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Continuity And Differentiability

Question

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Verify Rolle's Theorem for the function x² + 2 in [-2, 2]

Solution

Here space straight y equals straight f left parenthesis straight x right parenthesis equals straight x squared plus 2 This space is space straight a space polynomial space in space straight x left parenthesis straight i right parenthesis space Since space every space polynomila space in space straight x space is space continuous space for space all space straight x. therefore space space straight f space is space continuous space in space left square bracket negative 2 comma 2 right square bracket left parenthesis ii right parenthesis space dy over dx equals straight f apostrophe left parenthesis straight x right parenthesis equals 2 space straight x comma space which space exists space in space left parenthesis negative 2 comma 2 right parenthesis therefore space space straight f space is space differentiable space in space left parenthesis negative 2 comma 2 right parenthesis left parenthesis iii right parenthesis space straight f left parenthesis negative 2 right parenthesis equals left parenthesis negative 2 right parenthesis squared plus 2 equals 4 plus 2 equals 6 straight f left parenthesis 2 right parenthesis equals left parenthesis 2 right parenthesis squared plus 2 equals 4 plus 2 equals 6 therefore space straight f left parenthesis negative 2 right parenthesis equals straight f left parenthesis 2 right parenthesis therefore space space straight f space satisfies space all space the space conditions space of space rolle apostrophe straight s space theorem. therefore space space there space exists space at space least space one space value space straight c space of space straight x space such space that space straight f apostrophe left parenthesis straight c right parenthesis equals 0 where space minus 2 less than straight c less than 2. Now space straight f apostrophe left parenthesis straight c right parenthesis equals 0 space gives space us space 2 space straight c equals 0 therefore space space straight c equals 0 element of left parenthesis negative 2 comma space 2 right parenthesis therefore space space Rolle apostrophe straight s space Theorem space is space verified.

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