Posted by **Steve** on Monday, November 26, 2012 at 4:28pm.

use f(x)=square root of 5-x and g(x)=x^2 + 2x

a-find f(g(x)), state the domain

b-find g(f(x)),state the domain

c-find g(f(-1))

d-find f(x)/g(x), state the domain, then evaluate for x=3

- algebra -
**Reiny**, Monday, November 26, 2012 at 4:58pm
f(x) = √(5-x) , g(x) = x^2 + 2x

f(g(x)

= √(5 - (x^2+2x) )

= √(5 - x^2 - 2x)

for this to be real, 5 - x^2 - 2x ≥ 0

x^2 + 2x - 5 ≤ 0

solve x^2 + 2x - 5 = 0 ,

the domain will be all values of x **between** the two solution values

g(f(x)

= (√(5-x))^2 + 2√(5-x)

= 5 - x + 2√(5-x)

domain is :

5-x ≥ 0

-x ≥ -5

x ≤ 5

you try c and d

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