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Continuity And Differentiability

Question
CBSEENMA12035485

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

Solution

 Here f(x) = x (x2-4) = x3-4 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 [-2, 2]
(ii) f'(x) = 3 x2-4, which exists in (-2, 2)
∴ f(x) is derivable in (-2, 2)
(iii) f(-2) = 0, f(2) = 0
∴ f(-2) = f(2)
∴ f(x) satisfies all the conditions of Rolle's Theorem.
∴ there must exist at least one real value of c such that f'(c) = 0 where - 2 < c < 2
Now space straight f apostrophe left parenthesis straight c right parenthesis equals 0 space given space us space 3 straight c squared minus 4 equals 0
therefore space straight c squared equals 4 over 3 space space space rightwards double arrow space straight c equals plus-or-minus fraction numerator 2 over denominator square root of 3 end fraction equals plus-or-minus fraction numerator 2 square root of 3 over denominator 3 end fraction equals plus-or-minus fraction numerator 2 left parenthesis 1.7 right parenthesis over denominator 3 end fraction equals plus-or-minus fraction numerator 3.4 over denominator 3 end fraction equals plus-or-minus space 1.13 element of left parenthesis negative 2 comma space 2 right parenthesis
∴ Rolle's Theorem is verified.

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