Posted by Steve on .
use f(x)=square root of 5x and g(x)=x^2 + 2x
afind f(g(x)), state the domain
bfind g(f(x)),state the domain
cfind g(f(1))
dfind f(x)/g(x), state the domain, then evaluate for x=3

algebra 
Reiny,
f(x) = √(5x) , 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)
= (√(5x))^2 + 2√(5x)
= 5  x + 2√(5x)
domain is :
5x ≥ 0
x ≥ 5
x ≤ 5
you try c and d