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

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 Multiple Choice QuestionsShort Answer Type

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


Let f(x) = x2 - 5 x + 4
(a)    Since every polynomial is a continuous function for every value of x
∴ f is continuous in [1, 4]
(b)    f'(x) = 2 x - 5, which exists in (1, 4)
∴ f is derivable in (1, 4)
(c)    f(1) = 1 - 5 + 4 = 0
f(4) = 16 - 20 + 4 = 0
∴ f(1) = f(4)
∴ f satisfies all the condition of Rolle's Theorem
∴ there must exist at least one value c of x such that f'(c) = 0 where 1 < c < 4
Now space straight f apostrophe left parenthesis straight c right parenthesis equals 0 space gives space us space 2 straight c minus 5 equals 0 space or space straight c equals 5 over 2 element of left parenthesis 1 comma space 4 right parenthesis
∴ Rolle's theorem is verified.

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590.
Verify Rolle's Theorem for the function f(x) = x (x2-4) in the interval [-2, 2].
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