📖 generic · CBSE Class 12th English Medium · MATHEMATCS PART-1 · Page 23question

A Note The reader may note that in Example 35, we have used first derivative · Part 5

Chapter 6: APPLICATION OF DERIVATIVES · MATHEMATCS PART-1

( ) – ( ) + ( ) + = f ( ) = ( ) – ( ) + ( ) + = f ( ) = ( ) – ( ) + ( ) + = Thus, we conclude that absolute maximum value of f on [ , ] is , occurring at x = , and absolute minimum value of f on [ , ] is which occurs at x = . Example Find absolute maximum and minimum values of a function f given by [ , ] f x ∈− Solution We have f ( x ) = f ′ ( x ) = ( ) Thus, f ′ ( x ) = gives x = . Further note that f ′ ( x ) is not defined at x = . So the critical points are x = and x = .

Now evaluating the value of f at critical points x = , and at end points of the interval x = – and x = , we have f (– ) = ( ) ( ) f ( ) = ( ) – ( ) = f   = f ( ) = ( ) ( ) Hence, we conclude that absolute maximum value of f is that occurs at x = – and absolute minimum value of f is − that occurs at x = Example An Apache helicopter of enemy is flying along the curve given by y = x + . A soldier, placed at ( , ), wants to shoot down the helicopter when it is nearest to him. Find the nearest distance. Solution For each value of x , the helicopter’s position is at point ( x , x + ).

Therefore, the distance between the helicopter and the soldier placed at ( , ) is ) ) , i.e., ) Let f ( x ) = ( x

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