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| Maxima and Minima |
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| A function f(x) is said to have a local maximum at x = a, if $ is a neighbourhood I of 'a', such that |
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f(a)  f(x) for all x  I |
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| The number f(a) is called the local maximum of f(x). The point a is called the point of maximum. |
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| Note that when 'a' is the point of local maxima, f(x) is increasing for all values of x < a and f (x) is decreasing for all values of x > a in the given interval. |
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| At x = a, the function ceases to increase. |
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| A function f(x) is said to have a local minimum at x = a, if $ is a neighbourhood I of 'a', such that |
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f(a)  f(x) for all x  I |
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| Here, f(a) is called the local minimum of f(x). The point a is called the point of minimum. |
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| Note that, when a is a point of local minimum f (x) is decreasing for all x < a and f (x) is increasing for all x > a in the given interval. At x = a, the function ceases to decrease. |
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| If f(a) is either a maximum value or a minimum value of f in an interval I, then f is said to have an extreme value in I and the point a is called the extreme point. |
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| A function is said to be monotonic if it is either increasing or decreasing but not both in a given interval. |
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| Consider the function |
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| The given function is increasing function on R. Therefore it is a monotonic function in [0,1]. It has its minimum value at x = 0 which is equal to f (0) =1, has a maximum value at x = 1, which is equal to f (1) = 4. |
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| Here we state a more general result that, 'Every monotonic function assumes its maximum or minimum values at the end points of its domain of definition.' |
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| Note that 'every continuous function on a closed interval has a maximum and a minimum value.' |
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| (First Derivative Test) |
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| Let f (x) be a real valued differentiable function. Let a be a point on an interval I such that f '(a) = 0. |
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| (a) a is a local maxima of the function f (x) if |
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i) f (a) = 0 |
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ii) f (x) changes sign from positive to negative as x increases through a. |
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That is, f (x) > 0 for x < a and |
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f (x) < 0 for x > a |
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| (b) a is a point of local minima of the function f (x) if |
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i) f (a) = 0 |
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ii) f (x) changes sign from negative to positive as x increases through a. |
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That is, f (x) < 0 for x < a |
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f (x) > 0 for x > a |
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| Let f (x) be the real valued differentiable function. |
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| Step 1: Find f '(x) |
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| Step 2: Solve f '(x) = 0 to get the critical values for f (x). Let these values be a, b, c. These are the points of maxima or minima. |
| Arrange these values in ascending order. |
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| Step 3: Check the sign of f'(x) in the immediate neighbourhood of each critical value. |
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| Step 4: Let us take the critical value x= a. Find the sign of f '(x) for values of x slightly less than a and for values slightly |
| greater than a. |
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| (i) If the sign of f '(x) changes from positive to negative as x increases through a, then f (a) is a local maximum value. |
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| (ii) If the sign of f '(x) changes from negative to positive as x increases through a, then f (a) is local minimum value. |
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| (iii) If the sign of f (x) does not change as x increases through a, then f (a) is neither a local maximum value not a minimum value. In this case x = a is called a point of inflection. |
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| Example: |
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| Find the local maxima or local minima, if any, for the following function using first derivative test |
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| f (x) = x3 - 6x2 + 9x + 15 |
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| Solution: |
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| f (x) = x3 - 6x2 + 9x + 15 |
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| f ' (x) = 3x2 -12x + 9 |
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| = 3(x2- 4x + 3) |
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| = 3 (x - 1) (x - 3) |
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| Thus x = 1 and x = 3 are the only points which could be the points of local maxima or local minima. |
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| Let us examine for x=1 |
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| When x<1 (slightly less than 1) |
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| f '(x) = 3 (x - 1) (x - 3) |
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| = (+ ve) (- ve) (- ve) |
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| = + ve |
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| When x >1 (slightly greater than 1) |
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| f '(x) = 3 (x -1) (x - 3) |
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| = (+ ve) (+ ve) (- ve) |
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| = - ve |
