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

Solution

Here f(x) = x²This is a polynomial in x
(a)    Since every polynomial in x is continuous for all x
∴ f(x) is continuous in [- 1, 1].
(b)    f'(x) = 2 x, which exists in (- 1, 1)
∴ f(x) is derivable in (- 1, 1).
(c)    f(-1) = (-1)² =1, f(1) = (1)² = 1
∴ f(-1) = f(1)
∴ f(x) satisfies all the conditions of Rolle's Theorem.
∴ there exists at least one value c of x such that f'(c) = 0 , where - 1 < c < 1
Now f'(c) = 0 gives us 2 c = 0
∴ c = 0 ∈ (-1, 1)    ⇒ Rolle's Theorem is verified.

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Continuity And Differentiability

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

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Continuity And Differentiability

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

Some More Questions From Continuity and Differentiability Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

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