`f(x) = x^(3)-2x^(2)-x+3` in `[0,1]`

Answer

N/a

Explanation

We have, `f(x) = x^(3)-2x^(2)-x+3` in `[0,1]`
(i) Since `f(x)` is a polynomial function.
Hence, `f(x)` is continuous in `[0,1]`.
(ii) `f'(x) = 3x^(2)-4x-1`, which exists in `(0,1)`.
Hence, `f(x)` is differentiable in `(0,1)`
Therefore, by mean value theorem `3x in (0, 1)` such that,
`f(c) = (f(1) - f(0))/(1-0)`
`rArr 3c^(2) - 4c - 1 = ([1-2-1+3]-[0+3])/(1-0)`
`rArr 3c^(2) - 4c -1=(-2)/(1)`
`rArr 3c^(2)-4c + 1 = 0`
`rArr 3c^(2) -3c-c+1 = 0`
`rArr 3c(c-1)-1(c-1) = 0`
`rArr (3c-1)(c-1) = 0`
`rarr c = 1//3, 1`, where `1/3 in (0,1)`
Hence, the mean value theorem has been verified.