Question #ddb1d

1 Answer
May 7, 2017

Answer:

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]}