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Continuity And Differentiability

Question
CBSEENMA12034666
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Examine the derivability of the following functions :
|x| at x = 0

Solution
Let f(x) = |x|
straight L. straight H. straight D equals Lt with straight x rightwards arrow 0 to the power of minus below fraction numerator straight f left parenthesis straight x right parenthesis minus straight f left parenthesis 0 right parenthesis over denominator straight x minus 0 end fraction equals Lt with straight x rightwards arrow 0 to the power of minus below fraction numerator open vertical bar straight x close vertical bar minus 0 over denominator straight x minus 0 end fraction space space space space space space space space space space space equals Lt with straight x rightwards arrow 0 to the power of minus below fraction numerator open vertical bar straight x close vertical bar over denominator straight x end fraction equals Lt with straight h rightwards arrow 0 below fraction numerator open vertical bar 0 minus straight h close vertical bar over denominator 0 minus straight h end fraction space space left square bracket Put space straight x equals 0 minus straight h comma space straight h greater than 0 space so space that space straight h rightwards arrow 0 space as space straight x rightwards arrow 0 to the power of minus right square bracket space space space space space space space space space space space equals Lt with straight h rightwards arrow 0 below fraction numerator open vertical bar negative straight h close vertical bar over denominator negative straight h end fraction equals Lt with straight x rightwards arrow 0 below fraction numerator straight h over denominator negative straight h end fraction equals negative 1 straight R. straight H. straight D equals Lt with straight x rightwards arrow 0 to the power of plus below fraction numerator straight f left parenthesis straight x right parenthesis minus straight f left parenthesis 0 right parenthesis over denominator straight x minus 0 end fraction equals Lt with straight x rightwards arrow 0 to the power of plus below fraction numerator open vertical bar straight x close vertical bar minus 0 over denominator straight x minus 0 end fraction space space space space space space space space space space space equals Lt with straight x rightwards arrow 0 to the power of plus below fraction numerator open vertical bar straight x close vertical bar over denominator straight x end fraction equals Lt with straight h rightwards arrow 0 below fraction numerator open vertical bar 0 plus straight h close vertical bar over denominator 0 plus straight h end fraction space space left square bracket Put space straight x equals 0 plus straight h comma space straight h greater than 0 space so space that space straight h rightwards arrow 0 space as space straight x rightwards arrow 0 to the power of plus right square bracket space space space space space space space space space space space equals Lt with straight h rightwards arrow 0 below fraction numerator open vertical bar straight h close vertical bar over denominator straight h end fraction equals Lt with straight x rightwards arrow 0 below straight h over straight h equals 1 therefore space straight L. straight H. straight D not equal to straight R. straight H. straight D. therefore space straight f left parenthesis straight x right parenthesis equals open vertical bar straight x close vertical bar space is space not space differentialbe space at space straight x equals 0.

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