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

Solution

Here f(x) = x(x - 1)² = x³ - 2 x² + x
It is a polynomial in x.
(i) Since every polynomial in x is a continuous function for every value of x.
∴ f(x) is continuous in [0, 1].
(ii) f'(x) = 3 x² - 4 x + 1, which existsyn (0, 1)
∴ f(x) is derivable in (0, 1).
(iii) f(0) = 0, f(1) = 0
∴ f(0) = f(1)
∴ f(x) satisfies all the conditions of Rolle's Theorem
∴ there must exist at least one value c of x such that
f'(c) = 0 where 0 < c 1.
$Now space straight f apostrophe left parenthesis straight c right parenthesis equals 0 space gives space us space 3 straight c squared minus 4 straight c plus 1 equals 0 therefore space space space space space straight c equals fraction numerator 4 plus-or-minus square root of 16 minus 12 end root over denominator 6 end fraction equals fraction numerator 4 plus-or-minus 2 over denominator 6 end fraction equals 1 comma 1 third Now space 1 not an element of left parenthesis 0 comma 1 right parenthesis comma space but space 1 third element of left parenthesis 0 comma 1 right parenthesis therefore space space Rolle apostrophe straight s space Theorem space is space verified$

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Continuity And Differentiability

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

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

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