The absolute maxima and absolute minima are referred to describing the largest and smallest values of a function on an interval.
Let x0 be a number in the domain D of a function f(x) . Then f (x0) is the absolute maximum value of f (x) on D,
if f(x0) ≥ f(x) for all x ∈ D and f(x0) is the absolute minimum value of f(x) on D if f(x0) ≤ f(x)
A procedure for finding the absolute extrema of a continuous function f (x) on closed interval [a,b] .
Step 1 :
Find the critical numbers of f (x) in (a, b)
Step 2 :
Evaluate f(x) at all the critical numbers and at the endpoints a and b
Step 3 :
The largest and the smallest of the values in step 2 is the absolute maximum and absolute minimum of f(x) respectively on the closed interval [a, b] .
Find the absolute extrema of the following functions on the given closed interval.
(i) f (x) = x3 − 12x + 10 ; [1, 2]
Solution :
f (x) = x3 − 12x + 10
f'(x) = 3x2-12(1)
f'(x) = 3x2-12
f'(x) = 0
3x2-12 = 0
x2 = 4
x = ±2
x = 2 ∈ [1, 2]
f(1) = 13 − 12(1) + 10 ==> -1
f(2) = 23 − 12(2) + 10
= 8-24+10
= -6
Absolute maximum is -1 and absolute minimum is -6.
(ii) f(x) = 3x4-4x3 ; [-1, 2]
Solution :
f(x) = 3x4-4x3
f'(x) = 12x3 - 12x2
f'(x) = 0
12x3 - 12x2 = 0
12x2(x-1) = 0
x = 0 and x = 1
f(x) = 3x4-4x3
f(-1) = 3(-1)4-4(-1)3 = 7
f(0) = 3(0)4-4(0)3 = 0
f(1) = 3(1)4-4(1)3 = -1
f(2) = 3(2)4-4(2)3 = 16
So, absolute maximum is 16 and absolute minimum is -1.
(iii) f(x) = 6x4/3 - 3x1/3 ; [-1, 1]
Solution :
f(x) = 6x4/3 - 3x1/3
f'(x) = 6 (4/3) x1/3 - 3(1/3)x-2/3
f'(x) = 8x1/3 - x-2/3
f'(x) = 0
8x1/3 - x-2/3 = 0
8x1/3 = x-2/3
x1/3 x2/3 = 1/8
x = 1/8
(iv) f(x) = 2 cos x + sin 2x ; [0, π/2]
Solution :
f(x) = 2 cos x + sin 2x
f'(x) = -2 sinx + 2 cos 2x
f'(x) = 0
-2 sinx + 2 cos 2x = 0
-2 sinx + 2 (1-2sin2x) = 0
-2 sinx + 2 - 4sin2x = 0
4 sin2x+2sinx-2 = 0
Let t = sinx
4t2+2t-2 = 0
2t2+t-1 = 0
(2t-1) (t+1) = 0
t = 1/2 and t = -1
sin x = 1/2 and sin x = -1
x = sin-1(1/2) x = π/6 |
x = sin-1(-1) x = 3π/2 it is not in [0, π/2] |
f(x) = 2 cos x + sin 2x
f(π/2) = 2 cos π/2 + sin 2(π/2)
f(π/2) = 0
f(0) = 2 cos 0 + sin 2(0)
f(0) = 2
f(π/6) = 2 cos π/6 + sin 2(π/6)
f(π/2) = 1+√3/2
So, absolute maximum is 2 and absolute minimum is 0.
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