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Continuity And Differentiability

Question
CBSEENMA12034667
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Examine the derivability of the following functions :
|x2| at x = 0

Solution
Let f(x) = |x|2
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 squared minus 0 over denominator straight x minus 0 end fraction space space space space space space space space space space space space space space space space space space left square bracket because space straight f left parenthesis 0 right parenthesis equals open vertical bar 0 close vertical bar squared equals 0 right square bracket 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 open vertical bar straight x close vertical bar squared over straight x equals Lt with straight h rightwards arrow 0 below fraction numerator open vertical bar 0 minus straight h close vertical bar squared over denominator 0 minus straight h end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 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 straight h squared over denominator negative straight h end fraction equals Lt with straight x rightwards arrow 0 below left parenthesis negative straight h right parenthesis equals 0 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 open vertical bar straight x close vertical bar squared over straight x 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 0 plus straight h close vertical bar squared 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 straight h squared over straight h equals Lt with straight x rightwards arrow 0 below straight h over straight h equals 0 therefore space straight L. straight H. straight D not equal to straight R. straight H. straight D. equals 0 therefore space straight f left parenthesis straight x right parenthesis equals open vertical bar straight x close vertical bar squared space is space derivalbe space at space straight x equals 0 space and space has space the space derivative space 0.

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