Relative Minimum and Maximum

 A relative maximum or minimum is a little different.  All that is necessary for a point to be a relative maximum or minimum is for that point to be a maximum or minimum in some interval of x’s around x=c. There may be better or lesser values of the function at some other place, but relative to x=c or local to x=c, f(c) is larger or smaller than all the other function values that are near it.

The small point of a particular section of a graph

Note: The initial derivative test and the second derivative test are common methods used to find minimum values of a function.

The biggest point in a particular section of a graph

Note: The initial derivative test and the second derivatives test are common methods used to find maximum values of a function.


Relative Maximum and Minimum Examples:

 

Relative Maximum and Minimum - Example 1:

 Find the maximum function value of y = -5x2 + x – 10

Sol :  Y= -5x2 + x – 10

   = -5(x2 – 1 / 5) x – 10

   = -5(x – 1/10)2 + 1/20 – 10

   = -5(x – 1/10)2 – 9 11/20

Maximum value is – 9 11/20


Relative Maximum and Minimum - Example 2:

For the parabola with equation:-

          `Y= -5x^2-20x-30`

(i)  Find out whether it has a maximum or a minimum value.

(ii) Find out the maximum or minimum value.

(iii)Find out the value of x where the maximum or minimum value occurs.

Sol :

(i)  Since -5 is less than zero, the parabola opens downward and has a maximum value.

(ii) Complete the square

                   `Y= -5x^2-20x-30`

                   Y= -5[x2-4x]-30

                   Y= -5[(x2-4x+4)-4]-30

                   Y= -5(x+2)2 +20-30

                   Y= -5(x+2)2-10  

(iii)           The maximum value of the function is -10.

(iv)           The maximum value occurs when x=-2.

 

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Relative Maximum and Minimum - Example 3:

Find the maximum value of y = -3x2 + x – 5

Sol :  Y= -3x2 + x – 5

   = -3(x2 –`1/3` ) x – 5

   = -3(x –`1/6` )2 +`1/12` – 5

   = -3(x – `1/6` )2 – 4 `11/12`

Maximum value is – 4 `11/12.`

Find the relative maximum and relative minimum of the function f (x) = 2x3 - 21x2 +36x - 20. Find also the relative maximum and relative minimum values.

Sol :  f '(x) = 6x2 - 42x + 36

f '(x) = 0

 => 6x2 – 42x +36 = 0

=> 6(x2 – 7x +6) = 0

=> 6(x-1)(x-6) = 0

=> x = 1 and x = 6 are the critical values

f ''(x) =12x - 42

If x =1, f ''(1) =12 - 42 = - 30 < 0

=>x =1 is a point of relative maximum of f (x).

Maximum value = 2(1)3 - 21(1)2 + 36(1) - 20 = -3

If x = 6, f ''(6) = 72 - 42 = 30 > 0

=>x = 6 is a point of relative minimum of f (x)

Minimum value = 2(6)3 - 21 (6)2 + 36 (6)- 20

= -128