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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