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Application Of Derivatives

Question
CBSEENMA12035407

Find the points at which the function f given by f (x) = (x – 2)4 (x + 1 )3 has
(i) local maxima (ii) local minima (iii) point of inflexion .

Solution

Here,     straight f left parenthesis straight x right parenthesis space equals space left parenthesis straight x minus 2 right parenthesis squared space space space left parenthesis straight x plus 1 right parenthesis cubed
therefore space space space space straight f apostrophe left parenthesis straight x right parenthesis space equals space left parenthesis straight x minus 2 right parenthesis to the power of 4. space straight d over dx left parenthesis straight x plus 1 right parenthesis cubed space plus space left parenthesis straight x plus 1 right parenthesis cubed. space space straight d over dx left parenthesis straight x minus 2 right parenthesis to the power of 4
                 equals space left parenthesis straight x minus 2 right parenthesis to the power of 4. space 3 left parenthesis straight x plus 1 right parenthesis squared plus left parenthesis straight x plus 1 right parenthesis cubed. space 4 left parenthesis straight x minus 2 right parenthesis cubed
equals space left parenthesis straight x minus 2 right parenthesis cubed space left parenthesis straight x plus 1 right parenthesis squared space left square bracket 3 left parenthesis straight x minus 2 right parenthesis space plus space 4 left parenthesis straight x plus 1 right parenthesis right square bracket space equals space left parenthesis straight x minus 2 right parenthesis cubed space left parenthesis straight x plus 1 right parenthesis squared space left parenthesis 7 straight x minus 2 right parenthesis
        straight f apostrophe left parenthesis straight x right parenthesis space equals space 0 space space space gives space us space left parenthesis straight x minus 2 right parenthesis cubed space left parenthesis straight x plus 1 right parenthesis squared space left parenthesis 7 straight x minus 2 right parenthesis space equals space 0
therefore space space space space space space space straight x space equals space 2 comma space minus 1 comma space space 2 over 7
When x < 2 slightly, straight f apostrophe left parenthesis straight x right parenthesis space equals space left parenthesis negative right parenthesis thin space left parenthesis plus right parenthesis thin space left parenthesis plus right parenthesis space equals negative ve
When x > 2 slightly, straight f apostrophe left parenthesis straight x right parenthesis space equals space left parenthesis plus right parenthesis thin space left parenthesis plus right parenthesis thin space left parenthesis plus right parenthesis space equals space plus ve
therefore space space space space space at space straight x space equals space 2 comma space space straight f apostrophe left parenthesis straight x right parenthesis space changes space from space minus ve space to space plus ve
therefore space space straight f left parenthesis straight x right parenthesis space has space local space minima space at space straight x space equals space 2
When x < -1 slightly  f ' (x) = (-) (+) (-) = +ve
When x > -1 slightly, f ' (x) = (-) (+) (-) = +ve
therefore    at  x = -1,   f ' (x) does not change sign
therefore          x = -1 is a point of inflexion.
When space straight x thin space less than space 2 over 7 space space slightly comma space space space straight f space apostrophe left parenthesis straight x right parenthesis space equals space left parenthesis negative right parenthesis thin space left parenthesis plus right parenthesis thin space left parenthesis negative right parenthesis space equals space plus ve
When space straight x thin space greater than space 2 over 7 space slightly space comma space space straight f space apostrophe left parenthesis straight x right parenthesis space equals space left parenthesis negative right parenthesis thin space left parenthesis plus right parenthesis thin space left parenthesis plus right parenthesis space equals space minus ve
therefore space space space space at space straight x space equals space 2 over 7 comma space space space straight f space apostrophe left parenthesis straight x right parenthesis space changes space form space plus ve space to space minus ve
therefore space space space straight f left parenthesis straight x right parenthesis space has space local space maxima space at space straight x space equals space 2 over 7