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 ∨ 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