Rolles theorem

Asked by  | 18th Jan, 2010, 11:28: PM

Expert Answer:

P(x) =X3 -2x2-x+2 

Interval = [1,2]

P(x) is a polynomial function so it is continuous and differentiable  everywhere hence in particular in [1,2] and (1,2) respectively

P(1) =  0

P(2)= 0

So the conditions of Rolles theorem are satisfied and there exists a c in the interval (1,2) such that

P'(c)=0

P'(x)= 3x2-4x-1

P('c)= 3c2-4c+1= 0

gives c=1/3 or c =1

1/3 is in (a,b)

Answered by  | 19th Jan, 2010, 10:18: AM

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