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Application Of Derivatives

Question
CBSEENMA12035339

Find the intervals in which the function straight f left parenthesis straight x right parenthesis space equals space straight x over 2 plus 2 over straight x is increasing or decreasing.

Solution

Here,  straight f left parenthesis straight x right parenthesis space equals space straight x over 2 plus 2 over straight x comma space space space space space straight x not equal to space 0
therefore space space space space straight f apostrophe left parenthesis straight x right parenthesis space equals space 1 half minus 2 over straight x squared space equals space fraction numerator straight x squared minus 4 over denominator 2 straight x squared end fraction space equals space fraction numerator left parenthesis straight x minus 2 right parenthesis thin space left parenthesis straight x plus 2 right parenthesis over denominator 2 straight x squared end fraction
(i) For f(x) to be increasing,  straight f apostrophe left parenthesis straight x right parenthesis space greater than space 0
therefore space space space space space fraction numerator left parenthesis straight x minus 2 right parenthesis thin space left parenthesis straight x plus 2 right parenthesis over denominator 2 straight x squared end fraction greater than 0 space space space space space space space space space space rightwards double arrow space space space space left parenthesis straight x minus 2 right parenthesis thin space left parenthesis straight x plus 2 right parenthesis thin space greater than space 0
rightwards double arrow space space space space space space either space straight x space less than negative 2 space space space space space space space space space space space space space or space space space space straight x space greater than space 2
therefore space space space space space space space straight f left parenthesis straight x right parenthesis space is space increasing space for space straight x space greater than 2 space space space or space space straight x space less than space minus space 2.
(ii) For f(x) to be decreasing, straight f apostrophe left parenthesis straight x right parenthesis space less than space 0
therefore space space space space space space fraction numerator left parenthesis straight x minus 2 right parenthesis thin space left parenthesis straight x plus 2 right parenthesis over denominator 2 straight x squared end fraction less than 0 space space space space space space space space space space space rightwards double arrow space space space space left parenthesis straight x minus 2 right parenthesis thin space left parenthesis straight x plus 2 right parenthesis thin space less than 0
rightwards double arrow space space space space minus 2 less than straight x less than 2
therefore space space space straight f left parenthesis straight x right parenthesis space decreases space for space minus 2 less than straight x less than 2.