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Continuity And Differentiability

Question
CBSEENMA12035489
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Verify Roll's Theorem for the function :f(x) = (x2 - 1) (x - 2) in [-1, 2]

Solution

Let f (x) = (x2 - 1) (x - 2) = x3 - 2 x2 - x + 2
It is a polynomial in x
(a)    Since every polynomial in x is a continuous function for every value of x.
∴ f(x) is continuous in [-1, 2].
(b)    f'(x) = 3 x2 - 4 x - 1, which exists in (-1, 2)
∴ f is derivable in (-1, 2).
(c)    f(-1) = (1 - 1) (-1 -2) = 0
f(2) = (4 - 1) (2 - 1) (2 - 2) = 0
∴ f(-1) = f(2)
∴ f satisfies all the conditions of Rolle's Theorem
∴ there must exist at least one value c ofx such that f'(c) = 0 where - 1 < c < 2.
Now f'(c) = 0 gives us 3 c2 - 4 c - 1 = 0
or space straight c equals fraction numerator 4 plus-or-minus square root of 16 plus 12 end root over denominator 6 end fraction equals fraction numerator 4 plus-or-minus 2 square root of 7 over denominator 6 end fraction equals fraction numerator 2 plus-or-minus square root of 7 over denominator 3 end fraction Now space minus 1 less than fraction numerator 2 minus square root of 7 over denominator 3 end fraction less than fraction numerator 2 plus square root of 7 over denominator 3 end fraction less than 2 therefore space straight c equals fraction numerator 2 minus square root of 7 over denominator 3 end fraction space and space fraction numerator 2 plus square root of 7 over denominator 3 end fraction space both space lie space in space left parenthesis negative 1 comma space 2 right parenthesis
∴ Rolle's Theorem is verified.

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