Question

$Verify space Rolle apostrophe straight s space theorem space for space the space function straight f left parenthesis straight x right parenthesis equals fraction numerator 8 straight x squared over denominator 3 end fraction minus 2 straight x comma space straight x element of open square brackets 0 comma 3 over 4 close square brackets$

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Solution

Here space straight f left parenthesis straight x right parenthesis equals fraction numerator 8 straight x squared over denominator 3 end fraction minus 2 straight x This space is space straight a space polynomial space in space straight x left parenthesis straight i right parenthesis space since space every space polynomial space in space straight x 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 open square brackets 0 comma 3 over 4 close square brackets left parenthesis ii right parenthesis space straight f apostrophe left parenthesis straight x right parenthesis equals fraction numerator 16 space straight x over denominator 3 end fraction minus 2 comma space which space exists space in space open parentheses 0 comma 3 over 4 close parentheses therefore space straight f left parenthesis straight x right parenthesis space is space derivable space in space open parentheses 0 comma 3 over 4 close parentheses. left parenthesis iii right parenthesis space straight f left parenthesis 0 right parenthesis equals 0 minus 0 equals 0 comma space straight f open parentheses 3 over 4 close parentheses equals 8 over 3 cross times 9 over 16 minus 2 cross times 3 over 4 equals 3 over 2 minus 3 over 2 equals 0 therefore space straight f left parenthesis 0 right parenthesis equals straight f open parentheses 3 over 4 close parentheses 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 Theorem. 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 over 4. Now space straight f apostrophe left parenthesis straight c right parenthesis equals 0 space gives space us space fraction numerator 16 straight c over denominator 3 end fraction minus 2 equals 0 space space space rightwards double arrow space straight c equals 3 over 8 element of space open parentheses 0 comma 3 over 4 close parentheses therefore space Rolle apostrophe straight s space theorem space is space verified.

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