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 The sign of f '(x) changes from +ve to -ve as x increases through 1. |
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 x = 1 is a point of local maxima and |
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| f (1) = 13 - 6 (1)2 + 9 (1) +15 |
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| = 1- 6 + 9 + 15 =19 is local maximum value. |
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| Similarly, it can be examined that f '(x) changes its sign from negative to positive as x increases through the point x = 3. |
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| \ x = 3 is a point of minima and the minimum value is |
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| f (3) = (3)3- 6 (3)2+ 9(3) + 15 |
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| = 15 |
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Let f be a differentiable function on an interval I and let a  I. Let f "(a) be continuous at a. Then |
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| i) 'a' is a point of local maxima if f '(a) = 0 and f "(a) < 0 |
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| ii) 'a' is a point of local minima if f '(a) = 0 and f "(a) > 0 |
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| iii) The test fails if f '(a) = 0 and f "(a) = 0. In this case we have to go back to the first derivative test to find whether 'a' is a point of maxima, minima or a point of inflexion. |
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| Step 1 |
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| For a differentiable function f (x), find f '(x). Equate it to zero. Solve the equation f '(x) = 0 to get the Critical values of f (x). |
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| Step 2 |
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| For a particular Critical value x = a, find f "'(a) |
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| (i) If f ''(a) < 0 then f (x) has a local maxima at x = a and f (a) is the maximum value. |
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| (ii) If f ''(a) > 0 then f (x) has a local minima at x = a and f (a) is the minimum value. |
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(iii) If f ''(a) = 0 or  , the test fails and the first derivative test has to be applied to study the nature of f(a). |
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| Example: |
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| Find the local maxima and local minima of the function f (x) = 2x3 - 21x2 +36x - 20. Find also the local maximum and local minimum values. |
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| Solution: |
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| f '(x) = 6x2 - 42x + 36 |
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| f '(x) = 0 |
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 x = 1 and x = 6 are the critical values |
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| f ''(x) =12x - 42 |
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| If x =1, f ''(1) =12 - 42 = - 30 < 0 |
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 x =1 is a point of local maxima of f (x). |
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| Maximum value = 2(1)3 - 21(1)2 + 36(1) - 20 = -3 |
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| If x = 6, f ''(6) = 72 - 42 = 30 > 0 |
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 x = 6 is a point of local minima of f (x) |
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| Minimum value = 2(6)3 - 21 (6)2 + 36 (6)- 20 |
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| = -128 |
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| Let f (x) be a real valued function with its domain D. |
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| (i) f(x) is said to have absolute maximum value at x = a if
f(a) ³ f(x) for all x Î
D. |
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| (ii) f(x) is said to have absolute minimum value at x = a if
f(a) £ f(x) for all x Î
D. |
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| The following points are to be noted carefully with the help of the diagram. |
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| Let y = f (x) be the function defined on (a, b) in the graph. |
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| (i) f (x) has local maximum values at |
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| x = a1, a3, a5, a7 |
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| (ii) f (x) has local minimum values at |
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| x = a2, a4, a6, a8 |
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| (iii) Note that, between two local maximum values, there is a local minimum value and vice versa. |
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| (iv) The absolute maximum value of the function is f(a7)and absolute minimum value is f(a). |
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| (v) A local minimum value may be greater than a local maximum value. |
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| Clearly local minimum at a6 is greater than the local maximum at a1. |
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| Let f be a continuous function on an interval I = [a, b]. Then, f has the absolute maximum value and f attains it at least once in I. Also, f has the absolute minimum value and attains it at least once in I. |
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| Let f be a differentiable function on I and let x0 be any interior point of I. Then |
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| (a) If f attains its absolute maximum value at x0, then f ' (x0)= 0 |
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| (b) If f attains its absolute minimum value at x0, then f '(x0) = 0. |
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| In view of the above theorems, we state the following rule for finding the absolute maximum or absolute minimum values of a function in a given interval. |
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| Step 1: Find all the points where f ' takes the value zero. |
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| Seep 2: Take the end points of the interval. |
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| Step 3: At all the points calculate the values of f. |
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| Step 4: Take the maximum and minimum values of f out of the values calculated in step 3. These will be the absolute maximum or absolute minimum values. |
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