`f(x) = x(x-1)^(2)` in `[0,1]`

Answer

N/a

Explanation

We have, `f(x)=x(x-1)^(2)` in `[0,1]`
(i) Since, `f(x) = x(x-1)^(2)` is a polynomial function.
So, it is continuous in `[0,1]`
Now, `f'(x) = x.(d)/(dx)(x-1)^(2)+(x-1)^(2)(d)/(dx)x`
`=x.2(x-1).1+(x-1)^(2)`
`=2x^(2)-2x+x^(2)+1-2x`
`= 3x^(2)-4x+1` which exists in `(0,1)`.
So, `f(x)` is differentiable in `(0,1)`.
Now, `f(0) = 0` and `f(1) = 0 rArr f(0) = f(1)`
f satisfies the above conditions of Rolle's theorm
Hence, by Rolle's therorem `3c in(0,t)` such that
`f'(c)=0`
`rArr 3c^(2)-4c+1=0`
` rArr 3c^(2)-3c-c+1=0`
` rArr 3x(x-1)-1(c-1)=0`
`rArr (3c-1)(c-1)=0`
`rArr c= 1/3, 1rArr (1)/(3)in(0,t)`
Thus, we see that three exists a real number c in the open interval `(0,t)`.
Hence, Rolle's theorem has been verified.