If f(x) is differentiable on the interval, except possibly at c, then f (c) can be classified as follows:
(when moving across the interval I from left to right)
(i) If f′(x) changes from negative to positive at c , then f (x) has a local minimum f (c) .
(ii) If f′(x) changes from positive to negative at c , then f (x) has a local maximum f (c) .
(iii) If f′(x) is positive on both sides of c or negative on both sides of c , then f (c) is neither a local minimum nor a local maximum.
Find the intervals of monotonicities and hence find the local extremum for the following functions:
(i) f(x) = 2x3 + 3x2 −12x
Solution :
f(x) = 2x3 + 3x2 −12x
f'(x) = 6x2+6x-12
f'(x) = 0
6x2+6x-12 = 0
6(x2+x-2) = 0
(x+2)(x-1) = 0
x = -2 and x = 1
f'(x) changes from positive to negative when it passes through x = -2, so it has local maximum at x = -2.
Local maximum at x = -2 :
f(-2) = 2(-2)3 + 3(-2)2 −12(-2)
f(-2) = 2(-8) + 3(4) + 24
f(-2) = -16 + 12 + 24
f(-2) = 20
f'(x) changes from negative to positive when it passes through x = -2, so it has local maximum at x = 1.
Local minimum at x = 1 :
f(1) = 2(1)3 + 3(1)2 −12(1)
f(1) = 2+3-12
f(1) = -7
So, local maximum is 20 and local minimum is -7.
(ii) f(x) = x/(x-5)
Solution :
f(x) = x/(x-5)
u = x and v = x-5
u' = 1 and v' = 1
f'(x) = [(x-5)(1) - x(1)]/(x-5)2
f'(x) = -5 < 0
It is strictly decreasing, so there is no local extrema.
(iii) f(x) = ex/(1-ex)
Solution :
f(x) = ex/(1-ex)
u = ex and v = 1-ex
u' = ex and v' = -ex
f'(x) = [(1-ex)ex - ex(-ex)]/(1-ex)2
f'(x) = ex/(1-ex)2 > 0
It is strictly increasing, so there is no local extrema.
(iv) f(x) = (x3/3) - log x
Solution :
f(x) = (x3/3) - log x
f'(x) = (3x2/3) - (1/x)
f'(x) = x2 - (1/x)
f'(x) = (x3 - 1)/x
f'(x) = (x3 - 1)/x
x3 - 1 = 0
x3 = 1
x = 1
The domain of the given function is (0, ∞).
So, the intervals are (0, 1) and (1, ∞).
The function is strictly decreasing at (0, 1). f'(x) changes its sign from negative to positive, so it will have local minimum at x = 1.
Local minimum at x = 1 :
f(1) = (13/3) - log 1
f(1) = 1/3
(v) f(x) = sinx cosx + 5, x ∈ (0, 2π)
Solution :
f(x) = (2/2)sinx cosx + 5
f(x) = (1/2) sin 2x + 5
f'(x) = (2/2) cos 2x
f'(x) = cos 2x
f'(x) = 0
cos 2x = 0
2x = cos-1(0)
2x = π/2, 3π/2, 5π/2, 7π/2
x = π/4, 3π/4, 5π/4, 7π/4
So, the intervals are
(0, π/4), (π/4, 3π/4), (3π/4, 5π/4), (5π/4, 7π/4)
Local Minimum at x = π/4 :
f(π/4) = (1/2) sin 2(π/4) + 5 ==> 11/2
Local Maximum at x = 3π/4 :
f(3π/4) = (1/2) sin 2(3π/4) + 5
= (1/2) (-1) + 5
= (-1/2)+5
= 9/2
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