# Question ddb1d

May 7, 2017

range: $\left[\sqrt{3} , 2 \sqrt{1.5}\right]$

#### Explanation:

First find the domain:

For sqrt(x); " "x >= 0#

For $\sqrt{3 - x} : \text{ " 3-x >= 0; " " -x >= -3; " } x \le 3$

Domain: $0 \le x \le 3$

To find the range, find the minimum $y$ value and the maximum $y$ value.

The minimum $y$ value is found when $x = 0 , \text{ and } x = 3$:

$y = \sqrt{0} + \sqrt{3 - 0} = \sqrt{3}$

$y = \sqrt{3} + \sqrt{3 - 3} = \sqrt{3}$

To find the maximum $y$ value you need to have a graphing calculator or know Calculus.

From a TI graphing calculator the maximum occurs at $x = 1.5$ using 2ND CALC 4 and selecting two points on either side of the maximum, plus a guess point.

$y = \sqrt{1.5} + \sqrt{3 - 1.5} = \sqrt{1.5} + \sqrt{1.5} = 2 \sqrt{1.5} \approx 2.4495$

graph{sqrt(x) + sqrt(3-x) [-4.457, 5.543, -0.88, 4.12]}