FINDING ABSOLUTE MAX AND MIN OF A FUNCTION
What is absolute maximum ?
An absolute maximum point is a point where the function obtains its greatest possible value.
What is absolute minimum ?
An absolute minimum point is a point where the function obtains its least possible value.
Working rule :
Step 1 :
Let us assume the given function as f(x). Find the first derivative, that is f'(x).
Step 2 :
Equate the first derivative to 0. That is f'(x) = 0 and solve for the variable.
Step 3 :
The values will lie in between the given interval. These values are called critical numbers.
Apply the critical numbers and values we find in the closed interval in f(x).
- For what value of x, we get the maximum value of y, that can be fixed as absolute maximum.
- For what value of x, we get the minimum value of y, that can be fixed as absolute minimum.
Problem 1 :
y = x3 - 3x2 - 3; (0, 3)
Solution :
Let y = f(x) = x3 - 3x2 - 3
f'(x) = 3x2 - 6x
f'(x) = 0
3x2 - 6x = 0
3x(x - 2) = 0
3x = 0 and x - 2 = 0
x = 0 and x = 2
Put x = 0
f(0) = (0)3 - 3(0)2 - 3
f(0) = -3
Put x = 3
f(3) = (3)3 - 3(3)2 - 3
f(3) = 27 - 27 - 3
f(3) = -3
Put x = 2
f(2) = (2)3 - 3(2)2 - 3
f(2) = 8 - 12 - 3
f(2) = -7
Absolute minimum is at (2, -7) and there is no absolute maximum.
Problem 2 :
y = (5x + 25)1/3 ; [-2, 2]
Solution :
Let y = f(x) = (5x + 25)1/3
f'(x) = 1/3 (5x + 25)-2/3 ⋅ 5
5x + 25 = 0
5x = -25
x = -25/5
x = -5
Put x = -2
f(-2) = ∛(5(-2) + 25)
= ∛(-10 + 25)
= ∛15
Put x = 2
f(2) = ∛(5(2) + 25)
= ∛(10 + 25)
= ∛35
Put x = -5
f(-5) = ∛(5(-5) + 25)
= ∛(-25 + 25)
= ∛0
Absolute minimum is at (-2, ∛15) and absolute maximum is at (2, ∛35).
Problem 3 :
y = x3 - 3x2 + 6; [0, ∞)
Solution :
Let y = f(x) = x3 - 3x2 + 6
f'(x) = 3x2 - 6x
f'(x) = 0
3x2 - 6x = 0
3x(x - 2) = 0
3x = 0 and x - 2 = 0
x = 0 and x = 2
Put x = 0
f(0) = (0)3 - 3(0)2 + 6
f(0) = 6
Put x = 2
f(2) = (2)3 - 3(2)2 + 6
f(2) = 8 - 12 + 6
f(2) = 2
Absolute minimum is at (2, 2) and there is no absolute maximum.
Problem 4 :
y = x4 - 2x2 - 3; (0, ∞)
Solution :
Let y = f(x) = x4 - 2x2 - 3
f'(x) = 4x3 - 4x
f'(x) = 0
4x3 - 4x = 0
4x(x2 - 1) = 0
4x = 0 and x2 - 1 = 0
x = 0 and x = 1
Put x = 1
f(1) = 14 - 2(1)2 - 3
= 1 - 2 - 3
= -4
Absolute minimum is at (1, -4) and there is no absolute maximum.
Problem 5 :
Solution :
f(x) = 4(x2 + 2)-1
f'(x) = 4(-1) (x2 + 2)-2 (2x)
-8x = 0 and (x2 + 2)2 = 0
x = 0 and √(x2 + 2)2 = 0
x2 + 2 = 0
x2 = - 2
x = ±√-2
Absolute maximum is at (-2, 2/3) and there is no absolute minimum.
Problem 6 :
y = (x + 2)2/3; [-4, -2]
Solution :
Let y = f(x) = (x + 2)2/3
f'(x) = 2/3 (x + 2)-1/3 ⋅ 1
x + 2 = 0
x = -2
Put x = -4
f(-4) = ∛((-4) + 2)2
= ∛(-2)2
= ∛4
Put x = -2
f(-2) = ∛(-2 + 2)2
= 0
Absolute minimum is at (-2, 0) and absolute maximum is at (-4, ∛4).
Problem 7 :
Solution :
f'(x) = 0
|
3x2 - 12x = 0 3x(x - 4) = 0 x - 4 = 0 x = 4 |
(3x - 6)2 = 0 3x - 6 = 0 3x = 6 x = 6/3 x = 2 |
Absolute minimum is at (4, 8/3) and absolute maximum is at (3, 3), (6, 3).
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