MAXIMA AND MINIMA (2nd DERIVATIVE TEST)

Created: Thursday, 17 November 2011 08:39 | Published: Thursday, 17 November 2011 08:39 | Written by Super User | Print | Email

Second derivative test


Let ‘f’ be a function defined on an interval I and c Є I.  Let ‘f’ be twice differentiable at ‘c’. Then
•    x = c is a point of local maxima if f’(c) = 0 and f”(c) < 0. The value   f(c) is local maximum value of ‘f’.
•    x = c is a point of local minima if f’(c) = 0 and f”(c) > 0. The value    f(c) is local minimum value of ‘f’.
•    The test fails, if f’(c) = 0 and f”(c) = 0. In this case, we have to go for 1st derivative test and find whether ‘c’ is a point of local maxima, local minima or a point of inflection.

Let’s understand the concept with the help of following example:
Find the local maximum and local minimum of the function:                    

f(x) = 3x4 - 4x3 -12x2 + 12
We have f(x) = 3x4 - 4x3 - 12x2 + 12
f’(x) = 12x3 - 12x2 - 24x
f’(x) = 12x (x - 1)(x + 2)
f”(x) = 36x2 + 24x – 24 = 12(3x2 + 2x - 1)
f”(0) = -12 < 0
f”(1) = 48 > 0
f”(-2)= 84 > 0
Hence by second derivative test, x = 0 is a point of local maxima and local maximum value is f(0) = 12, while x = 1 and x = -2 are the points of local minima and local minimum values of ‘f’ are 7 and -20 respectively.


Maximum and minimum values of a function in a closed interval

Let f be a continuous function of an interval I = [a, b].  The f has absolute minimum value and absolute maximum value in I.
Working Rule:
Step I: Find all critical points of ‘f’ in the interval
Step II: Take the end points of the interval
Step III: At all these points calculate the values of ‘f’
Step IV: Identify the maximum and minimum values of ‘f’ out of the values calculated in Step 3.  The maximum will be the absolute maximum value of f and the minimum value will be the absolute minimum value of f.
Let’s understand the concept with the help of following example:
Find the absolute maximum and minimum values of a function f given by f(x) = 2x3 - 15x2 + 36x + 1 on the interval [1, 5]
Given f(x) = 2x3 - 15x2 + 36x + 1
f’(x) = 6x2 - 30x + 36
        = 6(x - 3) (x - 2)
f’(x) = 0 gives x = 2, 3
f(1) = 24
f(2) = 29
f(3) = 28
f(5) = 56
Hence absolute maximum value of f is 56 at x = 5 and absolute minimum value of f is 24 at x = 1.

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