FIND LOCAL MAXIMUM AND LOCAL MINIMUM OF A FUNCTION

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)

Find the intervals of monotonicities and hence find the local extremum for the following functions:

Problem 1 :

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.

Problem 2 :

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.

Problem 3 :

f(x)  =  e^x/(1-e^x)

Solution :

f(x)  =  e^x/(1-e^x)

u  =  e^x and v  =  1-e^x

u'  =  e^x and v'  =  -e^x

f'(x)  =  [(1-e^x)e^x - e^x(-e^x)]/(1-e^x)²

f'(x)  =  e^x/(1-e^x)²  > 0

It is strictly increasing, so there is no local extrema.

Problem 4 :

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

Problem 5 :

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)

• At x = π/4, f'(x) changes from positive to negative. So, local maximum at x = π/4.
• At x = 3π/4, f'(x) changes from negative to positive.  So, local minimum at x = 3π/4.
• At x = 5π/4, f'(x) changes from positive to negative. So, local maximum at x = 5π/4.
• At x = 7π/4, f'(x) changes from negative to positive.  So, local minimum at x = 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

Problem 6 :

Let f be a function defined on the closed interval −5 ≤ x ≤ 5 with f(1) = 3 . The graph of f', the derivative of f, consists of two semicircles and two line segments, as shown above.

(a) For −5 < x < 5 , find all values of x at which f has a relative maximum. Justify your answer.

(b) For −5 < x < 5 , find all values of x at which f has a point of inflection. Justify your answer.

(c) Find all intervals on which the graph of f (not shown) is concave up. Justify your answer.

(d) Find all intervals on which the graph of f (not shown) has a positive slope. Justify your answer.

Solution :

a)

• In the interval [-3, 1] the graph of the first derivative is below the x-axis. Then it is decreasing in this interval.
• In the interval [-5, -3] and [1, 4], the graph is above the x-axis. Then it is increasing in this interval.
• In the interval [4, 5], the graph is below the x-axis. Then it is decreasing in this interval.

So,  f has a relative maximum at x =−3 and x = 4 because f'(x) changes signs from positive to negative.

b) f has a point of inflection at 4, -1, and 2 because f" changes signs

c) f is concave up on (-5, -4) and (-1, 2) because f' is increasing or f " > 0.

d) f has a positive slope on (-5, -3) and (1, 4) because f'(x) > 0.

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