\pi

  1. Critical points in (0,2π)(0,2\pi)

    f′(x)=cos⁡x+sin⁡xf'(x)=\cos x+\sin x⁠. Setting it to zero: sin⁡x=−cos⁡x\sin x=-\cos x⁠.

    Could cos⁡x=0\cos x=0 here? Then sin⁡x=±1\sin x=\pm1⁠, and sin⁡x=−cos⁡x\sin x=-\cos x would say ±1=0\pm1=0⁠, impossible. So cos⁡x≠0\cos x\ne0 at any solution, and dividing gives tan⁡x=−1\tan x=-1⁠. In (0,2π)(0,2\pi)⁠, tan⁡x=−1\tan x=-1 at

    x=3π4andx=7π4.x=\tfrac{3\pi}{4}\quad\text{and}\quad x=\tfrac{7\pi}{4}.
    tan⁡\tan is negative where sine and cosine have opposite signs: the second and fourth quadrants. The reference angle for tan⁡=1\tan=1 is π4\frac{\pi}{4}⁠, giving π−π4\pi-\frac{\pi}{4} and 2π−π42\pi-\frac{\pi}{4}⁠. Checking cos⁡x=0\cos x=0 first ensures we lose no solution by dividing.
    Signs of sin and cos by quadrant. Basics: signs of sin and cos

  2. Second derivative test

    f′′(x)=−sin⁡x+cos⁡xf''(x)=-\sin x+\cos x⁠.

    • At 3π4\frac{3\pi}{4}⁠: sin⁡=22\sin=\frac{\sqrt2}{2}⁠, cos⁡=−22\cos=-\frac{\sqrt2}{2}⁠, so f′′=−2<0f''=-\sqrt2\lt0⁠: local maxima, value f=22+22=2f=\frac{\sqrt2}{2}+\frac{\sqrt2}{2}=\sqrt2⁠.
    • At 7π4\frac{7\pi}{4}⁠: sin⁡=−22\sin=-\frac{\sqrt2}{2}⁠, cos⁡=22\cos=\frac{\sqrt2}{2}⁠, so f′′=2>0f''=\sqrt2\gt0⁠: local minima, value f=−22−22=−2f=-\frac{\sqrt2}{2}-\frac{\sqrt2}{2}=-\sqrt2⁠.
    f(x)=sin⁡x−cos⁡xf(x)=\sin x-\cos x⁠: at 3π4\frac{3\pi}{4} we subtract a negative cosine, which adds; at 7π4\frac{7\pi}{4} both terms are negative.
    Theorem 4. Lesson: tests
x y O 3π/4 7π/4 2π −1 1 local max √2 local min −√2
f(x)=sin⁡x−cos⁡xf(x)=\sin x-\cos x on (0,2π)(0,2\pi) (open circles: end points excluded): local maximum 2\sqrt2 at 3π4\frac{3\pi}{4}⁠, local minimum −2-\sqrt2 at 7π4\frac{7\pi}{4}⁠.

Part (v): f(x)=x3−6x2+9x+15f(x)=x^3-6x^2+9x+15

  1. Critical points

    f′(x)=3x2−12x+9=3(x2−4x+3)=3(x−1)(x−3).\begin{aligned} f'(x)&=3x^2-12x+9\\ &=3(x^2-4x+3)\\ &=3(x-1)(x-3). \end{aligned}

    So f′(x)=0f'(x)=0 at x=1x=1 and x=3x=3⁠.

    x2−4x+3x^2-4x+3⁠: two numbers with product 33 and sum −4-4 are −1-1 and −3-3⁠.
    Factorising a quadratic. Basics: factorising

  2. Second derivative test

    f′′(x)=6x−12f''(x)=6x-12⁠.

