Find the range of the following functions
(i) `f(x) = log_(2)(sqrt(x-4) + sqrt(6-x)), 4 le x le 6`
(ii) `f(x) = 9^(x) - 3^(x) + 1`

Explanation

(i) We have
`f(x) = log_(2) (sqrt(x-4) + sqrt(6-x)), 4 le x le 6`
In the domain indicated the function is well defined. To find the range,
let `y = log_(2)(sqrt(x-4) + sqrt(6-x))`
`rArr 2^(y) = sqrt(x-4) + ssqrt(6-x) = t`, say (let `2^(y) = t gt 0`)
`rArr t^(2) = (x-4) + (6-x) + 2 sqrt((x-4)(6-x))`
`rArr t^(2) = 2+2sqrt(-x^(2) + 10 x - 24) rArr t^(2) - 2 = sqrt(1-(x-5)^(2))`
We observe that `0 le sqrt(1-(x-5)^(2)) le 1`, we have
`0 le t^(2) - 2 le 2`
`rArr 2 le t^(2) le 4`
`rArr -2 le t le -sqrt(2)` or `sqrt(2) le t le 2`
But `t = 2^(y) = +ve` always, hence
`sqrt(2) le t le 2`
`rArr 2^(1//2) le 2^(y) le 2^(1)`
`rArr (1)/(2) le y le 1` (`because 2^(y)` is an increasing function)
`rArr R_(f) = [(1)/(2), 1]`
(ii) We have,
`f(x) = 9^(x) - 3^(x) + 1 = (3^(x))^(2) - 2 xx 3^(x) xx (1)/(2) + (1)/(4) + (3)/(4) = (3)/(4) + (3^(x) - (1)/(2))^(2) ge (3)/(4), AA x in D_(f)`
Let `y = t^(2) - t + 1, t = 3^(x) gt 0`
`rArr t^(2) - t + (1-y) = 0`
`rArr t = (1 +-sqrt(1-4(1-y)))/(2) = (1 +- sqrt(4y - 3))/(2)`
For any `y in [(3)/(4), oo)` there always exists one value of t, viz, `t = (1+sqrt(4y-3))/(2)` (observe that t has to be positive) and consequently for eveery `y in [(3)/(4), oo)` there exists one value of `x in R`. Here the range of function is `[(3)/(4), oo)`.