9.2 Maximum and Minimum

Before we describe the critical pionts let us discuss the behavior of increasing and decreasing functions.

Increasing and Decreasing Functions

Consider the function on an open interval.

Definition
A function f is increasing on an interval (a, b) if for any two numbers x1 and x2 in (a, b),
x2 > x1 implies f(x2) > f(x1),
namely, an increase in x produces an increase in y.
fig 9.2-1

fig 9.2-2

A function f is decreasing on an interval (a, b) if for any two numbers x1 and x2 in (a,b),
x2 > x1   implies   f(x2) < f(x1),
namely, an increase in x produces a decrease in y.
We can determine whether a function is increasing or decreasing at a given interval by looking into its graph. Sometimes we can get the same information by using the first derivative.
Test for Increasing and Decreasing Functions
Let y = f(x) be differentiable on the open interval (a, b).
1. If f'(x) > 0 for each x ∈ (a, b), then f is increasing on (a, b).
2. If f'(x) < 0 for each x ∈ (a, b), then f is decreasing on (a, b).

Example 4.
Show that the function f(x) = x2 + 1 is decreasing on the interval x < 0 and increasing on the interval x > 0.
Solution
f(x) = x2 + 1,
f'(x) = 2x.
On x < 0, f;(x) = 2x < 0. Thus f is decreasing on x < 0
On x > 0, f;(x) = 2x > 0. Thus f is increasing on x > 0

Example 5.
Consider the function f(x) = x3 - 3x + 1.
(a) Find the open intervals on which f is increasing and those on which f is decreasing.
(b) Find the points at which the graph has a horizontal tangent.
Solution
(a) f(x) = x3 - 3x + 1.
f'(x) = 3x2 - 3
= 3(x - 1)(x + 1).
So f'(x > 0, when x - 1 and x + 1 have the same sign. Similarly, f'(x) < 0, when x - 1 and x + 1 ahve the opposite signs.
Interval x < -1 -1 < x < 1 x > 1
x + 1 - + +
x - 1 - - +
sign of f'(x) + - +
Behavior of f   increasing   decreasing   increasing  
Thus f(x) is increasing on x < -1 and x > 1, and decreasing on -1 < x < 1.
(b) The graph has a horizontal tangent when f'(x) = 0.
3(x - 1)(x + 1) = 0.
x = 1 or x = -1.
So f(-1) = 3, f(1) = -1.
The horizontal tangents are at the points (-1, 3) and (1, -1).

Critical Point

We now describe how to decide the maximum and minimum points of the function by determining the values of its derivatives. It is one of the most important application of differentiation. We first introduce the notion of critical point.

Definition
A critical point of a function f(x) is a point in the domain of the function such that either f'(x) = 0 or f'(x) is undefined.
f '(x) = 0
f '(x) = 0
f '(x) = 0
f '(x) is undefined
f '(x) is undefined
Test for Increasing and decreasing function on an Interval
Let f(x) be a function.
Step 1 Find f '(x).
Step 2 Find the critical points of f.
Step 3 Find the open intervals determining by the critical points.
Step 4 Test the signs of f '(x) for each interval.
Step 5 Decide whether f is increasing or decreasing on each interval.

Example 6
Find the critical points of:
(a) f(x) = x3 - 3x + 2
(b) f(x) = (x - 1) 2⁄ 3
(c) f(x) = 1⁄x
Solution
(a) curve: f(x) = x3 - 3x + 2
f '(x) = 3x2 - 3.
For critical points, f ' (x) = 0 so
3x2 - 3 = 0.
Therefore x = ± 1.
When x = 1,
f '(1) = 13 - 3(1) + 2 = 0.
When x = -1,
f(-1) = (-1)3 - 3(-1) + 2 = 4.
Therefore the critical points are (1, 0) and (-1, 4).

(b) curve: f(x) = (x - 1) 2⁄ 3
f '(x) = 2⁄3(x - 1)- 1⁄3
= 2⁄3(x - 1)1⁄3 .
It is observed that f ' (x) is undefined when x = 1. Therefore the critical point is (1, 0).

(c) curve: f(x) = 1⁄x
f ' (x) = - 1⁄x2.
It is observed that f ' (x) is undefined when x = 0.
However, x = 0 is not in the domain of the function.
Therefore, there are no critical points.