    • f′′(1)=−6<0f''(1)=-6\lt0⁠: local maxima at x=1x=1⁠; value f(1)=1−6+9+15=19f(1)=1-6+9+15=19⁠.
    • f′′(3)=6>0f''(3)=6\gt0⁠: local minima at x=3x=3⁠; value f(3)=27−54+27+15=15f(3)=27-54+27+15=15⁠.
    The sign chart below reaches the same verdict by the first derivative test: f′f' goes +,−,++,-,+⁠, so it changes from ++ to −- at 11 and from −- to ++ at 33⁠.
    Theorems 3 and 4. Lesson: tests
1 3 x x − 1 − + + 0 x − 3 − − + 0 f′(x) + − + 0 0 f(x) rises falls rises local max local min
Sign chart for f′(x)=3(x−1)(x−3)f'(x)=3(x-1)(x-3) (not to scale). The factor 33 is positive and does not affect the signs.

Part (vi): g(x)=x2+2x, x>0g(x)=\dfrac{x}{2}+\dfrac{2}{x},\ x\gt0

  1. Critical points with x>0x\gt0

    Writing 2x=2x−1\frac{2}{x}=2x^{-1}⁠, the power rule gives

    g′(x)=12−2x2=x2−42x2.g'(x)=\frac12-\frac{2}{x^2}=\frac{x^2-4}{2x^2}.

    This is zero when x2=4x^2=4⁠, i.e. x=2x=2 or x=−2x=-2⁠. Only x=2x=2 is in the domain x>0x\gt0⁠.

    Over the common denominator 2x22x^2⁠: 12=x22x2\frac12=\frac{x^2}{2x^2} and 2x2=42x2\frac{2}{x^2}=\frac{4}{2x^2}⁠. The denominator is never zero for x>0x\gt0⁠, so gg is differentiable on the whole domain.
    Discarding solutions outside the domain. Basics: interval notation

  2. Second derivative test

    g′′(x)=4x3g''(x)=\frac{4}{x^3} (the derivative of −2x−2-2x^{-2} is 4x−34x^{-3}⁠), so g′′(2)=48=12>0g''(2)=\frac48=\frac12\gt0⁠. Hence x=2x=2 is a point of local minima, with local minimum value

    g(2)=22+22=2.g(2)=\tfrac22+\tfrac22=2.

    It is the only critical point, so there is no local maximum.

    An alternative view: by the AM–GM idea, x2+2x≥2x2⋅2x=2\frac{x}{2}+\frac{2}{x}\ge2\sqrt{\frac{x}{2}\cdot\frac{2}{x}}=2 for x>0x\gt0⁠, with equality at x=2x=2⁠; the calculus agrees.
    Theorem 4(ii). Lesson: tests

Part (vii): g(x)=1x2+2g(x)=\dfrac{1}{x^2+2}

  1. Critical points

    Write g(x)=(x2+2)−1g(x)=(x^2+2)^{-1}⁠. By the chain rule (outer u−1u^{-1}⁠, inner u=x2+2u=x^2+2⁠, u′=2xu'=2x⁠),

    g′(x)=−2x(x2+2)2.g'(x)=-\frac{2x}{(x^2+2)^2}.

    The denominator is always positive (x2+2≥2x^2+2\ge2⁠), so g′(x)=0g'(x)=0 only at x=0x=0⁠.

    gg is defined and differentiable for every real xx because x2+2x^2+2 is never zero.
    Chain rule; x2+2≥2>0x^2+2\ge2\gt0⁠.

  2. First derivative test (and a second-derivative check)

    The sign of g′(x)g'(x) is the sign of −2x-2x⁠: positive for x<0x\lt0⁠, negative for x>0x\gt0⁠. So g′g' changes from ++ to −- at 00⁠, and by Theorem 3(i) x=0x=0 is a point of local maxima, with local maximum value g(0)=10+2=12g(0)=\frac{1}{0+2}=\frac12⁠.

    Check with Theorem 4: g′′(x)=6x2−4(x2+2)3g''(x)=\dfrac{6x^2-4}{(x^2+2)^3}⁠, so g′′(0)=−48=−12<0g''(0)=\frac{-4}{8}=-\frac12\lt0⁠. ✓ There is no local minimum.

