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Application Of Derivatives

Question
CBSEENMA12035396

Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:
straight f left parenthesis straight x right parenthesis space equals space 4 straight x minus 1 half straight x squared comma space space space straight x space element of space open square brackets negative 2 comma space space 9 over 2 close square brackets


Solution

Let straight f left parenthesis straight x right parenthesis space equals space 4 straight x minus 1 half straight x squared
It is differentiable for all x in open square brackets negative 2 comma space space space 9 over 2 close square brackets
   straight f apostrophe left parenthesis straight x right parenthesis space equals space 4 minus straight x
straight f apostrophe left parenthesis straight x right parenthesis space equals space 0 space space space rightwards double arrow space space space 4 minus straight x space equals space 0 space space space space space rightwards double arrow space space space straight x space equals space 4 space element of space space open square brackets negative 2 comma space space space 9 over 2 close square brackets
space space space straight f left parenthesis 4 right parenthesis space equals space 16 minus 1 half cross times 16 space equals space 16 minus 8 space equals space 8
space straight f left parenthesis negative 2 right parenthesis space equals space minus 8 minus 1 half left parenthesis negative 2 right parenthesis squared space equals space minus 8 minus 2 space equals space minus 10
straight f open parentheses 9 over 2 close parentheses space equals space 4 space open parentheses 9 over 2 close parentheses space minus space 1 half open parentheses 9 over 2 close parentheses squared space equals space 18 minus 81 over 8 space equals space 63 over 8 space equals space 7 7 over 8
therefore space space space absolute space max. space value space space equals space 8 space space and space aboslute space min. space value space space equals space minus 10