Question

$Verily space Rolle apostrophe straight s space theorem space for space the space function straight f left parenthesis straight x right parenthesis equals straight x cubed over 3 minus fraction numerator 5 straight x squared over denominator 3 end fraction plus 2 straight x comma space straight x element of left square bracket 0 comma 3 right square bracket$

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Solution

Here space space straight f left parenthesis straight x right parenthesis equals straight x cubed over 3 minus fraction numerator 5 straight x squared over denominator 3 end fraction plus 2 straight x This space is space straight a space polynomial space in space straight x space left parenthesis straight i right parenthesis space since space every space polynomial space in space straight x space is space continuous space for space all space straight x. therefore space straight f left parenthesis straight x right parenthesis space is space continuous space in space left square bracket 0 comma space 3 right square bracket left parenthesis ii right parenthesis space straight f apostrophe left parenthesis straight x right parenthesis space equals straight x squared minus fraction numerator 10 space straight x over denominator 3 end fraction plus 2 comma space which space is space exists space in space left parenthesis 0 comma 3 right parenthesis therefore space straight f left parenthesis straight x right parenthesis space is space derivable space in space left parenthesis 0 comma space 3 right parenthesis left parenthesis iii right parenthesis space straight f left parenthesis 0 right parenthesis equals 0 minus 0 plus 0 equals 0 straight f left parenthesis 3 right parenthesis equals 27 over 3 minus 45 over 3 plus 6 equals 9 minus 15 plus 6 equals 0 therefore space straight f left parenthesis 0 right parenthesis equals straight f left parenthesis 3 right parenthesis therefore space straight f left parenthesis straight x right parenthesis space satisfies space all space the space conditions space of space Rolle apostrophe straight s space theroem. therefore 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 comma space where space 0 less than straight c less than 3. Now space straight f apostrophe left parenthesis straight c right parenthesis equals 0 space given space us space straight c squared minus fraction numerator 10 space straight c over denominator 3 end fraction plus 2 equals 0 therefore space 3 straight c squared minus 10 straight c plus 6 equals 0 therefore space straight c equals fraction numerator 10 plus-or-minus square root of 100 minus 72 end root over denominator 6 end fraction equals fraction numerator 10 plus-or-minus square root of 28 over denominator 6 end fraction equals fraction numerator 10 plus-or-minus 5.2 over denominator 6 end fraction comma space fraction numerator 4.8 over denominator 6 end fraction equals 2.53 comma space 0.8 element of left parenthesis 0 comma space 3 right parenthesis therefore space Rolle apostrophe straight s space theorem space is space verified.

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