-->

Application Of Derivatives

Question
CBSEENMA12035387

Find the absolute maximum value and the absolute minimum value of 
straight f left parenthesis straight x right parenthesis space equals space 1 third straight x cubed minus 3 straight x squared plus 5 straight x plus 8 space in space left square bracket 0 comma space 4 right square bracket.

Solution

The function straight f left parenthesis straight x right parenthesis space equals space 1 third straight x cubed minus 3 straight x squared plus 5 straight x plus 8 is differentiable for all x in open square brackets 0 comma space 4 close square brackets
and straight f apostrophe left parenthesis straight x right parenthesis space equals space straight x squared minus 6 straight x plus 5
Now,  straight f apostrophe left parenthesis straight x right parenthesis space equals space 0 space space space space space space when space straight x squared minus 6 straight x plus 5 space equals space 0
i.e.,    when (x - 5)  (x - 1) = 0
i.e.,   when x = 1     or    x = 5
But only 1 space element of space space space open square brackets 0 comma space 4 close square brackets
Now,    straight f left parenthesis 1 right parenthesis space equals space 1 third left parenthesis 1 right parenthesis cubed space minus space 3. space space left parenthesis 1 right parenthesis squared space plus space 5 space left parenthesis 1 right parenthesis space plus space 8 space equals space 31 over 3
           straight f left parenthesis 0 right parenthesis space equals space 8 comma space space space space space straight f left parenthesis 4 right parenthesis space equals space 1 third left parenthesis 4 right parenthesis cubed space minus space 3 space left parenthesis 4 right parenthesis squared space plus space 5 space left parenthesis 4 right parenthesis space plus space 8 space equals space 4 over 3.
Hence absolute maximum value is 31 over 3 and absolute minimum value is 4 over 3. These are attained at 1 and 4 respectively.