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Continuity and Differentiability

Short Answer Type

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

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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582. Verify Rolle's Theorem for the function x² + 2 x - 8 in [-4, 2]

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583. Verify Rolle's Theorem for the function f(x) = x(x - 1)² in [0, 1].

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584.

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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585.

Verily space Rolle apostrophe straight s space theorem space for space the space function space straight f left parenthesis straight x right parenthesis equals 8 straight x minus straight x squared space in space left square bracket 0 comma space 8 right square bracket

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586. Verify Rolle's Theorem for the function :f(x) = x² in the interval [- 1, 1]

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587. Verify Rolle's Theorem for the function : f(x) = x² - 1 in the interval [- 1, 1]

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588. Verify Rolle's Theorem for the function :f(x)= x² - 4 x + 3 in the interval 1 ≤ x ≤ 3.

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589. Verify Rolle's Theorem for the function : x² - 5 x + 4 on [1, 4]