📖 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 4

Chapter 6: APPLICATION OF DERIVATIVES · MATHEMATCS PART-1

Let f be a differentiable function on a closed interval I and let c be any interior point of I. Then (i) f ′ ( c ) = if f attains its absolute maximum value at c . (ii) f ′ ( c ) = if f attains its absolute minimum value at c . In view of the above results, we have the following working rule for finding absolute maximum and/or absolute minimum values of a function in a given closed interval [ a , b ].

Working Rule Step : Find all critical points of f in the interval, i.e., find points x where either x = ′ or f is not differentiable. Step : Take the end points of the interval. Step : At all these points (listed in Step and ), calculate the values of f . Step : Identify the maximum and minimum values of f out of the values calculated in Step .

This maximum value will be the absolute maximum (greatest) value of f and the minimum value will be the absolute minimum (least) value of f . Fig . Example Find the absolute maximum and minimum values of a function f given by f ( x ) = x – x + x + on the interval [ , ]. Solution We have f ( x ) = x – x + x + f ′ ( x ) = x – x + = ( x – ) ( x – ) Note that f ′ ( x ) = gives x = and x = .

We shall now evaluate the value of f at these points and at the end points of the interval [ , ], i.e., at x = , x = , x = and at x = . So f ( ) = ( ) – ( ) + ( ) + = f ( ) =

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