Example 7.
Find the open intervals on which the following functions are increasing or decreasing.
(a) f(x) = 2x3 - 6x + 5
Solution
curve: f(x) = 2x3 - 6x + 5
f ' (x) = 6(x2 - 1) = 6(x + 1)(x - 1).
For critical points, f ' (x) = 0, namely x = ±1.
Interval x < -1 -1 < x < 1 x > 1
x + 1 - + +
x - 1 - - +
sign of f '(x) + - +
Behavior of f   increasing   decreasing   increasing 

Therefore, on x < -1 and x > 1, the given function is increasing, and on -1 < x < 1, the given function is decreasing.

(b) f(x) = (x2 - 9) 4⁄ 5
curve: f(x) = (x2 - 9) 4⁄ 5
f '(x) = 8⁄5 x⁄(x2 - 9)1⁄5
For critical points, f '(x) = 0, namely x = 0.
f '(x) is undefined, namely x = -3 and x = 3.
Interval x < -3 -3 < x < 0 0 < x < 3 x > 3
x + 3 - + + +
x - - + +
x - 3 - - - +
sign of f '(x) - + - +
Behavior of f   decreasing   increasing   decreasing   increasing  
Therefore, on x < -3 and 0 < x < 3, the given function is decreasing, and on -3 < x < 0 and x > 3the given function is increasing.

(c) f(x) = 2x - 3⁄x - 1
curve: f(x) = 2x - 3⁄x - 1
f ' (x) = 1⁄(x - 1)2.
It is observed that x = 1 does not belong to the domain of the function.
Interval x < 1 x > 1
(x - 1)2 + +
sign of f ' (x) + +
Behavior of f   increasing   increasing  
Therefore, on both x < 1 and x > 1, the function is increasing.

First Derivative Test

Consider the following graph which has a restricted domain of -5 ≤ x ≤ 5.
fig 9.2-3

In the figure, points A and C are minimum, and points B and D are maximum. The point A is called a global minimum because f (x) has a minimum value on the entire domain. Similarly the point D is called a global maximum because f (x) has a maximum value on the entire domain. B is a local maximum because it is a turning point on the graph and f (x) has a maximum value at B compared to the neighboring points. C is a local minimum because it is a turning point on the graph and f (x) has a minimum value at C compared to the neighboring points.
Test for Global Maximum and Global Minimum on a Closed Interval
Let f (x) be a continuous function on a closed interval.
Step 1 Find the critical points of f.
Step 2 Find f (x) at all critical points and endpoints of the interval.
Step 3 Take the smallest value as the global minimum and the largest value as the global maximum.

Example 8.
Find the global minimum and global maximum values of each of the following function:
(a) f (x) = x2 + 5x - 3     on [-4, 5],
(b) f (x) = x⁄x2 + 1     on [-3, 3].
Solution
(a) curve: f (x) = x2 + 5x - 3     on [-4, 5],
f ' (x) = 2x + 5.
For critical points, f ' (x) = 0, namely x = - 5⁄2.
Therefore, at the critical point f (- 5⁄2) = -9 1⁄4.
At the end points f (-4) = -7 and f (5) = 47.
Hence, the global minimum value is -9 1⁄4 and the global maximum value is 47.

(b) curve: f (x) = x⁄x2 + 1     on [-3, 3]
For critical point, f ' (x) = 0,
1 - x2 ⁄(x2 + 1)2 = 0,
1 - x2 = 0.
x = ±1.
At the critical points, f (1) = 1⁄2 and f (-1) = - 1⁄2.
At the end points, f (-3) = - 3⁄10 and f (3) = 3⁄10.
Hence, the global minimum value is - 1⁄2 and the global maximum value is 1⁄2.
First Derivative Test for Local Maximum and Local Minimum
Suppose that c is a critical point of a continuous function f, and that differentiable at every point in some interval containing c except possibly at c itself.
  1. If f ' changes from negative to positive at c , then f has a local minimum at c .
  2. If f ' changes from positive to negative at c, then f has a local minimum at c .
  3. If f ' does not change sign at c (f ' is positive on both sides of c or negative on both sides of c ), then f has no local maximum or local minimum at c
    4. .
Example 9.
Find the critical points of f (x) = x3 - 3x + 1.
Find the open intervals on which the function is increasing or decreasing. Identify the function's local maximum and minimum.
Solution
f (x ) = x3 - 3x + 1.
f ' (x) = 3x 2 - 3 = 3(x + 1)(x - 1).
For critical point, f ' (x) = 0, namely, x = ±1.
Interval x < -1 -1 < x < 1 x > 1
x + 1 - + +
x - 1 - - +
sign of f ' (x) + - +
Behavior of f   increasing   decreasing   increasing 
Thus the local maximum at x = -1 and local minimum at x = 1.

