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

Solution

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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Continuity And Differentiability

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

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Continuity And Differentiability

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

Some More Questions From Continuity and Differentiability Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

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