Q. 965.0( 1 Vote )

# For the function

Mean Value Theorem states that, Let f : [a, b] R be a continuous function on [a, b] and differentiable on (a, b). Then there exists some c in (a, b) such that

We have,

Since, f(x) is a polynomial function it is continuous on [1,3] and differentiable on (1,3).

Now, as per Mean value Theorem, there exists at least one c (1,3), such that

3(c2 – 1) = 2c2

3c2 – 2c2 = 3

c2 = 3

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