Application of Derivatives

Question

Find the local maxima or local minima, if any, of following functions using the first derivative test only. Find also the local maximum and the local minimum values, as the case may be:
$straight g left parenthesis straight x right parenthesis space equals space fraction numerator 1 over denominator straight x squared plus 2 end fraction space equals space left parenthesis straight x squared plus 2 right parenthesis to the power of negative 1 end exponent$

Solution

Let $straight g left parenthesis straight x right parenthesis space equals space fraction numerator 1 over denominator straight x squared plus 2 end fraction space equals left parenthesis straight x squared plus 2 right parenthesis to the power of negative 1 end exponent$
$therefore space space straight g apostrophe left parenthesis straight x right parenthesis space equals space minus left parenthesis straight x squared plus 2 right parenthesis space squared space space left parenthesis 2 straight x right parenthesis space equals space minus fraction numerator 2 straight x over denominator left parenthesis straight x plus 2 right parenthesis squared end fraction$
$straight g apostrophe left parenthesis straight x right parenthesis space equals space 0 space space space space space rightwards double arrow space space space space space minus fraction numerator 2 straight x over denominator left parenthesis straight x squared plus 2 right parenthesis squared end fraction space equals space 0 space space space space rightwards double arrow space space space space straight x space equals space 0$
$When space straight x space less than 0 space space space slightly comma space space then space space straight g apostrophe left parenthesis straight x right parenthesis space equals space minus 2 fraction numerator left parenthesis negative ve right parenthesis over denominator plus ve end fraction space equals space plus ve$
When $straight x greater than 0 space slightly comma space space space then space straight f apostrophe left parenthesis straight x right parenthesis space equals space minus 2 fraction numerator left parenthesis plus ve right parenthesis over denominator plus ve end fraction space equals space minus ve$
$therefore space space space straight f apostrophe left parenthesis straight x right parenthesis space changes space from space plus ve space to space minus ve space as space straight x space passes space through space 0. therefore space space space straight x space equals space 0 space is space straight a space point space of space local space maxima space and space this space local space max space value space space equals space fraction numerator 1 over denominator 0 plus 2 end fraction equals 1 half.$

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