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

May 30, 2017

$1 \le x < 2$
$f \left(x\right) \le f \left(1.318\right)$

#### Explanation:

We have $f : x \mapsto \sqrt{x - 1} - \frac{1}{\sqrt{2 - x}}$

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

$\sqrt{x - 1} \ge 0$

$x - 1 \ge 0$

$x \ge 1$

For the second square root, we need to be more careful, because we have $1 \text{/} \sqrt{2 - x}$. This means that $\sqrt{2 - x} > 0$.

$2 - x > 0$

$2 > x$

So our domain is $1 \le x < 2$

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

$f ' \left(x\right) = \frac{1}{2} \left(\frac{1}{\sqrt{x - 1}} - \frac{1}{2 - x} ^ \left(\frac{3}{2}\right)\right)$

Since $1 \text{/"(2-x)^(3/2) > 1"/} \sqrt{x - 1}$, $f ' \left(x\right) < 0$ 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 $f ' \left(x\right) = 0$

$\frac{1}{2} \left(\frac{1}{\sqrt{x - 1}} - \frac{1}{2 - x} ^ \left(\frac{3}{2}\right)\right) = 0$

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

Therefore, $f \left(x\right) \le f \left(1.318\right)$