Absolute maximum and minimum of functions of single variable

ABSOLUTE MAXIMUM AND MINIMUM

Definition: Let c be the number in the domain D of a function f.

Then f(c) is the,

a) Absolute maximum value of f on D if f(c) ≥ f(x) for all x ∈ D

b) Absolute minimum value of f on D if f(c) ≤ f(x) for all x ∈ D

An absolute maximum or minimum values is called a global maximum or minimum. The maximum and minimum values of ƒ are called extreme values of f.

The above diagram shows the graph of a function f absolute maximum at d and absolute minimum at a. Note that, (d, f(d)) is the highest point on the graph and (a, (ƒ(a)) is the lowest point. If we consider only value of x near b then f(b) is the largest of those values of f(x) and is called a local maximum value of f. f(c) is called a local minimum value of ƒ because ƒ (c) ≤ f(x) for x near c. The function ƒ also has a local minimum at e.

Definition: Critical point

A number c is said to be critical point of f (x) on D, If f'(c) = 0

Procedure for finding the absolute minima and the absolute maxima

To find the absolute maxima or absolute minima:

• Let f (x) be a function defined on the interval [a, b].

• Take the critical points which lies in [a, b]. Let them be x1, x2,...xn.

• Find f(a), f(b), f(x1), ƒ (x2), ... ƒ (xn)

• From the above, the value minimum value is absolute minima and the maximum value is absolute maxima

Example 130. Find the critical points for (x) = 5x3‒6x+1.

Solution:

The given function

f(x) = 5x3 ‒ 6x + 1

f '(x) = 15x2 ‒ 6

To find the critical point:

f '(x) = 0

15x2‒6=0

3(5x2‒2) = 0

⇒ 5x2‒2=0

⇒ x = ±√(2/5)

Example 131. Find the absolute maximum and minimum for f(x) = 3x4 ‒ 4x3‒ 12x2 + 1 on [−2,3]

Solution: The given function

 f(x)=3x4‒4x3‒12x2+1

is a polynomial, which is continuous on [‒2,3].

 f '(x) = 12x3‒12x2 ‒ 24x

= 12x(x2‒x‒2)

= 12x(x‒2)(x+1)

To find the critical point:

f '(x) = 0

12x(x‒2)(x+1)= 0

⇒x=0,2,‒1

Each of this critical points lies in the interval [‒2,3].

The value of the function at these critical points are

f(0) = 3(0) ‒ 4(0) – 12(0) + 1 = 1

f(2) = 3(2)4 ‒ 4(2)3 ‒ 12(2)2 + 1

= 48‒32‒48+1= ‒31

f(‒1)=3(‒1)4 — 4(‒1)3 ‒ 12(‒1)2 + 1

= 3+4 ‒ 12+ 1 = −4

The value of the function at the end points of the interval [‒2,3] are

f(‒2) = 3x4 ‒ 4x3 ‒ 12x2 + 1

= 3(‒2)4 ‒ 4(‒2)3 ‒ 12(‒2)2 + 1

= 48+32‒48 + 1 = 33

f(3) = 3x4 ‒ 4x3 – 12x2 + 1

= 3(3)4 ‒ 4(3)3 ‒ 12(3)2 + 1

= 3(81) ‒ 4(27) ‒ 12(9) + 1

= 243‒108‒108+1 = 28

Comparing these five values, we see that

The absolute minimum is f(2) = ‒31

The absolute maximum of ƒ (3) = 33

Example 132. Find the absolute maximum and minimum for f(x) = x3‒3x2+1 on 1 on [‒1/2,4]

Solution: The given function

f(x) = x3‒3x2 + 1

is a polynomial, which is continuous on [‒1/2,4]

f '(x) = 3x2‒6x

= 3x(x‒2)

To find the critical point:

f '(x) = 0

3x(x‒2)=0

⇒x=0, x=2 are critical points

Each of this critical points lies in the interval [‒1/2,4].

The value of the function at these critical points are

f(0) = 1

f(2) = (2)3‒3(2)2 + 1

=8‒12+1

=9‒12

= ‒3

The value of the function at the end points of the interval [‒1/2, 4] are

f(‒1/2)=(‒1/2)3‒3(‒1/2)2+1

= ‒ 1/8 ‒ 3/4 ‒ 1

= [‒1‒6‒8] / 8 = ‒ 15/8

ƒ(4) = (4)3 − 3(4)2 + 1

= 64 ‒ 3(16) + 1 = 65‒48

= 17

Comparing these four values, we see that

The absolute minimum of f(x) is f(2) = −3

The absolute maximum of f(x) is ƒ(4) = 17

Example 133. Find the absolute maximum and minimum for f(x) = (x2 ‒ 1)3 on [‒1,2].

Solution: The given function

f(x) = (x2 ‒ 1)3

is a polynomial, which is continuous on [‒1,2].

 f '(x)=3(x2 ‒ 1)2 2x

= 6x(x2‒1)2

= 6x[(x + 1)(x − 1) ]2

To find the critical point:

 f '(x) = 0

6x[(x + 1)(x‒1)]2 = 0

 ⇒ x=0,‒1,1 are critical points

Each of this critical points lies in the interval [‒1,2].

