FINDING ABSOLUTE MAXIMUM AND MINIMUM ON A CLOSED INTERVAL
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.
For each problem, find all points of absolute minima and maxima on the given closed interval.
Problem 1 :
y = −x3 − 6x2 − 9x + 3; [−3, −1]
Solution :
Lat f(x) = −x3 − 6x2 − 9x + 3
f('x) = -3x2 - 12x - 9
f'(x) = 0
-3x2 - 12x - 9 = 0
-3(x2 + 4x + 3) = 0
(x + 1)(x + 3) = 0
Critical numbers :
x = -1 and x = -3
Finding absolute maximum and minimum :
Critical numbers are -1 and -3
When x = -3
f(x) = −x3 − 6x2 − 9x + 3
f(-3) = −(-3)3 − 6(-3)2 − 9(-3) + 3
= 27 - 54 + 27 + 3
= 3 (minimum)
When x = -1
f(x) = −x3 − 6x2 − 9x + 3
f(-1) = −(-1)3 − 6(-1)2 − 9(-1) + 3
= 1 - 6 + 9 + 3
= 13 - 6
= 7 (maximum)
So, absolute minimum is at (-3, 3) and absolute maximum is at (-1, 7).
Problem 2 :
Solution :
Critical number is x = 0
Finding absolute maximum or minimum :
So, absolute minimum is at (5, 8/29) and absolute maximum is at (0, 2).
Problem 3 :
f(x) = x3 + 6x2 + 9 x + 3 ; [-4, 0]
Solution :
f(x) = x3 + 6x2 + 9 x + 3
f'(x) = 3x2 + 12x + 9
f'(x) = 0
3x2 + 12x + 9 = 0
3(x2 + 4x + 3) = 0
(x + 1)(x + 3) = 0
x = -1 and x = -3
Critical numbers are -3 and -1.
Finding maximum and minimum :
Applying x = -4 in f(x):
f(-4) = (-4)3 + 6(-4)2 + 9(-4) + 3
= -64 + 96 - 36 + 3
f(-4) = 1
Applying x = -3 in f(x):
f(-3) = (-3)3 + 6(-3)2 + 9(-3) + 3
= -27 + 54 - 27 + 3
f(-3) = 3
Applying x = -1 in f(x):
f(-1) = (1)3 + 6(1)2 + 9(1) + 3
= 1 + 6 + 9 + 3
f(-1) = 19
Applying x = 0 in f(x):
f(0) = (0)3 + 6(0)2 + 9(0) + 3
= 3
f(0) = 3
Absolute maxima are at (0, 3) and (-3, 3)
Absolute minimum are at (-4, 1) and (-1, 19)
Problem 4 :
f(x) = x4 - 3x2+ 4; [-1, 1]
Solution :
f(x) = x4 - 3x2+ 4; [-1, 1]
f'(x) = 4x3 - 6x + 0
f'(x) = 2x(x2 - 3)
Finding critical numbers :
f'(x) = 0
2x(x2 - 3) = 0
x = 0 and x2 = 3
x = ±√3
So, critical numbers are 0, √3 and -√3.
Finding absolute maximum and minimum :
Applying x = -1 in f(x) :
f(x) = x4 - 3x2+ 4
f(-1) = (-1)4 - 3(-1)2+ 4
= 1 - 3 + 4
= 2
Applying x = 0 in f(x) :
f(x) = x4 - 3x2+ 4
f(0) = 04 - 3(0)2+ 4
= 4
Applying x = 1 in f(x) :
f(x) = x4 - 3x2+ 4
f(1) = 14 - 3(1)2+ 4
= 2
Absolute minima are at (-1, 2) and (1, 2).
Absolute maximum is at (0, 4).
Problem 5 :
Solution :
Critical numbers are x = -3, -1 and 1.
Finding maximum and minimum :
Here x = 1 is not in the given interval.
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