`f(x) = 1/(4x-1)` in `[1,4]`

Answer

N/a

Explanation

We have `f(x) = (1)/(4x-1)` in `[1,4]`
(i) `f(x)` is continuous in `[1,4]`
Also, at `x = (1)/(4), f(x)` is discountinuous
Hence, `f(x)` is continous in `[1,4]`.
(ii) `f'(x)= - (4)/((4x-1)^(2))`, which exists in `(1,4)`.
Since, conditions of mean values theorem are satisfied.
Hence, there exists a real number `c in`] `1,4` [such that
`f'(c) = (f(4)-f(1))/(4-1)`
`rArr = (-4)/((4x-1)^(2)) = ((1)/(16-1)-(1)/(4-1))/(4-1)=(1/15-1/13)/(3)`
`rArr (-4)/((4x-1)^(2)) = (1-5)/(45) = (-4)/(15)`
`rArr (4x-1)^(2)=45`
`rArr 4c-1=+-3sqrt(5)`
`rArr c = (3sqrt(5)+1)/(4) in (1,4)` , [neglective `(-ve)` value]
Hence, mean value theorem has been verified.