Example 10.
Show that f(x) = x 3 is increasing on both sides of critical point.
Solution
curve: f (x) = x 3,
f ' (x) = 3x 2
For critical point, f ' (x ) = 0, namely x = 0.
Interval x < 1 x > 1
x2 + +
sign of f ' (x) + +
Behavior of f   increasing   increasing  
Thus the function is increasing on both sides of critical pint.
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Exercise 9.2
1. Consider the following graphs.
fig 9.2-4
fig 9.2-5
(a) State all open intervals where the function is increasing.
(b) State all open intervals where the function is decreasing.
Solution

  1. (i) The function is increasing on (a, 0) and (c, d).
    (ii) The function is increasing on (a, b), (c, d) and (e, f).
  2. (i) The function is decreasing on (0, c).
    (ii) The function is decreasing on (b, c), (d, e) and (f, g).


    3. 2. Find the critical points:
    (a) f (x) = x2 - 2x + 5
    (b) f (x) = x3 + 6x2 + 3x + 10
    (c) f (x) = 1 ⁄ x - 3
    (d) f (x) = (x2 - 2x) 2 ⁄ 3
    (e) f (x ) = 1 ⁄ x 2 - 2x + 1
    (f) f (x ) = x 2 ⁄ x 2 + 2
    Solution
    (a) f (x) = x2 - 2x + 5
    f ' (x) = 2x - 2
    For critical point,
    f '(x) = 0
    2x - 2 = 0
    2x = 2
    x = 1
    f (1) = 1 - 2 + 5 = 4
    The critical point is (1, 4).

    (b) f (x) = x3 + 6x2 + 3x + 10
    f ' (x = 3x2 + 12x + 3
    For critical point,
    f ' (x) = 0
    3x2 + 12x + 3 = 0
    x2 + 4x + 1 = 0
    x2 + 4x = -1
    x2 + 4x + 4 = -1 + 4
    (x + 2)2 = -1 + 4
    x + 2 = ± √ 3  
    x = -2 + √ 3   or x = -2 - √ 3  
    x = - 0.268 or x = - 3.732
    f (-0.268) = 9.608 or f (-3.732) = 30.392
    The critical points are (-0.268, 9.608) and (-3.732, 30.392).

    (c) f (x) = 1 ⁄ x - 3 = (x - 3)-1
    f ' (x) = -1(x - 3)-2 = - 1⁄(x - 3)2
    It is observed that x = 3 does not belong to the domain of function.
    Hence the given function has no critical points.

    (d) f (x) = (x2 - 2x) 2 ⁄ 3
    f ' (x) = 2⁄3 (x2 - 2x)- 1⁄3 (2x - 2)
    = 2(2x - 2)⁄3(x2 -2x) 1⁄3
    = (4x - 4)⁄3(x2 -2x) 1⁄3
    For critical points,
    f ' (x) = 0,
    4x - 4 = 0
    x = 1
    f ' (x) is undefined,
    x2 - 2x = 0
    x(x - 2) = 0
    x = 0 or x = 2
    x = 1 ⇒ f (1) = 1
    x = 0 ⇒ f (0) = 0
    x = 2 ⇒ f (2) = 0
    The critical points are (1, 1), (0, 0) and (2, 0).

    (e) f (x ) = 1 ⁄ x 2 - 2x + 1 = 1⁄(x - 1)2 = (x - 1)2
    f ' (x = -2(x - 1)-3 = -2⁄(x - 1)3
    It is observed that x = 1 does not belong to the domain of the function.
    Hence the given function has no critical points.

    (f) f (x ) = x 2 ⁄ x 2 + 2
    f ' (x) = (x2 + 2) 2x - x2(2x)⁄(x2 + 2)2
    = 2x3 + 4x - 2x3⁄(x2 + 2)2
    = 4x⁄(x2 + 2)2
    For critical point,
    f ' (x) = 0
    4x⁄(x2 + 2)2 = 0
    4x = 0
    x = 0
    f (0) = 0⁄ 2 = 0
    The critical point is (0, 0).

    3. Find the global minimum and global maximum values of the following functions:
    (a) f (x) = 3x - x3 + 2 on [-5, 5]
    (b) f (x) = sin 2x on [- π ⁄ 4 , π ⁄ 4 ].
    Solution
    (a) f (x) = 3x - x3 + 2 on [-5, 5]
    f ' (x) = 3 - 3x2
    For critical points,
    f ' (x) = 0
    3 - 3x2 = 0
    3x2 = 3
    x2 = 1
    x = ±1
    f (1) = 3 - 1 + 2 = 4
    f (-1) = -3 + 1 + 2 = 0
    f (-5) = -15 + 125 + 2 = 112
    f (5) = 15 - 125 + 2 = -108
    The global minimum value is -108 and the global maximum value is 112.

