Expert: Sherman D. - 8/30/2007

Question
I need some help with my Precalculus homework. I'm tryng to find the range for g(x)=5+ the square root of 4-x

Answer
g(x) = 5 + sqrt(4 - x)
y = 5 + sqrt(4 - x) or 5 - sqrt(4 - x)

switch the x and y, then solve for y

x = 5 ± sqrt(4 - y)
x - 5 = ±sqrt(4 - y)
±(x - 5) = sqrt(4 - y)
(±(x - 5)^2) = 4 - y
-y = ±(x - 5)^2 - 4
y = ±(x - 5)^2 + 4

y = ±(x - 5)^2 + 4
y = ±(x^2 - 10x + 25) + 4
y = ((x^2 - 10x + 25) or (-x^2 + 10x - 25)) + 4
y = (x^2 - 10x + 29) or (-x^2 + 10x - 21)

x = (-(-10))/(2(1)) or (-10)/(2(-1))
x = (10/2)
x = 5

y = 5^2 - 10(5) + 29
y = 25 - 50 + 29
y = -25 + 29
y = 4

but since we switched the x and y, the true answer is

y =< 5
or
y >= 5

for a graph, go to http://www.calculator.com/calcs/GCalc.html

type in 5 + sqrt(4 - x) and 5 - sqrt(4 - x), and click enter for each one, and you will get your graph which will show you the range.