The value of the function at these critical points are

ƒ(0) = ((0)2 ‒ 1)3

= (0‒1)3 = (‒1)3 = ‒1

f(‒1)=((‒1)2 ‒ 1)3 = (1‒1)3 = 0

f(1) = ((1)2 ‒ 1)3 = (1‒1)3 = 0

The value of the function at the end points of the interval [‒1, 2] are

ƒ(‒1) = ((‒1)2 ‒ 1)3 = (1‒1)3 = 0

f(2) = ((2)2 ‒ 1)3 = (4‒1)3 = 27

Comparing these four values, we see that

The absolute minimum of f(x) is f(0) = ‒1

The absolute maximum of f (x) is f(2) = 27

Example 134. Find the absolute maximum and minimum for f(x) = x + 1/x on [0.2,4]

Solution: The given function

f(x) = x + 1/x

is continuous on [0.2, 4].

f(x) = x+x‒1

f '(x) = 1‒x‒2

= 1 ‒ 1/x2

To find the critical point:

 f '(x) = 0

1 ‒ 1/x2 = 0

1 = 1/x2

x2 = 1

x = ±1

The critical point x = 1 lies in the interval [0.2, 4].

The value of the function at the critical point is

f(1) = 1+ 1/1 =2

The value of the function at the end points of the interval [0.2, 4] are

 f(0.2) = 0.2 + 1/0.2 = 0.2 + 5 = 5.2

 f(4) = 4 + ¼ = 4.25

Comparing these three values

The absolute minimum of f(x) is f(1) = 2

The absolute maximum of f(x) is f(0.2) = 5.2

Example 135. Find the absolute maximum and minimum for f(x) = x ‒ 2 sin x on [0,2π]

Solution: The given function

f(x) = x‒2 sin x

is continuous on [0,2π]

 f '(x) = 1 ‒ 2cos x

To find the critical point:

 f '(x) = 0

1‒2cosx = 0

2 cos x = 1

⇒ cosx = 1/2

x = cos‒1(1/2)

x = π/3, 2π ‒ π/3

= π/3, 5π/3

Each of this critical points lies in the interval [0,2π].

The value of the function at these critical points are

 f(π/3) = π/3 ‒ 2sin(π/3)

= π/3 ‒ 2 × √3/2

= π/3 ‒  √3

= 3.14/3  ‒  √3

= ‒ 0.684

f(5π/3) = 5π/3 ‒ 2sin(5π/3)

= 5π/3 ‒ 2 × (‒√3/2)

= 5.23 ‒ 2(‒0.866)

= 6.968

The value of the function at the end points of the interval [0, 2π] are

f(0) = 0 ‒ 2(sin 0) = 0

f(2π) = 2π ‒ 2sin(2π) = 2π‒0 = 2 × 3.14

= 6.28

Comparing these four values

The absolute minimum of ƒ(x) is ƒ(π/3) = −0.68

The absolute maximum of ƒ(x) is ƒ(5π/3) = 6.968

Example 136. Find the absolute maximum and minimum for f(x)= x ‒ log x on [1/2,2].

Solution: The given function

f(x) = x ‒ logx

is continuous on [1/2,2]

ƒ'(x) = 1 – 1/x

To find the critical point:

 f '(x) = 0

1 ‒ 1/x = 0

1/x = 1

x = 1

The critical point x = 1 lies in the interval [1/2,2].

The value of the function at the critical point is

 f(1) = 1 ‒ log(1) = 1

The value of the function at the end points of the interval [1/2,2] are

f(1/2) = 1/2 ‒ log(1/2)

= 0.5+ (0.693) = 1.193

ƒ(2) = 2 – log(2) = 2 ‒ (0.693)

= 1.306

Comparing the three values

The absolute minimum of f(x) is f(1) = 1

The absolute maximum of f(x) is f(2) = 1.306

Example 137. Find the absolute maximum and minimum values of the function f(x) = log(x2 + x + 1) in [−1, 1].

Solution: The given function

f(x) = log (x2 + x + 1)

is continuous on [‒1,1]

f '(x) = 2x+1 / x2+x+1

To find the critical point:

 f '(x) = 0

2x+1 / x2+x+1  = 0

⇒2x+1=0

⇒ x = ‒1/2

The critical point x = ‒1/2 lies in the interval [−1, 1].

The value of the function at the critical point is

 f(‒1/2) = loge (1/4 ‒ 1/2 + 1) = loge(3/4)

= ‒ 0.288

The value of the function at the end points of the interval [‒1, 1] are

f(‒1)= loge(1‒1+1) = loge(1) = 0

f(1) = loge(1 + 1 + 1) = loge(3) = 1.099

Comparing the three values

The absolute minimum of f(x) is f(‒1/2) = ‒0.288

The absolute maximum of f(x) is f(1) = 1.099 

EXERCISE

34. Find the critical points for the following

a. f(x) = 2x3‒3x2 ‒ 36            Ans: 3,‒2

b. f(x) = 2x ‒ 3x3            Ans: √2/3

35. Find the absolute maximum and minimum for f(x) = 2x3 ‒ 3x2 ‒ 12x + 1 on

[‒2,3].     

Ans: The absolute minimum of f(x) is ƒ(−1) = −19

The absolute maximum of f(x) is f(2) = 8

36. Find the absolute maximum and minimum for f(x) = 3x4 ‒ 16x3 + 18x2 on [‒1,4].

Ans: The absolute minimum of f(x) is ƒ(3) = −27

The absolute maximum of f(x) is f(−1) = 37

37. Find the absolute maxima and absolute minima for the following

a. f(x) = x3 − 3x2 + 1 in [‒1/2,4]

Ans: Minimum value f(2) = ‒3, Maximum value f(4) = 17

b. f(x) = x3‒ 12x + 1 in [‒3,5]

Ans: Minimum value f(2) = ‒15, Maximum value ƒ(5) = 66

c. f(x) = xa(1‒x)b in [0,1]

Ans: Minimum value ƒ (0) = f(1) = 0, Maximum value f( a / a+b ) = aabb / (a+b)a+b