LET’S BRUSH UP WITH MAXIMA & MINIMA

Do problems on Maxima and Minima haunt you ? So please stop nightmaring it.
And go through the article.Try to get into the thorough concept.

Before starting doing sums related to maxima and minima. let’s have a brief tour over the basics of Maxima and Minima..

( I ) A function,f (x) attains its maximum value at point x = c
When f (c + h ) – f (c ) < 0 ,[ where h is a very small increment to c]
And f (c- h ) – F ( c ) < 0
( II) And f(x ) attains its minimum value at point x = c
When f (c+ h ) – f ( c ) > 0
And f (c – h ) – f ( c ) > 0

And now the problem is how to determine whether the function attains its maximum or mininimum value at point c and also how to determine the value of point c.
The steps involve to find out the above is described below

(1 ) 1st of all we have to find f ‘ (x) and f “ (x)
[ Where f ‘(x ) = dy/ dx , and f “ ( x ) = d²x / dy² ]

( 2 ) Then we have to equate f ‘ ( x ) = 0 and have to sove out the corresponding value(s ) of x ., Let they be c1 and c 2.

(3 ) Now have to find the f “ ( c1 ) and f ( c 2 )

( 4 ) now if f” ( c1 ) > 0 ;the function will attain the minimum value at x = c1

And if f “ ( c 2 ) < 0 ; the function will attain the maximum value = c2

Now we will try to relate the above when solving problems based on maxima and minima.

(a) Find the turning point(s) of the following function and atain it is maximum &\or minimum.
Y = x3 – 9x 2 + 15x + 11

To find the turning point(s) of the given function we have to follow the steps as shown below

Step : 1

The given function Y = f ( x) = x3 – 9x 2 + 15x + 11

1st of all We have to find ,dy/dx i.e . ( f ‘ (x ) ) and we also have to find f”(x)


f(x) = x3 – 9x 2 + 15x + 11

f ‘( x ) = 3 x ² - 18 x + 15

f “ ( x ) = 6 x -18

Step:2

Now equate f’(x) = 0

f’(x) = 3 x ² - 18 x + 15 = 0
Now solve for corresponding x from the above equation

3 x ² - 18 x + 15 = 0

Or, 3 ( x² - 6 x + 5 ) = 0

Or, ( x² - 6 x + 5 ) = 0 [ dividing both sides by 3 ]

Or , [ x ² - ( 5 + 1 ) x + 5 ] = 0

Or, [ x ² - 5 x – x + 5 ] = 0 [ using middle term factorization ]

Or , [ x ( x – 5 ) –1 ( x -5 ) ] = 0

Or , [ ( x – 5 ) ( x – 1 )] = 0

Either,( x – 5 ) = 0 , or ( x – 1 ) = 0 [ using zero factor theorem]

We get , x = 5

And , x = 1

So the turning points of the given function are at ( x = 5 , x = 1)
Step : 3

We have f” ( x ) = 6x – 18

f” ( 5 ) = 6 * 5 – 18

= 12

So , f “ ( 5 ) > 0 , and the function attains its minimum value at x = 5

Now , f “ ( 1 ) = 6 * 1 – 18
= - 12

So f “( 1 ) < 0 , and the function attains its maximum value at x = 1

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