    The first derivative test is quicker here because the denominator of g′g' is always positive, so only the numerator decides the sign. Common sense agrees: x2+2x^2+2 is smallest at x=0x=0⁠, so its reciprocal is largest there.
    Theorem 3(i), Theorem 4(i). Lesson: tests

Part (viii): f(x)=x1−x, 0<x<1f(x)=x\sqrt{1-x},\ 0\lt x\lt1

  1. Differentiate with the product and chain rules

    On (0,1)(0,1)⁠, 1−x>01-x\gt0⁠, so 1−x\sqrt{1-x} is differentiable, with derivative −121−x\frac{-1}{2\sqrt{1-x}} (chain rule, inner function 1−x1-x⁠). By the product rule,

    f′(x)=1−x−x21−x=2(1−x)−x21−x=2−3x21−x.\begin{aligned} f'(x)&=\sqrt{1-x}-\frac{x}{2\sqrt{1-x}}\\ &=\frac{2(1-x)-x}{2\sqrt{1-x}}=\frac{2-3x}{2\sqrt{1-x}}. \end{aligned}
    To combine the two terms, multiply 1−x\sqrt{1-x} top and bottom by 21−x2\sqrt{1-x}⁠: its numerator becomes 2(1−x)2(1-x)⁠.
    Product rule (uv)′=u′v+uv′(uv)'=u'v+uv' with u=xu=x⁠, v=1−xv=\sqrt{1-x}⁠.

  2. Critical point and the first derivative test

    f′(x)=0f'(x)=0 when 2−3x=02-3x=0⁠, i.e. x=23x=\frac23⁠, which is in (0,1)(0,1)⁠. The denominator 21−x2\sqrt{1-x} is positive on (0,1)(0,1)⁠, so the sign of f′f' is the sign of 2−3x2-3x⁠: positive on (0,23)\left(0,\frac23\right)⁠, negative on (23,1)\left(\frac23,1\right)⁠. So f′f' changes from ++ to −-⁠, and x=23x=\frac23 is a point of local maxima (Theorem 3(i)).

    f ⁣(23)=2313=233=239.f\!\left(\tfrac23\right)=\tfrac23\sqrt{\tfrac13}=\frac{2}{3\sqrt3}=\frac{2\sqrt3}{9}.
    Rationalise: 233⋅33=239\frac{2}{3\sqrt3}\cdot\frac{\sqrt3}{\sqrt3}=\frac{2\sqrt3}{9}⁠. The first derivative test avoids differentiating the quotient again; for the record, f′′ ⁣(23)=−332<0f''\!\left(\frac23\right)=-\frac{3\sqrt3}{2}\lt0⁠, which agrees.
    Theorem 3(i); sign of a quotient. Basics: sign of a product
0 2/3 1 x 2 − 3x + − 0 2√(1 − x) + + f′(x) + − 0 f(x) rises falls local max
Sign chart for f′(x)=2−3x21−xf'(x)=\dfrac{2-3x}{2\sqrt{1-x}} on (0,1)(0,1) (not to scale): the denominator is always positive, so f′f' has the sign of 2−3x2-3x⁠.
Result

(i) Local minimum at x=0x=0⁠, value 00⁠; no local maximum.

(ii) Local minimum at x=1x=1⁠, value −2-2⁠; local maximum at x=−1x=-1⁠, value 22⁠.

(iii) Local maximum at x=π4x=\frac{\pi}{4}⁠, value 2\sqrt2⁠; no local minimum.

(iv) Local maximum at x=3π4x=\frac{3\pi}{4}⁠, value 2\sqrt2⁠; local minimum at x=7π4x=\frac{7\pi}{4}⁠, value −2-\sqrt2⁠.

(v) Local maximum at x=1x=1⁠, value 1919⁠; local minimum at x=3x=3⁠, value 1515⁠.

