`f(x)=sqrt(25-x^(2))` in `[1,5]`

Answer

N/a

Explanation

We have, `f(x) = sqrt(25-x)` in `[1,5]`
(i) Since, `f(x) = (25-x^(2))^(1//2)`, where `25-x^(2) ge 0`
` rArr x^(2) le 5 rArr -5 le x le 5`
`rArr x^(2) le +- 5 rArr le x le 5`
Hence, `f(x)` is continuous in `[1,5]`
(ii) `f'(x) = 1/2 (25-x^(2))^(-1//2). -2x= (-x)/(sqrt(25-x^(2)))`, which exists i `(1,5)`.
Hence, `f(x)` is differentiable in `(1,5)`
Since, conditions of mean value theorem are satisfied.
By mean value theorem `3c in (1,5)` such that
` f'(c) = (f(5)-f(1))/(5-1) rArr (-c)/(sqrt(25-c^(2))) = (0-sqrt(24))/(4)`
`rArr (c^(2))/(25-c^(2)) = 24/16`
`rArr (c^(2))/(25-c^(2)) = (24)/(16)`
`rArr 16c^(2) = 600 - 24 c^(2)`
`rArr c^(2) = (600)/(40) = 15`
`rArr c = +- sqrt(15)`
Also. `c = sqrt(15) in (1,5)`
Hence, the mean value theorem has been verified.