Question

Rolle's theorem for f(x) = x2m-1(a – x)2n, find the value of x between o and a where f'(x) = 0.

Answer 1

Answer:

The given function f(x)=x

2

(1−x

2

) on [0,1] is continuous and differentiable on (0,1), since it is a polynomial of degree 4.

Again f(0)=0=f(1).

So f(x) satisfies all the conditions of Rolle's theorem.

According to Rolle's theorem there exists c∈(0,1) such that f

′

(c)=0.

Now f

′

(c)=0 gives

2c−4c

3

=0

or, 2c(1−2c

2

)=0

or, c=

2

1

[ Since c∈(0,1)]