(vi) Local minimum at x=2x=2⁠, value 22⁠; no local maximum.

(vii) Local maximum at x=0x=0⁠, value 12\frac12⁠; no local minimum.

(viii) Local maximum at x=23x=\frac23⁠, value 239\frac{2\sqrt3}{9}⁠; no local minimum.

(Textbook answer: the same in every part.)

Check it with numbers

(ii) g(−1.1)=1.969g(-1.1)=1.969 and g(−0.9)=1.971g(-0.9)=1.971⁠, both below 22⁠. ✓

(iii) h(0.7)≈0.644218+0.764842h(0.7)\approx0.644218+0.764842 =1.409060=1.409060⁠, below 2≈1.414214\sqrt2\approx1.414214⁠. ✓

(v) f(0.9)=18.969f(0.9)=18.969⁠, f(1.1)=18.971f(1.1)=18.971 (below 1919⁠); f(2.9)=15.029f(2.9)=15.029⁠, f(3.1)=15.031f(3.1)=15.031 (above 1515⁠). ✓

(vi) g(1.9)≈2.00263g(1.9)\approx2.00263⁠, g(2.1)≈2.00238g(2.1)\approx2.00238⁠, both above 22⁠. ✓ (vii) g(±0.1)=12.01≈0.49751<0.5g(\pm0.1)=\frac{1}{2.01}\approx0.49751\lt0.5⁠. ✓

(viii) f(0.6)≈0.379473f(0.6)\approx0.379473⁠, f ⁣(23)≈0.384900f\!\left(\frac23\right)\approx0.384900⁠, f(0.7)≈0.383406f(0.7)\approx0.383406⁠. ✓

Common mistakes
  1. Keeping x=−2x=-2 in (vi). The domain is x>0x\gt0⁠; a critical point must lie in the domain.
  2. Dividing by cos⁡x\cos x without checking it is non-zero in (iii) and (iv). Here it is safe, but say why.
  3. Missing 7π4\frac{7\pi}{4} in (iv). tan⁡x=−1\tan x=-1 has two solutions in (0,2π)(0,2\pi)⁠, one in the second quadrant and one in the fourth.
  4. Calling a local maximum "the maximum". In (ii), g(−1)=2g(-1)=2 is only highest nearby: g(3)=18g(3)=18 is larger.
  5. Product rule slip in (viii). The derivative of 1−x\sqrt{1-x} carries a minus sign from the inner function 1−x1-x⁠.
What to take away
  1. Recipe: differentiate, solve f′=0f'=0 inside the domain, test each critical point, evaluate.
  2. Use the second derivative test when f′′f'' is easy; use the first derivative test when f′f' is a product or quotient whose sign is easy to read.
  3. A local extreme value describes the function only near the point; the function may go higher or lower elsewhere.

Every fact used, and where it comes from

FactUsed inSource
Local maxima, local minima, local maximum and minimum valuesEvery partDefinition 4, Section 6.4 (Lesson: maxima and minima)
Local extremum in an open interval ⇒ f′(c)=0f'(c)=0 or not differentiable; critical pointsThe first step of every partTheorem 2 and the Note after it, Section 6.4
First derivative testParts (i), (v), (vii), (viii)Theorem 3, Section 6.4 (Lesson: tests)
Second derivative testParts (i)–(vii); check in (viii)Theorem 4, Section 6.4
Power, product, quotient and chain rulesThe first step of every partChapter 5 (Continuity and Differentiability)
sin⁡\sin⁠, cos⁡\cos⁠, tan⁡\tan at π4\frac{\pi}{4}⁠, 3π4\frac{3\pi}{4}⁠, 7π4\frac{7\pi}{4}⁠; signs by quadrantParts (iii), (iv)Class XI, Chapter 3 (Basics)
x2x^2 has minimum value 00 at 00Part (i)Example 14, Section 6.4
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