    (b) f (x) = sin 2x on [- π ⁄ 4 , π ⁄ 4 ].
    f ' (x) = cos 2x ⋅ 2 = 2 cos 2x
    For critical points,
    2 cos 2x = 0
    cos 2x = 0
    2 cos2 x - 1 = 0
    cos2 x = 1⁄2
    cos x = ± 1⁄ √ 2  
    x = ± π⁄4
    f (- π⁄4) = sin 2(- π⁄4) = sin (- π⁄2) = -1
    f (π⁄4) = sin (2 × π⁄4) = sin (π⁄2) = 1
    The global minimum value is -1 and the global maximum value is 1.

    4. Find the open intervals on which the following functions are increasing or decreasing. Identify the function's local maximum and minimum.
    (a) f (x) = x3 - 3x2 + 5
    (b) f (x) =(x2 - 2x) 2 ⁄ 5
    (c) f (x) = x2 + 5 ⁄x
    (d) f (x) = -x3 - 6x2 + 12
    (e) f(x) = x - 3 ⁄ x - 1
    (f) f (x) = x 2 ⁄ 3 (x2 - 16)
    Solution
    (a) f (x) = x3 - 3x2 + 5
    f ' (x) = 3x2 - 6x
    For critical points,
    f ' (x) = 0
    3x2 - 6x = 0
    x2 - 2x = 0
    x(x - 2) = 0
    x = 0 or x = 2
    0 2
    Interval x < 0 0 < x < 2 x > 2
    x - + +
    x - 2 - - +
    sign of f ' (x) + - +
    Behavior of f   increasing   decreasing   increasing  

    The local maximum at x = 0 and local minimum at x = 2.

    (b) f (x) =(x2 - 2x) 2 ⁄ 5
    f ' (x) = 2 ⁄ 5 (x2 - 2x)- 3 ⁄ 5 (2x - 2)
    = 2(2x - 2) ⁄ 5(x2 - 2x) 3 ⁄ 5
    For critical points,
    f ' (x) = 0
    2(2x - 2) = 0
    x = 1
    f ' (x) is undefined
    when x2 - 2x = 0
    x(x - 2) = 0
    x = 0 or x = 2
    0 1 2
    Interval x < 0 0 < x < 2 1 < x < 2 x > 2
    x - 1 - - + +
    x - + + +
    x - 2 - - - +
    sign of f ' (x) - + - +
    Behavior of f   decreasing   increasing   decreasing   increasing  

    The local minimum is at x = 0, x = 2 and the local maximum is at x = 1.

    (c) f (x) = x2 + 5 ⁄x = x + 5x-1
    f ' (x) = 1 - 5x-2 = x2 - 5⁄x2
    It is observed that x = 0 does not belong to the domain of the function.

    (d) f (x) = -x3 - 6x2 + 12
    f ' (x) = -3x2 - 12x
    = -3x (x + 4)
    For critical points,
    -3x2 - 12x = 0
    x2 + 4x = 0
    x(x + 4) = 0
    x = 0 or x = -4
    -4 0
    Interval x < -4 -4 < x < 0 x > 0
    -x + + -
    x + 4 - + +
    sign of f ' (x) - + -
    Behavior of f   decreasing   increasing   decreasing  

    The loca maximum is at x = 0 and the local minimum is at x = -4.

    (e) f(x) = x - 3 ⁄ x - 1
    f ' (x) = (x - 1)1 - (x -3)1 ⁄(x - 1)2 = x - 1 - x + 3 ⁄ (x - 1)2 = 2 ⁄ (x - 1)2
    It is observed that x = 1 does not belong to the domain of the function.

    (f) f (x) = x 2 ⁄ 3 (x2 - 16)
    = x 8 ⁄ 3 - 16 x 2 ⁄ 3
    f ' (x) = 8 ⁄3 x 5 ⁄ 3 - 32 ⁄ 3 x -1 ⁄ 3 = 8 ⁄ 3 x 2 ⁄ 3 (x - 4x-1)
    For critical points,
    8 ⁄ 3 x 2 ⁄ 3 (x - 4x-1) = 0
    x = 0 or x2 - 4 = 0
    x = 0 or x = ±2
    Interval x < -2 -2 < x < 0 0 < x < 2 x > 2
    x2⁄3 + + + +
    x- 4⁄x - + - +
    sign of f ' (x) - + - +
    Behavior of f   decreasing   increasing   decreasing   increasing  

    The local minimum is at x = ±2 and the local maximum is at x = 0.