Question

Question #ddb1d

Answer 1

range: `[sqrt(3), 2sqrt(1.5)]`

Explanation:

First find the domain:

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

For `sqrt(3-x): " " 3-x >= 0; " " -x >= -3; " " x <= 3`

Domain: `0 <= x <= 3`

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

The minimum `y` value is found when `x = 0, " 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) ~~2.4495`

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