Question

How do you find the domain and range of `f(x) =sqrt( x - 1) - 1/sqrt( 2-x)`?

Answer 1

`1 <= x< 2`
`f(x) <= f(1.318)`

Explanation:

We have `f:x|->sqrt(x-1)-1/sqrt(2-x)`

Whenever we have a function with a square root, we know that the square root has to be bigger than or equal to `0`.

`sqrt(x-1) >= 0`

`x-1 >= 0`

`x >= 1`

For the second square root, we need to be more careful, because we have `1"/"sqrt(2-x)`. This means that `sqrt(2-x) > 0`.

`2-x > 0`

`2 > x`

So our domain is `1 <= x < 2`

To find the range, we need to find the stationary point and find if the function is increasing or decreasing.

`f'(x)=1/2(1/sqrt(x-1)-1/(2-x)^(3/2))`

Since `1"/"(2-x)^(3/2) > 1"/"sqrt(x-1)`, `f'(x) < 0` so the function is decreasing. This means the function will have a maximum point, and all other `y`-values will be below that point.

Let `f'(x)=0`

`1/2(1/sqrt(x-1)-1/(2-x)^(3/2))=0`

Solving for `x`, we get `x=1.318`.

Therefore, `f(x) <= f(1.318)`